Class 10 Maths - Polynomials

Can $(x^2 -1)$ be the quotient on division of $x^6 + 2x^3 + x- 1$ by a polynomial in $x$ of degree 5?

No,

$x^2-1$ cannot be the quotient on division of

$x^6-2x^3+x-1$ by a

polynomial in degree

$5$ because the degree of the product of the

quotient and the divisor should be equal to the power of the dividend.

Here, the degree of the product of the quotient and the divisor is

$7$ . But the degree of

$x^6-2x^3+x-1$ is

$6$

Look at the graphs given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x).$

The graph intersects the x-axis at one point. Thus, the polynomial has one zero.

Let $f(x)=x^3$ . The graph of the polynomial is shown in the figure. Find the number of zeros of polynomial f(x).

The graph intersects the x-axis at only one point. Thus, number of zeros is one.

Look at the graph given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x).$

The graph intersects the x-axis at one points. Thus, the polynomial has one zeros

Look at the graph given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x)$ .

The graph intersects the x-axis at three points. Thus, the polynomial has three zeros

Look at the graph given. It is the graph of y = p(x), where p(x) is a polynomial. Find the number of zeros of p(x).

The graph intersects the x-axis at two points. Thus, the polynomial has two zeros

The graph of y = p(x) is given in fig below, for some polynomial p(x). Find the number of zeros of p(x).

The graph meet the x-axis at

$3$ points. Thus the number of zeros is

$3$

The graph of $y = p(x)$ is given in figure for some polynomial $p(x)$ . The number of zeros of $p(x)$ are

There will be no zeroes for

$y=p(x)$ .

As,

$y$ never touches or crosses X-axis.

So, no root or zero.

Let $f(x)=x^3$
The graph of the polynomial is shown in fig.
The co-ordinates of the points, at which the graph intersects the x-axis is (0,0)
If true then enter $1$ and if false then enter $0$

$f(x)=x^3$
To find, the co-ordinate of the point where the graph intersects the x-axis, we have

$f(x)=0$
Thus,

$x^3=0$

$=>x=0$
Thus the point is

$(0,0)$

The graph of $y = p(x)$ is given in the figure for some polynomial $p(x)$ . The number of zeros of $p(x)$ is:

Given function is

$y=p(x)$

Number of zeros is same as the number of points which intersects

$x-$ axis.

The given graph is a parabola and it intersects

$x-$ axis at

$1$ point only. Thus, the number of zeros is

$1$

Hence, the correct answer is

$1$ .

Construct a graph and represent the zeroes of the cubic polynomial for the equation: $y=x^3-6x^2+11x-6$

The zeroes of polynomial

$y={x^3-6x^2+11x-6}$ are

$1,2,3$

On dividing $3x^{3} + x^{2} + 2x + 5$ by a polynomial $g(x)$ , the quotient and remainder are $(3x - 5)$ and $(9x + 10)$ respectively. Find $g(x)$

If we divide

$f(x)=3x^3 + x^2 +2x +5$ by

$g(x)$ we get

$q(x)=(3x-5)$ as quotient and

$r(x)=(9x+10)$ as remainder.

By using division algorithm,

$f(x)=g(x)q(x)+r(x)$

$\implies 3x^3+x^2+2x+5=g(x)(3x-5)+(9x+10)$

$\implies g(x)=\dfrac{3x^3+x^2+2x+5-9x-10}{(3x-5)}$

$\implies g(x)=\dfrac{3x^3+x^2-7x-5}{(3x-5)}$

Hence,

$g(x)=x^2+2x+1$

A polynomial $p(x)$ is divided by $(2x - 1)$ . The quotient and remainder obtained are $(7x^{2} + x + 5)$ and $4$ respectively. Find $p(x)$

We know,

$p(x) = [g(x) \times q(x)] + r(x)$ [Division algorithm for polynomials]

$p(x) = (2x - 1)(7x^{2} + x + 5) + 4$

$p(x) = 14x^{3} + 2x^{2} + 10x - 7x^{2} - x - 5 + 4$

$p(x) = 14x^{3} - 5x^{2} + 9x - 1$

$\therefore$

$the dividend$

$p(x) = 14x^{3} - 5x^{2} + 9x - 1$ $

What must be subtracted from $6x^{4} + 13x^{3} + 13x^{2} + 30x + 20$ , so that the resulting polynomial is exactly divisible by $3x^{2} + 2x + 5$ ?

$p(x) = g(x)\times q(x) + r(x)$ [Division algorithm for polynomials]

$\Rightarrow p(x) - r(x) = g(x) \times q(x)$
It is clear the RHS of the above equation is divisible by

$g(x)$ . i.e., the divisor.

$\therefore$ LHS is also divisible by the divisor.
Therefore, if we subtract remainder

$r(x)$ from dividend

$p(x)$ , then it will be exactly divisible by the divisor

$g(x)$ ]
Let us divide,

$6x^{4} + 13x^{3} + 13x^{2} + 30x + 20$ by

$3x^{2} + 2x + 5$
we get, quotient

$q(x) = 2x^{2} + 3x - 1$ remainder

$r(x) = 17x + 25$

$\therefore$ If we subtract

$17x + 25$ from

$6x^{4} + 13x^{3} + 13x^{2} + 30x + 20$ , it will be exactly divisible by

$(3x^{2} + 2x + 5)$ .

Which type of polynomial, the given expression $x^{3} + x + 4$ is?

Given

$x^{3}+x+4$

$\Rightarrow x^{3}+x^{1}+4x^{0}$

Here the highest power is

$3$ .

Hence degree is

$3$ .

Then

$x^{3}+x+4$ is a cubic polynomial.

Based on degree, which type of polynomial, the given expression $5x^{2} + x - 7$ is?

Given polynomial is

$5x^{2}+x-7$

$= 5x^{2}+x^{1}-7x^{0}$

Here the highest power of

$x$ is

$2$ .

So, degree of the polynomial is

$2$ .

Hence, this is a

$Quadratic$ polynomial.

Which type of polynomial, the given expression $x - 1$ is?

Given

$x-1$

$\Rightarrow x^{1}-x^{0}$

Here the highest power is

$1$ .

Hence degree is

$1$ .

Therefore,

$x-1$ is a linear polynomial.

Which type of polynomial, the given expression $3p$ is?

Given expression is

$3p$

$\Rightarrow 3p^{1}$

Here the highest power is

$1$ .

Hence degree is

$1$ .

Therefore,

$3p$ is a linear polynomial.

Which type of polynomial, the given expression $x - x^{3}$ is?

Given,

$x-x^{3}$

$\Rightarrow x^{1}-x^{3}$

Here the highest power is

$3$ .

Hence degree is

$3$ .

Therefore,

$x-x^{3}$ is cubic polynomial.

Classify the following expression as linear, quadratic, or cubic polynomials.
$\pi r^{2}$

The polynomials are classified by the degree of the polynomial. The degree of the polynomial is the highest power of the variable of the polynomial. If the degree is 1, the polynomial is linear. If the degree is 2 the polynomial is quadratic and if the degree is 3 the polynomial is cubic.

Here, the expression is

$\pi r^{2}$

Therefore, the highest power is 2.

Also, degree is 2.

Hence, the expression is a quadratic polynomial.

Draw the graph of $y = x^2 - x - 6$ and find zeroes. Justify the answer.

Clearly graph of

$y=x^2-x-6$ cut the x-axis at

$x=-2,3$ .

$\Rightarrow x=-2,3$ are zeros of

$y=x^2-x-6$ .
Justification:
when

$x=-2;$

$y=4+2-6=0$
when

$x=3;$

$y=9-3-6=0$

Divide the polynomials $p(x)$ by the polynomials $q(x)$ . Find the quotient and remainder. $p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,\,\,\,\,\,\,\,\,\,\, q\left( x \right) = {x^2} - 2$

Consider the given polynomial.

$p\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x-3$

$q\left( x \right)={{x}^{2}}-2$

Therefore, the working steps are shown in figure,

Hence, the quotient is $(x-3)$ and remainder is $(7x-9)$ .

994343_1060380_ans_5682f46913114568aa478a88e34b71ea.PNG

Classify the following as linear, quadratic and cubic polynomials
$5{x}^{2}+x-7$

Solution:-

$5{x}^{2} + x - 7 = 5{x}^{2} + {x}^{1} - 7{x}^{0}$

Highest power

$= 2$

Hence, degree is

$2.$

Hence, the equation is quadratic.

Find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by
$\dfrac{-8}{3},\dfrac{4}{3}$

Let

$a$ and

$b$ be the zeros of the required polynomial.

Given,

$a+b=-\dfrac{8}{3}$ .....(1) and

$ab=\dfrac{4}{3}$ ......(2).

Now the polynomial whose zeros are

$a$ and

$b$ be

$x^2-(a+b)x+ab$

$=x^2+\dfrac{8}{3}x+\dfrac{4}{3}$ .[Using (1) and (2)].

Now solving (1) and (2) we get,

$a=-\dfrac{2}{3}$ or

$-2$ and

$b=-2$ or

$-\dfrac{2}{3}$ .

So,

$-\dfrac{2}{3}$ and

$-2$ are the zeros of the polynomial.

Classify the following as linear, quadratic and cubic polynomials
$x-1$

Solution:-

The given expression is,

$x - 1 = {x}^{1} - 1{x}^{0}$

Highest power of

$x= 1$

So, the degree of the polynomial is

$1.$

Hence, the equation is linear.

Classify the following as linear, quadratic and cubic polynomials
${x}^{2}+x+4$

Solution:-

${x}^{2} + x + 4 = {x}^{2} + {x}^{1} + 4{x}^{0}$

Highest power

$= 2$

Hence, degree is

$2.$

Hence, the equation is quadratic.

What is meant by division algorithm give example?

when we divide a number or polynomial by another number or polynomial then the relation Divident

$=$ divisor

$\times$ Quotient

$+$ Remainder is always satisfied

This is known as division algorithm

e.g If we divide polynomial

$2{x^2} + 3x + 1$ by polynomial

$x + 2$

then we get

Divident

$2{x^2} + 3x + 1$

Divisor

$=x+2$

Quotient

$=2x-1$

Remainder

$=3$

Now, Divisor

$\times$ Quotient

$+$ Remainder

$\left( {x + 2} \right)\left( {2x - 1} \right) + 3$

$= 2{x^2} - x + 4x - 2 + 3$

$= 2{x^2} + 3x + 1$

$=$ Dividend

Classify whether $x-x^3$ is linear, quadratic, or a cubic polynomial.

$x - {x}^{3} = {x}^{1} - {x}^{3}$

The highest power of the given polynomial is

$3$ so, the degree of the polynomial is also

$3$ , and a polynomial of degree

$3$ is known as a cubic polynomial.

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following.
$p(x)=x^{4}-3x^{2}+4x-3\quad g(x)=x^{2}-2$

Quotient

$=x^2-1$

Reminder

$=4x-5$

1161770_1090879_ans_4e37aee1929640dfa30f3c8146a40207.png

Find the remainder when $x^4 + 5x^2 + 6$ is divided by $x^2 + 1$ .

1121154_1140092_ans_8c6f91167507485193829f0aa084233c.jpeg

Divide $(y^{3} - 3y^{2} + 5y - 1)$ by $(y - 1)$ to find quotient and remainder.

$y^2-2y+3$

_______________

$y-1)y^3-3y^2+5y-1$

$y^3-y^2$

________________

$-2y^2+5y-1$

$-2y^2+2y$

______________

$3y-1$

$3y-3$

_________

$2$

Quotient

$Q=y^2-2y+3$ & Remainder

$R=2$ .

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following.
$p(x)=x^{4}-5x+6\quad g(x)=2-x^{2}$

Quotent

$=-x^2-2$

Reminder

$=-5x+10$

1161908_1090880_ans_d39dd18d10bd45d2a786bc7d82f2f30b.png

Divide $6x^2-31x+47$ by $2x-5$ and verify the division algorithm.

Let,

$p(x) = 6x^2 - 31x + 47$ ,

$g(x) = 2x - 5$

$q(x) = 3x - 8$

and

$r = 7$

$\therefore$ By division algorithm,

$p(x) = g(x) \times q(x) + r$

$= (2x - 5) (3x - 8) + 7$

$= 6x^2 - 15x - 16x + 40 + 7$

$= 6x^2 - 31x + 47$

$= p(x)$

Hence division algorithm is verified.

1188032_1142339_ans_7510d9a8aa344449988166d2f867c9a6.png

Classify the following as linear, quadratic and cubic polynomials
$3p$

Solution:-

$3p = 3 {p}^{1}$

Highest power

$= 1$

Hence, degree is

$1$ .

Hence, the equation is linear.

Divide and also verify the result ${x^2} + 3x + 2$ by $\left( {x + 2} \right)$

$x^2+3x+2 = x^2+2x+x+2$

$\implies (x+1)(x+2)$

On dividing

$x+2$ we get,

$(x+1)$

Give an example of a quadratic binomial.

A binomial is a variable expression with two terms. Example-

$3x+2$ or

$5x^{3}+1$

A quadratic binomial is a second degree binomial, such as

$3x^{2}+2$

Divide and write the quotient and remainder.
(a) $(y^2+10y+24) \div (y+4)$
(b) $(p^2+7p-5) \div (p+3)$

(a) Quotient

$=(y+6)$

Remainder

$=0$

(b) Quotient

$=(p+4)$

Remainder

$=-17$

1358568_1193786_ans_536fe6b4d33a4f0babcee6aecb55d019.PNG

Divide and write the quotient and remainder.
$(y^2+10y+24) \div (y+4)$

(a) Quotient

$=(y+6)$

Remainder

$=0$

(b) Quotient

$=(p+4)$

Remainder

$-17$

$=(4{x}^{2}+18x+75)$

Remainder

$=300$

1358359_1193884_ans_52d806344dfa4e60b5a7cee3a6ef6539.PNG

Divide:
$2{x^3} - {x^2} - 3x + 1\,\,\,\ by\,\,\ (x + 1)$

1123674_1197275_ans_ca5f4b72ceeb45338057982eaac1dcab.jpeg

Find the quotient and remainder.

(a) $\dfrac{{{a^2} + 8a - 9}}{{a - 7}}$

(b) $\dfrac{{{x^2} + 6x - 11}}{{x - 3}}$

(a)

$\dfrac{a^2+8a-9}{a-7}$

$(a+15)$

___________

$(a-7))a^2+8a-9$

$a^2-7a$

____________

$15a-9$

$15a-105$

___________

$96$

$\therefore$ Quotient

$=(a+15)$

remainder

$=96$

(b)

$\dfrac{x^2+6x-11}{x-3}$

$(x+9)$ ----Quotient

___________

$x-3)x^2+6x-11$

$x^2-3x$

____________

$9x-11$

$9x-27$

____________

$16$ --- remainder

Quotient

$=(x+9)$

Remainder

$=16$ .

1123685_1197307_ans_b38250dfa44e4b748a974959b4920647.jpeg

Fill in the blacks.
A polynomial of degree $1$ is called a _____ polynomial

A polynomial of degree

$1$ is know as linear polynomial.

E.g.- Degree of polynomial

$3x + 5$ is

$1$ , thus it is a linear polynomial.

Hence the required answer is linear polynomial.

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.
$-p$ .

$-p$ is a linear polynomial because it has a degree of

$1$

If the graph of a polynomial intersects the $x - asis$ at only one point, can it be a quadratic polynomial ?

Yes, because every quadratic has at the most two zeros.

Examine, seeing the graph of the polynomial given below, whether they are a linear or quadratic polynomial or neither linear nor quadratic polynomial:

The shape of the graph is neither a straight line nor a parabola.
So, the graph is not of a linear polynomial or a quadratic polynomial.

Classify the following as linear , quadratic and cubic polynomial :
$1 + x$

$1 + x$ is a linear polynomial as its degree is

$1$

Classify the following as linear , quadratic and cubic polynomial :
3t

$3t$ is a linear polynomial as its degree is

$1$

Classify the following as linear , quadratic and cubic polynomial :
$y + y^{2} + 4$

A linear polynomial is one in which the degree of the polynomial is

$1$ .

A quadratic polynomial is one in which the degree of the polynomial is

$2$ .

A cubic polynomial is one in which the degree of the polynomial is

$3$ .

$y + y^{2} + 4$ is a quadratic polynomial as its degree is

$2$

Classify the following polynomial as linear, quadratic and cubic polynomial .
$a^{2}$

The degree of the polynomial

$a^{2}$ is 2
So, the polynomial

$a^{2}$ is a quadratic polynomial

Classify the following as linear , quadratic and cubic polynomial :
$r^{2}$

$r^{2}$ is a quadratic polynomial as its degree is

$2$

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$3x^2 + 4x- 4$ .

Let

$f(x)=3x^{ 2 }+4x-4$ .
Comparing it with the standard quadratic polynomial

$ax^2+bx+c$ , we get,

$a=3$ ,

$b=4$ ,

$c=-4$ .
Now,

$3x^{ 2 }+4x-4$

$=3x^{ 2 }+6x-2x-4$

$=3x(x+2)-2(x+2)$

$=(x+2)(3x-2)$ .
The zeros of

$f(x)$ are given by

$f(x)=0$ .

$=>(x+2)(3x-2)=0$

$=>x+2=0, 3x-2=0$

$=>x=-2, x=\dfrac { 2}{ 3}$ .
Hence the zeros of the given quadratic polynomial are

$-2$ ,

$\dfrac { 2}{ 3}$ .

Verification of the relationship between the roots and the coefficients:
Sum of the roots

$=-2+\dfrac { 2}{ 3}$

$=\dfrac { -6+2}{ 3}$

$=\dfrac { -4}{ 3}$

$=\dfrac { -coefficient\quad of\quad x }{ coefficient\quad of\quad { x }^{ 2 } }$ .
Product of the roots

$= -2\times (\dfrac { 2 }{ 3 })$

$=\dfrac { -4 }{ 3 }$

$=\dfrac { constant\quad term }{ coefficient\quad of\quad { x }^{ 2 } }$ .

Therefore, hence verified.

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$4x^2 + 5 \sqrt{2}x -3$

$4x^2 + 5 \sqrt{2}x -3$

$=4x^2+6\sqrt{2}x -\sqrt{2}x -3$

$=2\sqrt{2}x (\sqrt{2}x +3)-(\sqrt{2}x +3)$

$=(\sqrt{2}x +3)(2\sqrt{2}x-1)$
So, the zeros of the polynomial are

$(\sqrt{2}x +3)=0$ and

$(2\sqrt{2}x-1)=0$

$x=\frac{-3}{\sqrt{2}}$ and

$x=\frac{1}{2\sqrt{2}}$

$x=\frac{-3\sqrt{2}}{2}$ and

$x=\frac{\sqrt{2}}{4}$
Now, the polynomial is

$4x^2 + 5 \sqrt{2}x -3$
Comparing it with

$ax^2+bx+c=0$ , we get

$a=4$ ,

$b=5\sqrt{2}$ and

$c=-3$
Sum of the zeros

$=\frac{-3\sqrt{2}}{2}+\frac{\sqrt{2}}{4}$

$=\frac{-6\sqrt{2}+\sqrt{2}}{4}$

$=\frac{-5\sqrt{2}}{4}$

$=\frac{-b}{a}$
Product of the zeros

$=(\frac{-3\sqrt{2}}{2})(\frac{\sqrt{2}}{4})$

$=(\frac{-3}{2})(\frac{2}{4})$

$=\frac{-3}{4}$

$=\frac{c}{a}$

If one root of the equation $x^{3}-13x^{2}+15x+189= 0$ exceeds other by $2$ then find its smallest root.

Let the roots be

$\displaystyle \alpha ,\alpha +2,\beta$

$\displaystyle S_{1}= \Sigma \alpha = 2\alpha +\beta +2= 13 \\ \displaystyle \Rightarrow \beta = 11-2\alpha \ \ ...(1)$

$\displaystyle S_{2}= \Sigma \alpha \beta \\ = \alpha \left ( \alpha +2 \right )+\left ( \alpha +2 \right )\beta +\beta \alpha \\ \Rightarrow \displaystyle \alpha ^{2}+2\alpha +2\left ( \alpha +1 \right )\beta = 15\ \ ...(2)$

$\displaystyle S_{3}= \alpha \beta \gamma\\ \Rightarrow \displaystyle \alpha \beta \left ( \alpha +2 \right )= -189$

Eliminating

$\beta$ between (1) and (2), we get

$\displaystyle \alpha ^{2}+2\alpha +2\left ( \alpha +1 \right )\left ( 11-2\alpha \right )= 15\\ \Rightarrow \displaystyle 3\alpha ^{2}-20\alpha -7= 0$

$\displaystyle \Rightarrow \left ( \alpha -7 \right )\left ( 3\alpha +1 \right )= 0\\ \Rightarrow \displaystyle \alpha = 7,\displaystyle -\frac{1}{3}\\ \Rightarrow \beta = -3, \dfrac{35}{3}$
Out of these values

$\displaystyle \alpha = 7, \beta = -3$ satisfy the third relation

$\displaystyle \alpha \beta \left ( \alpha +2 \right )= -189 \Rightarrow \displaystyle \left ( -21 \right )\left ( 9 \right )= -189$
Hence the roots are

$7, 7+2, -3 \Rightarrow 7,9,-3.$

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2 -8x + k$ such that $\alpha^2+\beta^2=40$ , then the value of $k$ is

$p(x) = { x }^{ 2 }-8x+k$

$a=1, b=-8, c=k$

$\alpha+\beta=\dfrac{-b} a=8$ ,

$\alpha\beta=\dfrac c a=k$ ,

${\alpha}^{2}+{\beta}^{2}=40$

${\alpha}^{2}+{\beta}^{2}={(\alpha+\beta)}^{ 2 }-2\alpha\beta$

$40 = { 8 }^{ 2 } - 2k$

$2k=24$

$k=12$

Prove that there do not exist natural numbers $x$ and $y,$ with $x > 1,$ such that $\displaystyle \frac {x^7-1}{x-1}=y^5+1$ .

Simple factorization gives

$y^5 = x(x^3 + 1)(x^2 + x + 1)$ .
The three factors on the right are mutually coprime and hence they all have to be fifth powers.
In particular,

$x = r^5$ for some integer

$r$ .
This implies

$x^3 +1 = r^{15} +1$ , which is not a fifth power unless

$r = -1$ or

$r = 0.$

$\therefore$ there are no solutions to the given equation.

Verify that $-5, \displaystyle \frac{1}{2}, \frac{3}{4}$ are zeros of cubic polynomial $8x^3 + 30x^2 -47x + 15$ . Also verify the relationship between the zeros and the coefficients.

Let

$f \left(x\right) = 8x^3+ 30x^2-47x + 15$

Here,

$a =8,b=30,c=-47,d=15$

$f\left(-5\right)=8 \left(-5\right)^3+ 30\left(-5\right)^2-47\left(-5\right)+15$

$=-1000+750+235+15=0$

$f \left( \cfrac{1}{2}\right) = 8 \left( \cfrac{1}{2} \right)^3 +30 \left( \cfrac{1}{2}\right)^2- 47 \left(\cfrac{1}{2} \right) + 15$

$=8 \times \cfrac{1}{7} + 30 \times \cfrac{1}{4} -\cfrac{47}{2} + 15$

$= \cfrac{8+60-188+120}{8} =0$

and

$f\left( \cfrac{3}{4} \right) = 8 \left(\cfrac{3}{4} \right)^3 + 30\left(\cfrac{3}{4} \right)^2 - 47 \left( \cfrac{3}{4}\right) + 15$

$= 8 \times \cfrac{27}{64} + 30 \times \cfrac{9}{17} - \cfrac{14}{4} +15$

$= \cfrac{27}{8} + \cfrac{135}{8} - \cfrac{141}{4} +15$

$\cfrac{27+135-282+120}{8}=0$

So,

$-5 , \cfrac{1}{2}$ and

$\cfrac{3}{4}$ are the zeros of the polynominal

$f(x)$ .

Let,

$\alpha = -5, \beta = \cfrac{1}{2}$ and

$y= \cfrac{3}{4}$

Sum of zeros

$=\alpha + \beta + \gamma$

$=-5 + \cfrac{1}{2}+ \cfrac{3}{4} + \cfrac{-15}{4}$

Sum of zeros

$= - \cfrac{\text {coefficient of}\, x^2}{\text {coefficient of} \, x^3}$

$= -\cfrac{b}{a} = \cfrac{-30}{8} =\cfrac{-15}{4}$

So, sum of zeros

$=\alpha + \beta +\gamma$

$= \cfrac{\text {coefficient of}\, x^2}{\text {coefficient of} \, x^3}$

Sum of product of zeros taken two at a time

$=\alpha \beta + \beta \gamma + \gamma \alpha$

$=(-5)( \cfrac{1}{2} + (\cfrac{1}{2}) (\cfrac{3}{4}) + (\cfrac{3}{4}) (-5)$

$= \cfrac{-5}{2}+ \cfrac{3}{8}-\cfrac{15}{4}$

$= \cfrac{-20+3-30}{8} = \cfrac{-47}{8}$

Also, sum of product of zeros taken two at a time

$= \cfrac{\text {coefficient of} \, x}{\text {coefficient of} \, x^3}= \cfrac{c}{a} = \cfrac{-47}{8}$

So, sum of product of zeros taken two at a time

$=\alpha \beta + \beta \gamma + \gamma \alpha = \cfrac{\text {coefficient of} \, x}{\text {coefficient of} \, x^3}$

Product of zeros =

$\alpha \beta \gamma$

$=\left(-5\right) \left( \cfrac{1}{2}\right) \left(\cfrac{3}{4}\right) = -\cfrac{15}{8}$

Also, product of zeros =

$-\cfrac{\text {cosntant term}}{\text {coefficient of} \, x^3}$

$= - \cfrac{d}{a} = -\cfrac{15}{8}$

So, product of zeros

$=\alpha \beta \gamma$

$= - \cfrac{\text {constant term}}{\text {coefficient of}\, x^3}$

$3x^{2}+2x-1=0$ is a equation of degree _____

We know that the degree of a polynomial is the value of the greatest exponent (except the constant). To find the degree all that you have to do is find the largest exponent in the polynomial.

The given polynomial

$3x^2+2x-1=0$ has three terms. The first one is

$3x^2$ , the second is

$2x$ , and the third is

$-1$ .

The exponent of the first term is

$2$ .

The exponent of the second term is

$1$ because

$2x = 2x^1$ .

The exponent of the third term is

$0$ because

$-1 = -1\times x^0=-1\times 1$ (anything with an exponent of

$0$ is always equal to

$1$ ).

Since the highest exponent is

$2$ ,

Hence, the degree of the polynomial

$3x^2+2x-1=0$ is

$2$ .

Classify the following as linear, quadratic and cubic polynomials:
(i) $x^2+x$
(ii) $x-x^3$
(Iii) $y+y^2+4$
(iv) $1+x$
(v) $3t$
(vi) $r^2$
(vii) $7x^3$

(i) The highest degree of

$x^{2}+x$ is

$2$ , so it is a quadratic polynomials.

(ii) The highest degree of

$x-x^{3}$ is

$3$ , so it is a cubic polynomials.

(iii)The highest degree of

$y+y^{2}+4$ is

$2$ , so it is a quadratic polynomials.

(iv) The highest degree of

$x$ in

$(1+x)$ is

$1$ , so it is a linear polynomials.

(v)The highest degree of

$t$ in

$3t$ is

$1$ , so it is a linear polynomials.

(vi)The highest degree of

$r^{2}$ is

$2$ , so it is a quadratic polynomials.

(vii)The highest degree of

$x$ in

$7x^{3}$ is

$3$ , so it is a cubic polynomials.

The graphs of $y = p(x)$ are given in the figure, for some polynomials $p(x)$ . Find the number of zeroes of $p(x)$ , in each case.

(i) The graph does not intersect the

$x$ -axis, so there are no zeros in

$p(x)$ .

(ii) The graph intersects the

$x$ -axis at one place, so here there is one zero in

$p(x)$ .

(iii) The graph intersects the

$x$ -axis at three places, so that there are three zeros in

$p(x)$ .

(iv) The graph intersects the

$x$ -axis at two places, so here there are two zeros in

$p(x)$ .

(v) The graph intersects the

$x$ -axis at four places, so there are four zeros in

$p(x)$ .

(vi) The graph intersects the

$x$ -axis at three places, so that there are three zeros in

$p(x)$ .

Give the geometric representations of $2x+9=0$ as an equation
(i) in one variable
(ii) in two variable

(i) In one variable

Given equation is

$2x+9=0$

$\Rightarrow 2x=-9$

$\Rightarrow x=\dfrac{-9}{2}=-4.5$

(ii) Given equation is

$2x+9=0$

$\Rightarrow 2x+0y+9=0$

$\Rightarrow 2x=-9$

$\Rightarrow x=\dfrac{-9}{2}=-4.5$

It is line parallel to

$y$ -axis at a negative distance.

Two points lying it are the points

$A(-4.5, 0), B(-4.5, 2).$

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$ (ii) $4s^2 - 4s + 1$ (v) $t^2 - 15$ (vi) $3x^2 - x - 4$

(i) $x^2 - 2x - 8$

Factorize the equation, we get

$(x + 2) (x - 4)$
So, the value of

$x^2 - 2x - 8$ is zero when

$x + 2 = 0, x - 4 = 0,$ i.e., when

$x = -2$ or

$x = 4$ .
Therefore, the zeros of

$x^2 - 2x - 8$ are -2 and 4.
Now,

$\Rightarrow$ Sum of zeroes

$= -2 + 4 = 2$ =

$-\dfrac{2}{1} = -\dfrac{Coefficient \space\ of \space\ x}{Coefficient \space\ of \space\ x^{2}}$

$\Rightarrow$ Product of zeros

$= (-2) \times (4) = -8$ =

$\dfrac{-8}{1}= \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ x^2}$

(ii) $4s^2 - 4s + 1$

Factorize the equation, we get

$(2s - 1) (2s - 1)$
So, the value of

$4s^2 - 4s + 1$ is zero when

$2s - 1 = 0, 2s - 1 = 0$ , i.e., when

$s = \dfrac{1}{2}$ or

$s = \dfrac{1}{2}$ .
Therefore, the zeros of

$4s^2 - 4s + 1$ are

$\dfrac{1}{2}$ and

$\dfrac{1}{2}$ .
Now,

$\Rightarrow$ Sum of zeroes =

$\dfrac{1}{2}+\dfrac{1}{2} = 1 =-\dfrac{-4}{4} = -\dfrac{Coefficient \space\ of \space\ s}{Coefficient \space\ of \space\ s^{2}}$

$\Rightarrow$ Product of zeros =

$\dfrac{1}{2}\times \dfrac{1}{2} = \dfrac{1}{4} = \dfrac{1}{4}= \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ s^2}$

(iii) $6x^2 - 3 - 7x$

Factorize the equation, we get

$(3x + 1)(2x - 3)$
So, the value of

$6x^2 - 3 - 7x$ is zero when

$3x + 1 = 0, 2x - 3 = 0$ , i.e., when

$x = -\dfrac{1}{3}$ or

$x = \dfrac{3}{2}$ .
Therefore, the zeros of

$6x^2 - 3 - 7x$ are

$-\dfrac{1}{3}$ and

$\dfrac{3}{2}$ .
Now,

$\Rightarrow$ Sum of zeroes =

$-\dfrac{1}{3}+\dfrac{3}{2} = \dfrac{7}{6} =-\dfrac{-7}{6} = -\dfrac{Coefficient \space\ of \space\ x}{Coefficient \space\ of \space\ x^{2}}$

$\Rightarrow$ Product of zeros =

$-\dfrac{1}{3}\times \dfrac{3}{2} = -\dfrac{1}{2} = \dfrac{-3}{6}=\dfrac{-1}{2}= \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ x^2}$

(iv) $4u^2 + 8u$

Factorize the equation, we get

$4u(u+ 2)$
So, the value of

$4u^2 + 8u$ is zero when

$4u = 0, u + 2 = 0$ , i.e., when

$u = 0$ or

$u = -2$ .
Therefore, the zeros of

$4u^2 + 8u$ are

$0$ and

$-2.$
Now,

$\Rightarrow$ Sum of zeroes =

$0 - 2 = -2 = -\dfrac{8}{4}= -2 = -\dfrac{Coefficient \space\ of \space\ u}{Coefficient \space\ of \space\ u^{2}}$

$\Rightarrow$ Product of zeros =

$-0 x -2 = 0 = \dfrac{0}{4}= 0 = \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ u^2}$

(v) $t^2 - 15$

Factorize the equation, we get

$t = \pm \sqrt{15}$
So, the value of

$t^2 - 15$ is zero when

$t + \sqrt{15} = 0, t - \sqrt{15} = 0$ , i.e., when

$t = \sqrt{15}$ or

$t = -\sqrt{15}$ .
Therefore, the zeros of

$t^2 - 15$ are

$\pm \sqrt{15}$ .
Now,

$\Rightarrow$ Sum of zeroes =

$\sqrt{15}-\sqrt{15} = 0 = -\dfrac{0}{1} = 0 = -\dfrac{Coefficient \space\ of \space\ t}{Coefficient \space\ of \space\ t^{2}}$

$\Rightarrow$ Product of zeros =

$\sqrt{15}\times \sqrt{-15}= -15 = \dfrac{-15}{1}= \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ t^2}$

(vi) $3x^2 - x - 4$

Factorize the equation, we get

$(x + 1) (3x - 4)$
So, the value of

$3x^2 - x - 4$ is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x =

$\dfrac{4}{3}$ .
Therefore, the zeros of

$3x^2 - x - 4$ are -1 and

$\dfrac{4}{3}$ .
Now,

$\Rightarrow$ Sum of zeroes =

$-1+\dfrac{4}{3} = \dfrac{1}{3} = -\dfrac{-1}{3} = -\dfrac{Coefficient \space\ of \space\ x}{Coefficient \space\ of \space\ x^{2}}$

$\Rightarrow$ Product of zeros =

$-1 \times \dfrac{4}{3}= -\dfrac{4}{3} = \dfrac{-4}{3}= \dfrac{Constant \space\ term}{Coefficient \space\ of \space\ x^2}$

If two zeroes of the polynomial $x^4- 6x^3 - 26x^2 + 138x -35$ are $2\pm \sqrt{3}$ , find the other zeroes.

The two zeroes of the polynomial is

$2+ \sqrt{3}, 2-\sqrt{3}$

Therefore,

$(x-2+ \sqrt{3})(x-2- \sqrt{3}) = x^2+4-4x-3$
=

$x^2-4x+1$ is a factor of the given polynomial.

Using division algorithm, we get

$x^4 -6x^3 -26x^2 + 138x -35 = (x^2-4x+1)(x^2-2x-35)$

So,

$(x^2-2x-35)$ is also a factor of the given polynomial.

$x^2-2x-35 = x^2-7x+5x-35$

$=x(x-7)+5(x-7)$

$= (x - 7)(x + 5)$

Hence,

$7$ and

$-5$ are the other zeros of this polynomial.

On dividing $x^3- 3x^2 + x + 2$ by a polynomial g(x), the quotient and remainder were $( x-2 )$ and $(-2x + 4)$ , respectively. Find $g(x).$

We know the formula,

Dividend = Divisor x quotient + Remainder

$p(x) = g(x) \times q(x) + r(x)$

Substitute the values, we get

$x^3 -3x^2 + x + 2 = g(x)\times (x - 2) - 2x + 4$

Add

$2x$ and subtract

$4$ both side,

$x^3 - 3x^2+x+2+2x-4=g(x)\times(x-2)$

Simplify and divide by

$x-2$

$\dfrac{x^3-3x^2+3x-2}{x-2}=g(x)$

$g(x)=\dfrac{(x-2)(x^2-x+1)}{x-2}$

So,

$g(x)=x^2-x+1$

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) $t^2 - 3, 2t^4 + 3t^3-2t^2-9t -12$
(ii) $x^2+3x+1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$
(iii) $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$

(i)

$t^2 - 3, 2t^4 + 3t^3-2t^2-9t -12$
Remainder is 0, hence

$t^2 -3$ is a factor of

$2t^4 + 3t^3-2t^2-9t -12$

(ii) $x^2+3x+1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$
Remainder is 0, hence

$x^2 + 3x +1$ is a factor of

$3x^4 + 5x^3 - 7x^2 + 2x + 2$

(iii) $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$
Remainder is 2, hence

$x^3-3x +1$ is not a factor of

$x^5 - 4x^3 + x^2 + 3x + 1$

Give examples of polynomials $p(x), g(x), q(x)$ and $r(x)$ , which satisfy the division algorithm and
(i) deg $p(x) =$ deg $q(x)$ (ii) deg $q(x) =$ deg $r(x)$ (iii) deg $r(x) = 0$

(i) deg p(x) = deg q(x)

We know the formula,

Dividend = Divisor x quotient + Remainder

$p(x) = g(x) \times q(x) + r(x)$

So here the degree of quotient will be equal to degree of dividend when the divisor is constant.

Let us assume the division of

$4x^2$ by

$2$ .
Here,

$p(x) =$

$4x^2$

$g(x) = 2$

$q(x) =$

$2x^2$ and

$r(x) = 0$

Degree of

$p(x)$ and

$q(x)$ is the same i.e.,

$2$ .

Checking for division algorithm,

$p(x) = g(x) \times q(x) + r(x)$

$4x^2 = 2(2x^2)$

Hence, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of

$x^3 + x$ by

$x^2$ ,

Here, p(x) =

$x^3 + x$ , g(x) =

$x^2$ , q(x) = x and r(x) = x

Degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

$p(x) = g(x) \times q(x) + r(x)$

$x^3 + x = x^2 \times x + x$

$x^3 + x = x^3+ x$
Hence, the division algorithm is satisfied.

(iii) deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.
Let us assume the division of

$x^4+1$ by

$x^3$
Here, p(x) =

$x^4+1$
g(x) =

$x^3$

$q(x) = x$ and

$r(x) = 1$

Degree of

$r(x)$ is

$0.$

Checking for division algorithm,

$p(x) = g(x) \times q(x) + r(x)$

$x^4 + 1 = x^3 \times x + 1$

$x^4 + 1 = x^4+ 1$
Hence, the division algorithm is satisfied.

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) $2x^3+x^2-5x+2; \dfrac{1}{2}, 1, -2$

(ii) $x^3-4x^2+5x-2; 2, 1, 1$

(i) $2x^3+x^2-5x+2; \dfrac{1}{2}, 1, -2$

$p(x) = 2x^3+x^2-5x+2$ .... (1)

Zeroes for this polynomial are

$\dfrac{1}{2}, 1, -2$

Substitute the

$x=\dfrac 12$ in equation (1)

$p\left(\dfrac{1}{2}\right)=2(\dfrac{1}{2})^3 +(\dfrac{1}{2})^2 -5(\dfrac{1}{2})+2$

$=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{5}{2}+2$

$= 0$

Substitute the

$x= 1$ in equation (1)

$p(1)=2\times1^3+1^2-5\times 1+2$

$= 2 + 1 - 5 + 2 = 0$

Substitute the

$x= -2$ in equation (1)

$p(-2)= 2(-2)^3+(-2)^2-5(-2)+2$

$= -16 + 4 + 10 + 2 = 0$

Therefore,

$\dfrac{1}{2}, 1, -2$ are the zeroes of the given polynomial.

Comparing the given polynomial with

$ax^3+bx^2+cx+d$ we obtain,

$a = 2, b = 1, c = -5, d = 2$

Let us assume

$\alpha = \dfrac{1}{2}$ ,

$\beta = 1$ ,

$\gamma = -2$
Sum of the roots =

$\alpha+\beta+\gamma = \dfrac{1}{2}+1=2 = \dfrac{-1}{2}=\dfrac{-b}{a}$

$\alpha\beta + \beta \gamma+\alpha \gamma = \dfrac{1}{2}+ 1(-2)+\dfrac{1}{2}(-2)=\dfrac{-5}{2}=\dfrac{c}{a}$

Product of the roots =

$\alpha \beta \gamma = \dfrac{1}{2}\times x\times (-2)=\dfrac{-2}{2}=\dfrac{d}{a}$

Therefore, the relationship between the zeroes and coefficient are verified.

(ii) $x^3-4x^2+5x-2; 2, 1, 1$

$p(x) =$

$x^3-4x^2+5x-2$ .... (1)
Zeroes for this polynomial are

$2, 1, 1$

Substitute

$x=2$ in equation (1)

$p(2)=2^3-4\times 2^2+5\times 2-2$

$= 8 - 16 + 10 - 2 = 0$

Substitute

$x=1$ in equation (1)

$p(1)=x^3-4x^2+5x-2$

$=1^3-4(1)^2+5(1)-2$

$=1 - 4 + 5 - 2 = 0$

Therefore,

$2, 1, 1$ are the zeroes of the given polynomial.

Comparing the given polynomial with

$ax^3+bx^2+cx+d$ we obtain,

$a = 1, b = -4, c = 5, d = -2$
Let us assume

$\alpha = 2$ ,

$\beta = 1$ ,

$\gamma = 1$

Sum of the roots =

$\alpha+\beta+\gamma = 2+1+1=4=-\dfrac{-4}{1}\dfrac{-b}{a}$

Multiplication of two zeroes taking two at a time=

$\alpha\beta + \beta \gamma+\alpha \gamma = (2)(1)+(1)(1)+(2)(1)=5=\dfrac{5}{1}=\dfrac{c}{a}$

Product of the roots =

$\alpha \beta \gamma = 2\times 1\times 1=2=-\frac{-2}{1}=\dfrac{d}{a}$

Therefore, the relationship between the zeroes and coefficient are verified.

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following:
(i) $p(x) = x^3 -3x^2 + 5x -3,\ g(x) = x^2 -2$
(ii) $p(x) = x^4 -3x^2 + 4x + 5,\ g(x) = x^2 + 1 -x$
(iii) $p(x) = x^4 -5x + 6,\ g(x) = 2- x^2$

(i)

$x-3$

$x^2-2)\overline{ x^3-3x^2+5x-3}$

$x^3$

$-2x$

________________

$-3x^2+7x-3$

$-3x^2$

$-6$

________________

$7x-9$

Quotient

$=x-3$

Remainder

$=7x-9$

(ii)

$x^2+x-3$

$x^2-x+1)\overline { x^4-3x^2+4x+5+0x^3}$

$x^4+x^2$

$-x^3$

______________________

$x^3+4x^2+4x+5$

$x^3-x^2 +x$

______________________

$-3x^2+3x+5$

$-3x^2+3x-3$

______________________

$8$

Quotient

$=x^2+x-3$

Remainder

$=8$

(iii)

$-x^2-2$

$2-x^2) \overline {x^4+0x^3+0x^2-5x+6}$

$x^4$

$-2x^2$

____________________

$0x^3+2x^2-5x+6$

$+2x^2$

$-4$

_____________________

$-5x+10$

Quotient

$=-x^2-2$

Remainder

$=-5x+10$ .

Sketch the graph and identify zeroes of a linear polynomial: $\dfrac{x + 10}{4}$ .

Let

$y = \dfrac{x + 10}{4}$

$4y = 10 + x$

$y = \dfrac{10+x}{4}$
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.
Plotting

$(0, 2.5), (-10, 0)$ and joining them, we get another straight line.
The graph of

$4y - x = 10$ intersect the x-axis at

$(-10, 0)$ . Hence the zero of given polynomial is

$(-10, 0)$ as it intersects at

$x$ -axis at

$(-10, 0)$ .

Construct a graph which represent the zeroes of the quadratic polynomial for an equation $y=x^2-10x$ .

Factor of

$y=x^2-10x = x(x-10)=0$ or

$x = 0, x = 10$
Plotting

$(0, 0) (5, -25) (10, 0)$ and joining them.
Here, the graph cuts

$x$ -axis at two distinct points

$A$ and

$A'$ .
The

$x$ -coordinates of

$A(10, 0)$ and

$A'(0, 0)$ are the two zeroes of the quadratic polynomial

$y=x^2-10x$ .

Graphically represent the zeros of the given linear polynomial: $y=x-12$

It is known that if the curve of

$y=f(x)$ cuts the

$x-$ axis at point

$(p,0)$ then

$p$ is said to be the zero of

$y=f(x)$ .

Consider

$y = x - 12$

Plotting

$(0, -12), (12, 0)$ and joining them, we get ]graph of

$y=x-12$ .

The graph of

$x - y = 12$ intersect the

$x$ -axis at

$(12, 0)$ . Hence the zero of

$x - 12$ is

$12$ .

Graphically represent the zeroes of the quadratic polynomial for an equation $y=2x^2-11x+5$ in graph.

Factor of

$y=2x^2-11x+5 = (x-0.5)(x-5)$
Plotting

$(0.5, 0) (1, -2) (3, -5) (5, 0)$ and joining them.
Here, the graph cuts

$x$ -axis at two distinct points A and A'.
The x-coordinates of

$A(0.5, 0)$ and

$A'(5, 0)$ are the two zeroes of the quadratic polynomial

$y=2x^2-11x+5$ .

What are the zeroes of the quadratic polynomial from the graph?

A give polynomial cuts the x-axis at two points. So, it must have two roots.
Here the x-coordinates of

$(-2, 0)$ and

$(4, 0)$ are the two zeroes of the quadratic polynomial.

Find the zeroes of the quadratic polynomial from the graph.

Given polynomial graph cuts the x-axis at

$2$ points. So, it must have two roots.
Here the

$x$ -coordinates of

$(-2, 0)$ and

$(3, 0)$ are the two zeroes of the quadratic polynomial.

Represent the zeroes of the quadratic polynomial for an equation $y=x^2+x-9$ in graph.

Factor of

$x^2+x-9 = (x-3)(x+3)$
Plotting

$(-3, 0) (-1, -8) (0, -9) (1, -8) (3, 0)$ and joining them.
Here, the graph cuts

$x$ -axis at two distinct points

$A$ and

$A'$ .
The x-coordinates of

$A(-3, 0)$ and

$A (3, 0)$ are the two zeroes of the quadratic polynomial

$x^2+x-9$ .

Sketch a graph and represent the zeroes of the quadratic polynomial for the equation $y = x^2+2x-3$ .

Factor of

$x^2+2x-3 = (x+3)(x-1)$
Plotting

$(-3, 0) (-2, -3) (-1, -4) (1, 0$ ) and joining them.
Here, the graph cuts

$x$ -axis at two distinct points

$A$ and

$A'$ .
The

$x$ -coordinates of

$A(-3, 0)$ and

$A'(1, 0)$ are the two zeroes of the quadratic polynomial

$y = x^2+2x-3$ .

Graphically represent the zeroes of the cubic polynomial: $y=x^3+4x^2+x-6$

Factor of

$y=x^3+4x^2+x-6 = (x+2)(x-1)(x+3)$
Plotting

$(1, 0), (-1, -4), (2, 0), (-3, 0)$ and joining them.
Here

$1, -2, -3$ are the zeroes of the cubic polynomial.
These co-ordinates are the

$x$ -coordinates of the three points where the graph of

$y=x^3+4x^2+x-6$ intersects the

$x$ -axis.

Using division algorithm, divide the polynomial $2x^3+2x^2-5x+1$ by $(x-1)$ .

We divide the polynomial

$2x^3+2x^2-5x+1$ by

$(x-1)$ as shown in the above image:

Hence, the quotient is

$2x^2+4x-1$ and the remainder is

$0$ .

1107449_469634_ans_5f6ea726df564c76b1639a9f7f929b5f.jpg

Graphically represent the zeroes of the cubic polynomial: $y=x^3-3x+2$

Factor of

$x^3-3x+2 = (x-1)(x-1)(x+2)$
Plotting

$(-2, 0), (-1, -4), (0, -2), (1, 0)$ and joining them.
Here

$-1, -1, 2$ are the zeroes of the cubic polynomial.
These co-ordinates are the

$x$ -coordinates of the three points where the graph of

$x^3-3x+2$ intersects the

$x$ -axis.

$y=x^3-3x^2-6x+8$ , Draw a graphical representation of the polynomial.

The zeroes of polynomial

$y={x^3-3x^2-6x+8}$ are

$1,-2,4$

Quadratic polynomial: $y=x^2-4x+4$ , draw a graph and represent the zeroes of the polynomial.

Given quadratic polynomial is

$y=x^2-4x+4$

To find the points on the graph, we will substitute few values of

$x$ and find the values of

$y$ .

$x=0$ , then

$y=0^{2}-4\times0+4=0-0+4=4$

$x=1$ , then

$y=1^{2}-4\times1+4=1-4+4=1$

$x=2$ , then

$y=2^{2}-4\times2+4=4-8+4=0$

So, the points are

$(0, 4), (1, 1), (2, 0)$ .
Plotting

$(0, 4), (1, 1), (2, 0)$ and joining them.
Here, the graph cuts

$x$ -axis at one distinct point

$\text{A}$ .
The

$x$ -coordinates of

$\text{A} (2, 0)$ is the one zeroes of the quadratic polynomial

$y=x^2-4x+4$

Sketch a graphical representation of the zeros of the given polynomial: $y=x^3-12x^2+44x-48$

The zeroes of polynomial

$y={x^3-12x^2+44x-48}$ are

$2,4,6$

Divide: $2x^4+x^3+8x^2+14x+5$ by $(2x+1)$ (use division algorithm method).

Solution:

Let

$p(x)=2x^4+x^3+8x^2+14x+5$

and

$g(x)=2x+1$

Using Division Algorithm Method

$p(x)=q(x).g(x)+r(x)$

or,

$p(x)=2x^4+x^3+8x^2+14x+5$

and

$g(x)=2x+1$

then,

$q(x)=x^3+4x+5$ and

$r(x)=0$

Use synthetic division to find the quotient Q and Remainder R when dividing
$\dfrac12{x}^{3}-\dfrac13{x}^{2}-\dfrac32x+\dfrac13$ by $x-\dfrac12$

Since the divisor

$=x-\dfrac{1}{2}$

The zero of

$x-\dfrac{1}{2}=0$

$\Rightarrow x=\dfrac{1}{2}$

$\therefore$ The quotient is

$: \dfrac{1}{2}x^2-\dfrac{1}{12}x -\dfrac{37}{24}$

$\Rightarrow$ Remainder

$=\dfrac { \left( \dfrac { -21 }{ 48 } \right) }{ \left( x-\dfrac { 1 }{ 2 } \right) } .$

Hence, solved.

Use synthetic division to find the quotient Q and Remainder R when dividing
$7{x}^{5}-2{x}^{4}-5{x}^{3}+{x}^{2}-3x+5$ by $x-2$

$\Rightarrow Q=7x^4+12x^3+19x^2+39x+75$

$\Rightarrow R=155.$

Hence, solved.

1175542_486302_ans_61b34608d3404ae4993260f7bad924b3.png

Use synthetic division to find the quotient Q and Remainder R when dividing
$4{x}^{3}+{x}^{2}-2x+3$ by $x+2$

$\Rightarrow Q=4x^2-7x+12$

$\Rightarrow R=-21$

Hence, solved.

1171505_486298_ans_4ea7ab598f5e4e5e88ad5c32206c7c91.png

$\dfrac{3x^4+x^3+6x^2-16x-6}{3x+1}$ = ? (Use division algorithm method).

$x^3+2x-6$

$3x+1)\overline{ 3x^4+x^3+6x^2+6x-6}$

$3x^4+x^3$

_____________

$6x^2-16x$

$6x^2+2x$

_____________

$-18x-6$

___________

$0$

$\dfrac{3x^4+x^3+6x^2-16x-6}{3x+1}=0$

Given $f(x) =$ $4x^4+6x^3+6x^2+6x+2$ , $g(x) =$ $2x+2$ . Divide $\dfrac{f(x)}{g(x)}$ (Use division algorithm method)

$f\left( x \right) =4x^4+6x^3+6x^2+6x+2$

$g\left( x \right) =2x+2$

$\dfrac{f\left( x \right)}{g\left( x \right)}$ using division algorithm method,

so, the quotient

$=2x^3+x^2+2x+1$

Hence, the answer is

$2x^3+x^2+2x+1.$

Find the quotient and remainder on dividing $p(x) = x^{3} - 6x^{2} + 15x - 8$ by $g(x) = x - 2$

$p(x) = x^{3} - 6x^{2} + 15x - 8$

$\therefore$ degree of

$p(x)$ is

$3$ .

$g(x) = x - 2$

$\therefore$ degree of

$g(x)$ is

$1$ .

$\therefore$ degree of quotient

$q(x) = 3 - 1 = 2$ ,

and degree of remainder

$r(x)$ is zero.

Let,

$q(x) = ax^{2} + bx + c$ (Polynomial of degree

$2$ ) and

$r(x) = k$ (constant polynomial)

By using division algorithm, we have

$p(x) = [g(x) \times q(x) + r(x)]\\ = x^{3} - 6x^{2} + 15x - 8 = (x - 2) (ax^{2} + bx + c) + k$

$= ax^{3} + bx^{2} + cx - 2ax^{2} - 2bx - 2c + k$

$\therefore x^{3} - 6x^{2} + 15x - 8 = ax^{3} + (b - 2a)x^{2} + (c - 2b) x - 2c + k$
We have cubic polynomials on both the sides of the equation.

$\therefore$ Let us compare the coefficients of

$x^{3}, x^{2}, x$ and

$k$ to get the values of

$a, b, c$ .

$1 = a$ , it is the coefficient of

$x^{3}$ on both sides

$-6 = b - 2a$ , it is the coefficient of

$x^{2}$ on both sides

$15 = c - 2b$ , it is the coefficient of

$x$ on both sides

$-8 = -2c + k$ , it is the constant term on both sides
Let us solve these equations to get the values of

$b, c$ , and

$k$ .

$b - 2a = -6$

$\therefore b = -6 + 2a$

$\therefore b = -6 + 2(1) = -4$

$c - 2b = 15$

$\therefore c = 15 + 2b$

$\therefore c = 15 + 2(-4) = 7$

$-2c + k = -8$

$\therefore k = -8 + 2c$

$\therefore k = -8 + 2(7) = 6$

Hence,

$q(x) = ax^{2} + bx + c$

$= (1)x^{2} + (-4)x + 7$

$= x^{2} - 4x + 7$

and

$r(x) = k = 6$

$\therefore$ the quotient is

$x^{2} - 4x + 7$ and the remainder is

$6$ .

Divide $(5x + x^{2} + 14 + 2x^{3})$ by $(x + 2)$

Quotient :

$2x^2-3x+11$

Remainder :

$-8.$

Hence, solve.

1156952_562630_ans_a2068726a0ad4990aab7c26447974953.png

A polynomial $p(x)$ is divided by $g(x)$ , the obtained quotient $q(x)$ and the remainder $r(x)$ are given in the table. Find $p(x)$ in each case.
Sl. $p(x)$ $g(x)$ $q(x)$ $r(x)$
i ? $x - 2$ $x^{2} - x + 1$ $4$
ii ? $x + 3$ $2x^{2} + x + 5$ $3x + 1$
iii ? $2x + 1$ $x^{3} + 3x^{2} - x + 1$ $0$
iv ? $x - 1$ $x^{3} - x^{2} - x - 1$ $2x - 4$
v ? $x^{2} + 2x + 1$ $x^{4} - 2x^{2} + 5x - 7$ $4x + 12$

A polynomial

$p\left( x \right)$ is defined as

$\Rightarrow p\left( x \right) =g\left( x \right) q\left( x \right) +r\left( x \right)$

where

$g\left( x \right) =$ divisor ;

$q\left( x \right) =$ quotient and

$r\left( x \right) =$ remainder

$\therefore$

$p\left( x \right)$ can be found by multiplying

$g\left( x \right)$ with

$q\left( x \right)$

$\&$ adding

$r\left( x \right)$ to the product.

$(i).\; g\left( x \right) =\left( { x-2 } \right) ;$

$q\left( x \right)=x^2-x+1;$

$r\left( x \right)=4$

$\therefore p\left( x \right)=\left( { x-2 } \right) \left[ { x^2-x+1 } \right]+4$

$=x^3-x^2+x-2x^2+2x-2+4$

$=x^3-3x^2+3x+2$

$(ii).\; g\left( x \right) =\left( { x+3 } \right) ;$

$q\left( x \right)=2x^2+x+5;$

$r\left( x \right)=3x+1$

$\therefore p\left( x \right)=\left( { x+3 } \right) \left[ { 2x^2+x+5 } \right]+\left(3x+1\right)$

$=2x^3+x^2+5x+6x^2+3x+15+3x+1$

$=2x^3+7x^2+11x+16$

$(iii).\; g\left( x \right) =\left( { 2x+1 } \right) ;$

$q\left( x \right)=x^3+3x^2-x+1;$

$r\left( x \right)=0$

$\therefore p\left( x \right)=\left( { 2x+1 } \right) \left[ { x^3+3x^2-x+1 } \right]+\left(0\right)$

$=2x^4+6x^3-2x^2+2x+x^3+3x^2-x+1$

$=2x^4+7x^3+x^2+x+1$

$(iv).\; g\left( x \right) =\left( { x-1 } \right) ;$

$q\left( x \right)=x^3-x^2-x-1;$

$r\left( x \right)=2x-4$

$\therefore p\left( x \right)=\left( { x-1 } \right) \left[ { x^3-x^2-x-1 } \right]+\left(2x-4\right)$

$=x^4-x^3-x^2-x-x^3+x^2+x+1+2x-4$

$=x^4-2x^3+2x-3$

$(v).\; g\left( x \right) =\left( { x^2+2x+1 } \right) ;$

$q\left( x \right)=x^4-2x^2+5x-7;$

$r\left( x \right)=4x+12$

$\therefore p\left( x \right)=\left( { x^2+2x+1 } \right) \left[ { x^4-2x^2+5x-7} \right]+\left(4x+12\right)$

$=x^6-2x^4+5x^3-7x^2+2x^5+4x^3+10x^2-14x+x^4-2x^2+5x-7+4x+12$

$=x^6-x^4+x^3+x^2+2x^5-5x+5$

$=x^6+2x^5-x^4+x^3+x^2-5x+5$

Hence, solve.

What must be added to the polynomial $p(x) = x^{4} + 2x^{3} - 2x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2x - 3$

We know,

$p(x) = [g(x) \times q(x)] + r(x)$

$\therefore p(x) - r(x) = g(x) \times q(x)$

$\therefore p(x) + \left \{-r(x) \right \} = g(x) \times q(x)$

It is clear that RHS is divisible by

$g(x)$ .

$\therefore LHS$ is also divisible by

$g(x)$

Thus, if we add

$-r(x)$ to

$p(x)$ , then the resulting polynomial is divisible by

$g(x)$ .

Let us divide

$p(x)=x^4+2x^3-2x^2+x-1$ by

$g(x)=x^2+2x-3$ to find the remainder

$r(x)$ .

$\therefore r(x) = -x + 2\Rightarrow \left \{-r(x)\right \} = x - 2$

Hence, we should add

$(x - 2)$ to

$p(x)=x^4+2x^3-2x^2+x-1$ so that the resulting polynomial is exactly divisible by

$x^2+2x-3$ .

Draw the graph of $p(x) = x^{2} + 3x - 4$ and find zeroes.
Verify the zeroes of the polynomials

The graph of

$p(x)$ is given above.
Clearly, from the graph

$x=-4,1$ are zeros of

$p(x)$ .
Verification:-

$p(-4)=(-4)^2+3\times-4-4=16-12-4=0$

$p(1)=(1)^2+3\times1-4=1+3-4=0$

By which polynomial we should divide $(4x^{4} - 5x^{3} - 39x^{2} - 46x - 2)$ to get quotient as $( x^{2} - 3x - 5 )$ and remainder as $( -5x + 8 )$ .

$f(x)=(4x^{4}-5x^{3}-39x^{2}-4x-2)$ be the divident

$g(x)$ be the divider

$q(x)=x^{2}-3x-5$ be the quotient and

$r(x)=-5x+8$ be the remainder

We know that

$f(x)=g(x)q(x)+r(x)$

$\Rightarrow f(x)-r(x)=g(x)q(x)$

$\dfrac{f(x)-r(x)}{q(x)}=g(x)$

$\dfrac{4x^{4}-5x^{3}-39x^{2}-46x-2-(-5x+9)}{x^{2}-3x-5}=g(x)$

$\Rightarrow 4x^{4}-5x^{3}-39x{2}-14x-10\div x^{2}-3x-5$

By doing division; we get

$\left. { x }^{ 2 }-3x-5 \right) 4{ x }^{ 4 }-{ 5x }^{ 3 }-{ 39x }^{ 2 }-41x-10\left( { 4x }^{ 2 }+7x+2 \right. \\ \qquad \qquad \quad { 4x }^{ 2 }-{ 12x }^{ 3 }-{ 20x }^{ 2 }\\ \qquad \qquad \quad\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\_ \_ \_ \_ \\ \qquad \qquad \qquad \quad { 7x }^{ 3 }-{ 19 }x^{ 2 }-41x-10\\ \qquad \qquad \qquad \quad { 7x }^{ 3 }-{ 21x }^{ 2 }-35x\\ \qquad \qquad \quad\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \qquad \qquad \qquad \quad { 2x }^{ 2 }-6x-10\\ \qquad \qquad \qquad \qquad \quad { 2x }^{ 2 }-6x-10\\ \qquad \qquad \quad\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \qquad \qquad \qquad \qquad 0$

$\therefore g(x)=4x^{2}+7x+2$

If $\alpha, \beta, \gamma$ are the zeroes of the cubic polynomial $x^3 + 4x + 2$ , then find the value of
$\dfrac{1}{\alpha + \beta} + \dfrac{1}{\beta + \gamma} + \dfrac{1}{\gamma + \alpha}$

We have,

$P(x)=x^3+4x+2$

Let,

$\alpha, \beta, \gamma$ are the zeroes of this polynomial.

Here,

$a=1,b=0,c=4$ and

$d=2$

Now,

$\alpha+\beta+ \gamma=-\dfrac{b}{a}=-\dfrac{0}{1}=0.......(1)$

$\alpha\beta+ \beta\gamma+\gamma \alpha=\dfrac{c}{a}=\dfrac{4}{1}=4......(2)$

$\alpha\beta \gamma=-\dfrac{d}{a}=-\dfrac{2}{1}=-2.......(3)$

Since,

$\dfrac{1}{\alpha+\beta}+\dfrac{1}{ \beta+\gamma}+\dfrac{1}{\gamma+\alpha}$

$= \dfrac{(\beta+\gamma)(\gamma +\alpha)+(\alpha+\beta)(\gamma+\alpha)+(\alpha+\beta)(\beta+\gamma)}{(\alpha+\beta)( \beta+\gamma)(\gamma+\alpha)}$

$= \dfrac{\beta\gamma+\alpha\beta+\gamma^2 +\alpha\gamma+\alpha\gamma+\alpha^2+\beta\gamma+\alpha\beta+\alpha\beta+\alpha\gamma+\beta^2+\beta\gamma}{(\alpha+\beta)( \beta+\gamma)(\gamma+\alpha)}$

$= \dfrac{\alpha^2+\beta^2+\gamma^2 +3(\alpha\beta+\beta\gamma+\gamma\alpha)}{(\alpha+\beta)( \beta+\gamma)(\gamma+\alpha)}$

$= \dfrac{\alpha^2+\beta^2+\gamma^2 +2(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha\beta+\beta\gamma+\gamma\alpha)}{(\alpha+\beta)( \beta+\gamma)(\gamma+\alpha)}$

$= \dfrac{(\alpha+\beta+\gamma)^2+(\alpha\beta+\beta\gamma+\gamma\alpha)}{(\alpha+\beta)( \beta+\gamma)(\gamma+\alpha)}$

$= \dfrac{(\alpha+\beta+\gamma)^2+(\alpha\beta+\beta\gamma+\gamma\alpha)}{(-\gamma)( -\alpha)(-\beta)}$ [from

$(1)$ ]

$= \dfrac{(\alpha+\beta+\gamma)^2+(\alpha\beta+\beta\gamma+\gamma\alpha)}{-\alpha\beta\gamma}$

$= \dfrac{0^2+4}{-(-2)}$

$= \dfrac{4}{2}$

$= 2$

Hence, the answer is

$2.$

When a polynomial $f(x)$ is divided by $(x - 1)$ , the remainder is $5$ and when it is divided by $(x - 2)$ , the remainder is $7$ . Find the remainder when it is divided by $(x -1) (x - 2)$ .

Using Division Algorithm here:-

$Dividend=Divisor\times Quotient+Remainder$

So, Applying it

$:-$

Let

$q(x),k(x)$ be quotient when

$f(x)$ is divided by

$x-1$ and

$x-2$ respectively

$\Rightarrow f(x)=(x-1)q(x)+5$

$\therefore f(1)=5$ .....

$(1)$

Also,

$f(x)=(x-2)k(x)+7$

$\therefore f(2)=7$ .....

$(2)$

Now, let

$ax+b$ be remainder when

$f(x)$ is divided by

$(x-1)(x-2)$ and

$g(x)$ be quotient.

$f(x)=(x-1)(x-2)g(x)+(ax+b)$

Using

$(1)$ and

$(2)$

$5=a+b$ ......

$(3)$

$7=2a+b$ ......

$(4)$

Solving

$(3)$ and

$(4)$ , we get

$a=2$ and

$b=3$

$\therefore 2x+3$ is remainder when

$f(x)$ is divided by

$(x-1)(x-2).$

Find $k$ so that $x^2+2x+k$ is a factor of $2x^4+x^3-14x^2+5x+6$ . Also, find the zeroes of the two polynomials.

Given,

$x^2+2x+k$ is a factor of

$2x^4+x^3-14x^2+5x+6$ .

Then we can write,

$2x^4+x^3-14x^2+5x+6=$

$(x^2+2x+k)(2x^2+ax+b)$

[Where we are to determine the values of

$k,a,b$ ]

or,

$2x^4+x^3-14x^2+5x+6=2x^4+x^3(a+4)+x^2(b+2a+2k)+x(2b+ka)+kb$

Comparing the both sides we get,

$a+4=1\Rightarrow a=-3$ ........(1) and

$b+2a+2k=-14$ ........(2) and

$2b+ka=5$ ......(3) and

$kb=6$ ......(4).

Using (1) equation (2) and (3) can be written as

$b+2k=-8$ and

$2b-3a=5$

Solving these two equations we get,

$b=-2$ and

$k=-3$ , which satisfies equation (4).

So,

$k=-3$ .

Now the factors are

$(x^2+2x-3)(2x^2-3x-2)=(x+3)(x-1)(2x+1)(x-2)$ .

So, zeros of the

$4^{th}$ degree polynomial are

$-3,1,-\dfrac{1}{2},2$ among which

$-3,1$ are the zeros of the given second order polynomial.

If $9a+5B$ is divisible by 14, then prove that $5B+9a$ is also divisible by $14$ .

Given that,

$9a+5B$ is divisible by 14

Let, and

Then,

$9a+5B=9\times 1+5\times 1=14$

Which is divisible by 14

Now,

Put $a=1$ and B=1 in $5B+9a$

$\Rightarrow 5B+9a=5\times 1+9\times 1=14$

Which is divisible by 14

Hence, proved

If $x = \dfrac{2}{3}$ and $x = -3$ are the roots of the equation $ax^2 \, + \, 7x \, + \, b \, = \, 0,$ find the value of $a^2+ b^2$ .

$\Rightarrow$

$x=\dfrac{2}{3}$ and

$x=-3$ are the roots of the equation, then

$\left(x-\dfrac{2}{3}\right)(x-(-3))=0$ be that quadratic equation.

$\Rightarrow$

$\left(x-\dfrac{2}{3}\right)(x-(-3))=0$

$\Rightarrow$

$\left(x-\dfrac{2}{3}\right)(x+3)=0$

$\Rightarrow$

$x(x+3)-\dfrac{2}{3}(x+3)=0$

$\Rightarrow$

$x^2+3x-\dfrac{2}{3}x-2=0$

$\Rightarrow$

$3x^2+9x-2x-6=0$

$\Rightarrow$

$3x^2+7x-6=0$

$\Rightarrow$

$3x^2+7x-6=0$ , comparing it with

$ax^2+7x+b=0$

$\therefore$

$a=3$ and

$b=-6$

$\Rightarrow$

$a^2+b^2=(3)^2+(-6)^2=9+36=45$

Use division algorithm to show that any positive odd integer is of the form $6q+1$ or $6q+3$ or $6q+5$ , where $q$ is some integer.

Let

$a$ be any positive integer and

$b=6.$

Then by division algorithm,

$a=6q+r$ where

$r=0,1,2,3,4,5$

So,

$a$ is of the form

$6q,\;6q+1,\;6q+2,\;6q+3,\;6q+3,\;6q+4,\;6q+5.$

Therefore if

$a$ is an odd integer then

$a$ is of the form

$6q+1$ or

$6q+3$

$6q+5.$

Hence, any positive odd integer is of the form

$6q+1$ or

$6q+3$ or

$6q+5.$

Check whether ${x^2} - x + 1\,$ is a factor in ${x^3} - 3{x^2} + 3x - 2$

Given polynomial is

$x^3-3x^2+3x-2$ .

To find out: Whether

$x^2-x+1$ is a factor of the given polynomial or not.

$x^2-x+1$ will be a factor of

$x^3-3x^2+3x-2$ when the remainder is zero.

Dividing

$x^3-3x^2+3x-2$ by

$x^2-x+1$ , as shown in the figure. We get:

Quotient as

$x-2$ .

and Remainder as

$0$ .

Since the remainder is

$0$ ,

$x^2-x+1$ is a factor of

$x^3-3x^2+3x-2$ .

999931_1057647_ans_a03f6283bbc94a3daf90d7112cde5e72.png

Check $x^2-x+1$ is a factor of $x^3-3x^2-3x-2$

We divide

$x^3-3x^2-3x-2$ by

$x^2-x+1$ and if we get the remainder as

$0$ then it means that

$x^2-x+1$ is a factor of

$x^3-3x^2-3x-2$ .

From the above division, we have the quotient

$x-2$ and the remainder is

$0$ .

Hence,

$x^2-x+1$ is a factor of

$x^3-3x^2-3x-2$ .

How many zeros does cubic polynomial has?

A cubic polynomial will have

$3$ zeroes since its highest power (or degree) is

$3$

Find the reminder when $x^{3}+3x^{3}+3x+1$ is divided by
a) $5+2x$
b) $x-\dfrac{1}{2}$

Considering the question to be,

Find the reminder when

$x^3+3x^3+3x+1$ is divided by

(i)

$5+2x$

(ii)

$x-\dfrac{1}{2}$

Let

$p(x)=x^3+3x^3+3x+1$

(i)

$5+2x$

Put divisor =0

$5+2x=0$

$\Rightarrow x= -\dfrac{5}{2}$

Putting

$x= -\dfrac{5}{2}$ in

$x^3+3x^3+3x+1$

we get,

$p\left ( -\dfrac{5}{2} \right )=\left ( -\dfrac{5}{2} \right )^{3}+3\left (-\dfrac{5}{2} \right )^{3}+3\left ( -\dfrac{5}{2} \right )+1$

$=-\left ( \dfrac{125}{8} \right )-\left ( \dfrac{375}{8} \right )-\left ( \dfrac{15}{2} \right )+1$

$=\dfrac{-125-375-60+8}{8}$

$=\dfrac{-552}{8}=-69$

Thus, remainder of

$p\left ( -\dfrac{5}{2} \right )=-69$

(ii)

$x-\dfrac{1}{2}$

Put divisor =0

$$x\dfrac{1}{2}=0$$

$\Rightarrow x= -\dfrac{1}{2}$

Putting

$x= \dfrac{1}{2}$ in

$x^3+3x^3+3x+1$

we get,

$p\left ( \dfrac{1}{2} \right )=\left ( \dfrac{1}{2} \right )^{3}+3\left (\dfrac{1}{2} \right )^{3}+3\left ( \dfrac{1}{2} \right )+1$

$=\left ( \dfrac{1}{8} \right )+\left ( \dfrac{3}{8} \right )+\left ( \dfrac{3}{2} \right )+1$

$=\dfrac{1+3+12+8}{8}$

$=\dfrac{24}{8}=3$

Thus, remainder of

$p\left ( \dfrac{1}{2} \right )=3$

If $\alpha$ and $\beta$ are roots of given equation $8x^2-8\sqrt2 x+ 4=0$ and $\alpha\beta=\dfrac{1}{m}$ , then find $m$ .

$8x^2-8\sqrt{2}x+4=0$

comparing with equation

$ax^{2}+bx+c=0$

$a=8$ ,

$\quad b=-8\sqrt{2}$ ,

$\quad c=4$

$\alpha\beta=\cfrac{1}{m}=\cfrac{c}{a}=\cfrac{4}{8}$

$\therefore m=2$

Divide $6x^2-x-15$ by $(2x+3)$ and also verify the result.

$6x^2-x-15 = 6x^2 +9x -10x-15$

$\implies 3x(2x+3) -5(2x+3)$

$\implies (2x+3)(3x-5)$

Dividing by

$2x+3$ we get,

$=3x-5$

What must be subtracted from $4x^{4}-2x^{3}-6x^{2}+x-5$ so that result is exactly divisible by $2x^{2}+x-1$ ?

We divide

$4x^4−2x^3−6x^2+x−5$ by

$2x^2+x−1$ as shown in the below image:

From the division, we observe that the quotient is

$2x^2-2x-1$ and the remainder is

$-6$ .

Hence,

$-6$ should be subtracted from

$4x^4−2x^3−6x^2+x−5$ so that result is exactly divisible by

$2x^2+x−1$ .

Find the remainder when $4x^3-3x^2+4x-2$ divided by (i) $x-1$ (ii) $x-2$

Part

$1 :-$

$\left. x-1 \right) \overline { { 4x }^{ 3 }-{ 3x }^{ 2 }+4x-2 } \left( { 4x }^{ 2 }+x+5 \right. \\ \quad \quad \quad \quad { 4x }^{ 3 }-{ 4x }^{ 2 }\\ \quad \quad -\quad \quad \quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad { x }^{ 2 }+4x\\ \quad \quad \quad \quad \quad \ { x }^{ 2 }-x\\ \quad \quad \quad \quad -\quad \quad \quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad 5x-2\\ \quad \quad \quad \quad \quad \quad 5x-5\\ \quad \quad \quad \quad \quad -\quad \quad \quad \quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad 3\quad Remainder.\\$

Part

$2 :-$

$\left. x-2 \right) \overline { { 4x }^{ 3 }-{ 3x }^{ 2 }+4x-2 } \left( { 4x }^{ 2 }+5x+14 \right. \\ \quad \quad \quad \quad { 4x }^{ 3 }-{ 8x }^{ 2 }\\ \quad \quad -\quad \quad \quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad { 5x }^{ 2 }+4x\\ \quad \quad \quad \quad \quad { 5x }^{ 2 }-10x\\ \quad \quad \quad \quad \quad \quad -\quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad 14x-2\\ \quad \quad \quad \quad \quad \quad 14x-28\\ \quad \quad \quad \quad \quad -\quad \quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad 26\quad Remainder.$

Find the remainder when ${x}^{3} + {3x}^{2} + {3x} + 1$ is divided by
$X+1$

By remainder theorem

$x+1=0$

$x=-1$

$p(x)={ x }^{ 3 }+{ 3x }^{ 2 }-3x-1p(-1)$

$={ (-1) }^{ 3 }+3{ (-1) }^{ 2 }-3(-1)-1$

$=-1+3(1)+3-1$

$=-1+3+3-1$

$=6-2$

$=4$

Thus remainder is 4

Find the values of $a$ and $b$ so that the polynomial $({x}^{3}-10{x}^{2}+ax+b)$ is exactly divisible by $(x-1)$ as well as $(x-2)$

$x^{3}-10x^{2}+ax+b=f(x)$

$f(x)$ is divisible by

$(x-1)$ and

$(x-2)$

$\Rightarrow f(1)=0=f(2)$

$f(1)=1-10+a+b=0\Rightarrow a+b=9$

$f(2)=8-40+2a+b=0\Rightarrow 2a+b=32$

$\therefore a=23$

$b=-14$

Find :
$\dfrac{x^4-3x^2-2x-3}{(x-2)}$

We divide

$x^4-3x^2-2x-3$ by

$(x-2)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3+2x^2+x$ and the remainder is

$-3$ .

Hence, the quotient is

$x^3+2x^2+x$ and the remainder is

$-3$ .

1239639_1111019_ans_6266539ff3cd43fb995a955ad7500204.png

Divide using the long division method and check the answer.
$-11x+5x-4$ by $2x-1$

We divide

$-11x^2+5x-4$ by

$(2x-1)$ as shown in the above image:

From the division, we observe that the quotient is

$-\dfrac {11}{2}x-\dfrac {1}{4}$ and the remainder is

$-\dfrac {17}{4}$ .

We know that the division algorithm states that:

Dividend

$=$ Divisor

$\times$ Quotient

$+$ Remainder

We apply the division algorithm to check the answer as follows:

$Dividend=Divisor\times Quotient+Remainder\\ -11x^{ 2 }+5x-4=\left[ (2x-1)\left( -\dfrac { 11 }{ 2 } x-\dfrac { 1 }{ 4 } \right) \right] -\dfrac { 17 }{ 4 } \\ -11x^{ 2 }+5x-4=-11x^{ 2 }-\dfrac { 1 }{ 2 } x+\dfrac { 11 }{ 2 } x+\dfrac { 1 }{ 4 } -\dfrac { 17 }{ 4 } \\ -11x^{ 2 }+5x-4=-11x^{ 2 }+\dfrac { 10 }{ 2 } x-\dfrac { 16 }{ 4 } \\ -11x^{ 2 }+5x-4=-11x^{ 2 }+5x-4\\ Hence\quad verified.$

1248441_1127733_ans_1c6949d93b864c9290ddea54d99d2894.png

Find the remainder when $y^3 +y^2-2y+1$ is divided by $(y-a).$

By factor theorm ,

$y-a=0\\\implies y=a$

So the remiander is

$a^3+a^2-2a+1$

On dividing polynomial $x^3-3 x^2+x+2$ by a polynomial $g(x)$ , the quotient and remainder were $x-2$ and $-2x+4$ , respectively. Find $g(x)$ .

As we know that,

$\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} ..... \left( 1 \right)$

According to the question,

$\text{Dividend} = f{\left( x \right)} = {x}^{3} - 3{x}^{2} + x + 2$

$\text{Divisor} = g{\left( x \right)} = ?$

$\text{Quotient} = q{\left( x \right)} = x - 2$

$\text{Remainder} = r{\left( x \right)} = -2x + 4$

Now, from

${eq}^{n} \left( 1 \right)$ , we have

$f{\left( x \right)} = g{\left( x \right)} \times q{\left( x \right)} + r{\left( x \right)}$

${x}^{3} - 3{x}^{2} + x + 2 = g{\left( x \right)} \times \left( x - 2 \right) + \left( -2x + 4 \right)$

$\left( {x}^{3} - 3{x}^{2} + x + 2 \right) - \left( -2x + 4 \right) = g{\left( x \right)} \times \left( x - 2 \right)$

$\Rightarrow g{\left( x \right)} = \cfrac{\left( {x}^{3} - 3 {x}^{2} + 3x - 2 \right)}{\left( x - 2 \right)}$

As from the attached image, we have

$\left( {x}^{3} - 3 {x}^{2} + 3x - 2 \right) \div \left( x - 2 \right) = {x}^{2} - x + 1$

$\therefore g{\left( x \right)} = \cfrac{\left( {x}^{3} - 3 {x}^{2} + 3x - 2 \right)}{\left( x - 2 \right)} = \left( {x}^{2} - x + 1 \right)$

1060916_1073479_ans_e80e110762b946a9b15866b74e9d9dca.png

Divide using the long division method and check the answer.
$x^{2}-3x+2$ by $x-2$

$\left. x-2 \right) { x }^{ 2 }-3x+2\left( x-1 \right. \\ \quad \quad \quad { x }^{ 2 }-2x\\ \quad \quad \quad -\quad +\quad \\--------------\\ \quad \quad -x+2\\ \quad \quad -x+2\\ \quad \quad +\quad -\\ ---------------\\ \quad \quad \quad 0$

Ans :

$x-1$

Divide using the long division method and check the answer.
$3x^{2}+4x^{2}+x+7$ by $x^{2}+1$

We divide

$3x^3+4x^2+x+7$ by

$(x^2+1)$ as shown in the above image:

From the division, we observe that the quotient is

$3x+4$ and the remainder is

$-2x+3$ .

Now we will apply the division algorithm to check the answer as follows:

$\begin{aligned}{}{\text{Dividend}} &= {\text{Divisor}} \times {\text{Quotient}} + {\text{Remainder}}\\3{x^3} + 4{x^2} + x + 7 &= \left[ {({x^2} + 1)\left( {3x + 4} \right)} \right] - 2x + 3\\ &= (3{x^3} + 4{x^2} + 3x + 4) - 2x + 3\\& = 3{x^3} + 4{x^2} + 3x - 2x + 4 + 3\\3{x^3} + 4{x^2} + x + 7 &= 3{x^3} + 4{x^2} + x + 7\quad\quad\quad\quad[\text{Which is true}]\end{aligned}$

Hence verified

1248442_1127744_ans_017f9a43874746ce90837801ae532c16.png

Solve:
$\left( {{y^2} + 10y + 24} \right) \div \left( {y + 4} \right)$

$y^{2}+10y+24$

$=y^{2}+4y+6y+24$

$=y(y+4)+6(y+4)$

$=(y+6)(y+4)$

$\therefore \dfrac{y^{2}+10y+24}{y+4}=\dfrac{(y+6)(y+4)}{y+4}$

$=y+6$

Divide $2x^{2}+{3x}+{1}$ by $x+2$ and find quotient and remainder.

We divide

$2x^2+3x+1$ by

$(x+2)$ as shown in the above image:

From the division, we observe that the quotient is

$2x-1$ and the remainder is

$3$ .

Hence, the quotient is

$2x-1$ and the remainder is

$3$ .

1248963_1129487_ans_a68c2ad43f2248d9a57a1489db910353.png

$(2x^{4} + 3x^{3} + 4x - 2x^{2})\div (x + 3)$ .

Divide

$(2x^4+3x^3+4x-2x^2)$ by

$(x+3)$

$2x^3-3x^2+7x-17$

_________________

$x+3)2x^4+3x^3-2x^2+4x$

$2x^4+6x^3$

______________

$-3x^3-2x^2+4x$

$-3x^3-9x^2$

________________

$7x^2+4x$

$7x^2+21x$

_____________

$-17x$

$-17x-51$

__________

$51$

________

$=2x^3-3x^2+7x-17$

Remainder

$=51$ .

Divide $z^2+16z+64$ by $z+8$ .

${ z^{ 2 } }+16z+64\, \, by\, \, z+8 \\ { z^{ 2 } }=16z+64\left( { z+8 } \right) \to quotient \\ { \underline { z } ^{ 2 } }+\underline { 8 } z \\ \overline { \begin{matrix} \, \, \, \, \, \, \, \, 8z+64 \\ \, \, \, \, \, \, \, \, 8z+64 \\ \, \, \, \, \, \, \, \, \overline { 0\, \, \, \, \, \, \, \, \, \, \, 0 } \\ \end{matrix} } \, \, \, \, \, { { Remainder } }\, \, \, 0$

Divide. Write the quotient and the remainder.
(a) $(21{x}^{4}-14{x}^{2}+7x)\div 7{x}^{3}$
(b) $(6{x}^{5}-4{x}^{4}+8{x}^{3}+2{x}^{2})\div 2{x}^{2}$
(c) $(25{m}^{4}-15{m}^{3}+10m+8)\div 5{m}^{3}$

$\dfrac{21x^{4}-14x^{2}+7x}{7x^{3}}$

$Q=3x, R=-14x^{2}+7x$

$\dfrac{6x^{5}-4x^{4}+8x^{3}+2x^{2}}{2x^{2}}$

$Q=3x^{3}-2x^{2}+4x+1$

$\dfrac{25m^{4}-15m^{3}+10m+8}{5m^{3}}$

$Q: 5m-3$

$R: 10m+8$

$16x(9x+12)$ by $(3x-4)$

The given polynomial

$16x(9x+12)$ can be rewritten as

$144x^2+192x$ .

We now divide

$144x^2+192x$ by

$(3x-4)$ as shown in the above image:

From the division, we observe that the quotient is

$48x$ and the remainder is

$384x$ .

Hence, the quotient is

$48x$ and the remainder is

$384x$ .

1248445_1128594_ans_bc6c7e42b08b44c4b1e4e1d30a07a5ca.png

Rewrite the terms in proper order and then divide
$12+10y^{3}+10y^{2}+8y-2$ by $y^{2}-y+6$

The given polynomial

$12+10y^3+10y^2+8y-2$ can be rewritten as

$10y^3+10y^2+8y+10$ .

We now divide

$10y^3+10y^2+8y+10$ by

$(y^2-y+6)$ as shown in the above image:

From the division, we observe that the quotient is

$10y+20$ and the remainder is

$-32y-110$ .

Hence, the quotient is

$10y+20$ and the remainder is

$-32y-110$ .

1248965_1129565_ans_ba62436f366145f39a09f271f9f7e08e.jpg

Divide $10x^4+12x^3-62x^2+30x-3$ by $2x^2+7x-1$ .

1188035_1142345_ans_38e9f7eae9894132824114d70d85ce4b.JPG

Solve:
$x+6-{3x-5-2(x-5)}=2x-9-{2x+5-3(x+1)}$

$x+6-3x-5-2(x-5)=2x-9-2x+5-3(x+1)$

$\Rightarrow x+6-3x-5-2x+10=-9+5-3x-3$

$\Rightarrow -x+11=-7$

$\Rightarrow x=11+7$

$\Rightarrow x=18$ .

Divide $\left( { 12x }^{ 2 }-3x-450 \right) \div (x+6)$ .

$\therefore \left( 12{ x }^{ 2 }-3x-450 \right) \div \left( x+6 \right) =12x-75$
1381030_1244671_ans_2b7d7a9386e44156b198a7854b62449e.jpg

Solve :
$\left( {2{m^3} + {m^2} + m + 9} \right) \div \left( {2m - 1} \right)$

Given:

$(2m^3+m^2+m+9)\div (2m-1)$

$m^2+m+1$

__________________

$(2m-1))$

$2m^3+m^2+m+9$

$2m^3-m^2$

__________________

$2m^2+m+9$

$2m^2-m$

_______________

$2m+9$

$2m-1$

____________

$10$

$\therefore (2m^3+m^2+m+9)=(2m-1)(m^2+m+1)+10$ .

1118328_1195590_ans_3b59a1011ea24631b86b565a3b192249.jpeg

If $\alpha = 3 , \beta = 5 , \gamma = 2 ,$ find the cubic polynomial

$\alpha = 3 , \beta 5, \gamma =2$

Cubic polynomial

$\Rightarrow (x-\alpha )(x-\beta )(x-\gamma )$

$\Rightarrow (x-3)(x-5)(x-2)=0$

$\Rightarrow (x^2-8x+15)(x-2)=0$

$\Rightarrow x^3 - 8x^2+15 x - 2x^2 +16 x-30 =0$

$\Rightarrow x^3 - 10x^2 +31 x - 30 =0$

$\therefore$ Cubic polynomial is

$x^3 - 10x^2 +31 x-30=0$

1213144_1309897_ans_7ef7520be82d4b18b41650dc62728e9d.jpeg

Find values of $a$ and $b$ , if $(x^2 + 1)$ is a factor of the polynomial $x^2+ x^3 +8x^2 + ax + b$ .

1194184_1283993_ans_5fe4750300a643be90dd12ce9ba2e614.jpg

Using division algorithm, find the quotient and remainder on dividing.
$8x^{4}+14x^{3}-2x^{2}+8x-12$ by $4x^{2}+3x-2$

1183406_1240333_ans_d80e4375ee2b472d8cf5359ded150c30.png

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

1179171_1275843_ans_4843a79e27e0406796564b04a52ea709.jpg

Check whether $2{x}^{4}-3{x}^{3}-3{x}^{2}+6x-2$ is divisible by ${x}^{2}-2$

1211651_1362833_ans_7af865e4f91848238f184d8485bcab1f.JPG

If $a{x}^{n-1}+b{x}^{n-2}+c{x}^{n-3}$ is a cubic polynomial where $n\in N$ , then find the value of $n$

General form of a cubic polynomial is

$a{x}^{3}+b{x}^{2}+cx+d$

$\Rightarrow a{x}^{n-1}+b{x}^{n-2}+c{x}^{n-3}=a{x}^{3}+b{x}^{2}+cx+d$

$\Rightarrow n-1=3$ (on comparing the coefficients)

$\Rightarrow n=3+1=4$

$\therefore\,n=4$

Divide $(2x^{4}+3x^{3}+4x-2x^{2})\div (x+3)$

$x+3)2x^4+3x^3-2x^2+4x(2x^3-3x^2+7x-17$

$2x^4+6x^3$

_______________

$-3x^3-2x^2+4x$

$-3x^3-9x^2$

_________________

$7x^2+4x$

$7x^2+21x$

__________________

$-17x$

$-17x-51$

___________________

$51$

1214795_1396371_ans_96ab18190b0d45e28e4078ac4cadda17.jpg

For which value of a and b, the zeroes of $q\left(x\right)=x^{3}+2x^{2}+a$ are also the zeroes of the polynomial $p\left(x\right)=x^{5}-x^{4}-4x^{3}+3x^{2}+3x+b$ ?

$\Rightarrow 3-a-4=0,3a+3=0,b-2a=0$

$\Rightarrow -a-1=0,3\left(a+1\right)=0,b=2a$

$\Rightarrow -a=1,a=-1,b=2a$

$\therefore a=-1$ ,

$b=2a=2\times -1=-2$

$\therefore a=-1,b=-2$

1298634_1430988_ans_eab295a5e935499692b2df8802bf55c8.PNG

What is cubic polynomial ?

A polynomial of degree 3 is called a cubic polynomial.

Divide the polynomial ${ 6x }^{ 3 }+{ 11x }^{ 2 }-10x-7$ by the binomial $2x + 1$ . Write the quotient and the remainder.

From the above division, we can conclude that,

The quotient is

$3x^2+4x-7$ and the remainder is

$0.$

1306722_1447609_ans_a8b90d695e704d5dbfc7c337e37fbdf6.png

Find all zeroes of the polynomial $2x^3+x^2-6x-3$ if two of its zeroes are $-\sqrt{3}$ & $\sqrt{3}$

Given polynomial is

$2 x^3+x^2-6 x-3$

The two roots of polynomial are

$-\sqrt{3},\sqrt{3}$

Let the other root be

$a$

Product of the roots will be

$-\dfrac{(-3)}{2}$

$\implies (-\sqrt{3})(\sqrt{3})a=\dfrac{3}{2}$

$\implies a=-\dfrac{1}{2}$

So the other zero is

$-\dfrac{1}{2}$

The graph of a polynomial $y=p(x)$ is given below. By looking at the graph, find the number of zeroes of $p(x)$ .

Here graph of a polynomial cuts the

$x-axis$ at

$4$ points, so the number of zeros of

$p(x)$ is 4.

Given that $x - \sqrt 5$ is a factor of the cubic polynomial ${x^3} - 3\sqrt 5 {x^2} + 13x - 3\sqrt 5 ,$ Find all the zeros of the polynomial.

Given

polynomial is

$x^3-3\sqrt5x^2+13x-3\sqrt5$

And

$(x-\sqrt5)$ is a factor.

So, we use long division to find the other factor as shown in the image,

$\therefore x^3-3\sqrt5x^2+13x-3\sqrt5=(x-\sqrt5)(x^2-2\sqrt5x+3)$

Now using quadratic formula for

$x^2-2\sqrt5x+3$ , We get

$x=\sqrt5\pm\sqrt2$

Hence the roots are

$\sqrt5,\ \sqrt5+\sqrt2$ and

$\sqrt5-\sqrt2$

1655891_1440924_ans_ddee536dc1494ff9a1bee265cb6142d6.png

Find the zeroes of the given polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
${ 7y }^{ 2 }-\dfrac { 11 }{ 3 } y-\dfrac { 2 }{ 3 }$

Let

$f(y) = 7y^2 - \dfrac{11}{3}y - \dfrac{2}{3}$

For zeroes of

$f(y), f(y) =0$

$\Rightarrow 7y^2 - \dfrac{11}{3} y - \dfrac{2}{3} = 0$

$\Rightarrow 21y^2 - 11y - 2 =0$

$\Rightarrow 21y^2 - 14y + 3y - 2 = 0$

$\Rightarrow 7y(3y-2) + 1(3y-2)=0$

$\Rightarrow 3y - 2 = 0$ or

$7y +1 = 0$

$\Rightarrow y = \dfrac{2}{3}$ or

$y = \dfrac{-1}{7}$

Verification:

$\alpha = \dfrac{2}{3}, \beta = \dfrac{-1}{7}$ ,

$q = 7, b = -\dfrac{11}{3}, c = \dfrac{-2}{3}$

$\Rightarrow \alpha + \beta = \dfrac{-b}{a}$

$\Rightarrow \left( \dfrac{2}{3}\right) - \dfrac{1}{7} = \dfrac{+\dfrac{11}{3}}{7}$

$\Rightarrow \dfrac{14-3}{21} = \dfrac{11}{3}\times \dfrac{1}{7}$

$\Rightarrow \dfrac{11}{21} = \dfrac{11}{21}$

$\Rightarrow LHS = RHS$

Hence, verified.

$\alpha . \beta = \dfrac{c}{a}$

$\Rightarrow \left(\dfrac{2}{3}\right) \times \left(\dfrac{-1}{7}\right) = \dfrac{\dfrac{-2}{3}}{7}$

$\Rightarrow \dfrac{-2}{21} = \dfrac{-2}{3}\times \dfrac{1}{7}$

$\Rightarrow \dfrac{-2}{21} = \dfrac{-2}{21}$

$\Rightarrow LHS = RHS$

Hence, verified.

Given that the zeroes of the cubic polynomial $f(x)=x^3-6x^2+3x+10$ are of the form $a,a+b,a+2b$ for some real numbers $a$ and $b$ , find the values of $a$ and $b$ as well as the zeros of the given polynomial.

Given

Cubic polynomial is

$f(x)=x^3-6x^2+3x+10$

And the roots are

$a,\ a+b$ and

$a+2b$

Using the relationship between coefficient and Zeroes of the polynomial we get,

$\text{Sum of roots}=-\dfrac{\text{coefficient of } x²}{\text{ coefficient of } x³}\\$

$\Rightarrow (a)+(a+b)+(a+2b)=-\dfrac{(-6)}1\\$

$\Rightarrow 3(a+b)=6$

$\Rightarrow a+b=2\qquad...(i)$

And

$\text{Product of roots} = -\dfrac{\text{constant}}{\text{coefficient of }x³}\\$

$\Rightarrow a(a+b)(a+2b)=-\dfrac{10}1\\$

$\Rightarrow a(a+b)(a+b+b)=-10$

$\Rightarrow a(2)(2+b)=-10\qquad[$ Putting the value of

$a+b$ from

$(i)]$

$\Rightarrow 2a(2+2-a)=-10\qquad[$ Putting the value of

$b$ from

$(i)]$

$\Rightarrow a(4-a)=-5$

$\Rightarrow 4a-a^2+5=0$

$\Rightarrow a^2+4a-5=0$

Now solving obtained quadratic equation we get,

$a = 5 , -1$

when

$a = 5 ,\ b = -3$ ; Zeroes are

$5 , ( 5 + (-3) ) = 2 , ( 5 + 2(-3) ) = -1$

when

$a = -1 ,\ b = 3$ : Zeroes are

$-1 , ( -1 + 3 ) = 2 , ( -1 + 2(3) ) = 5$

Hence,

$a = 5\ \&\ b = -3$ or

$a = -1\ \&\ b = 3$ and Zeroes are

$-1 , 2 , 5.$

Identify constant, linear, quadratic and cubic polynomials from the following polynomials:
$r(x) = 3x^{3} + 4x^{2} + 5x - 7$ .

1432916_1668723_ans_c9d061a1d8164c7fbd87c4694d8b6a7d.jpg

Find the zeros of the following polynomial by factorization method and verify the relations between the zeroes and the coefficient of the polynomials :
$v^{ 2 }+4\sqrt { 3 } v-15$

Let

$f(v) = v^2 + 4v\sqrt{3} - 15$

For zeroes of

$f(v), f(v) = 0$

$\Rightarrow v^2 + 4 \sqrt{3}v - 15 = 0$

$\Rightarrow v^2 + 5 \sqrt{3}v - 1\sqrt{3}v - 15 = 0$

$\begin{bmatrix} 15=5\times 3 \\= 1\times 5 \times \sqrt{3}\times \sqrt{3} \end{bmatrix}$

$\Rightarrow v(v+5\sqrt{3}) - \sqrt{3}(v+5\sqrt{3}) = 0$

$\Rightarrow (v + 5 \sqrt{3})(v - \sqrt{3}) = 0$

$\Rightarrow (v + 5\sqrt{3}) = 0$ or

$(v - \sqrt{3}) = 0$

$\Rightarrow v = -5\sqrt{3}$ or

$v = \sqrt{3}$

verification:

$\alpha = -5\sqrt{3}, \ \beta = \sqrt{3}$

$a = 1, \ b = 4 \sqrt{3}$ , and

$c = -15$

$\alpha + \beta = \dfrac{-b}{a}$

$\Rightarrow -5 \sqrt{3} + \sqrt{3} = \dfrac{-4 \sqrt{3}}{1}$

$\Rightarrow -4 \sqrt{3} = -4 \sqrt{3}$

$\Rightarrow LHS = RHS$

Hence, verified.

$\alpha . \beta = \dfrac{c}{a}$

$\Rightarrow (-5\sqrt{3})(\sqrt{3}) = \dfrac{-15}{1}$

$\Rightarrow -5 \times 3 = -15$

$\rightarrow -15 = -15$

$\Rightarrow LHS = RHS$

Hence, verified.

Obtain all the zeroes of the polynomial $2x^4+x^3-14x^2-19x-6$ . If two zeroes are $-2$ and $-1$ .

1313519_1534390_ans_0d5c12e2d6b940f081ca316d66bfe535.jpeg

Identify constant, linear, quadratic and cubic polynomials from the following polynomials:
$g(x) = 2x^{3} - 7x + 4$ .

$g(x)=2x^3-7x+4$

Here, we can see highest degree of variable

$x$ is

$3$ .

We know that,

A cubic polynomial is a polynomial of degree

$3.$

$\therefore$

$g(x)=2x^3-7x+4$ is a cubic polynomial.

What must be added to $x^{3} - 3x^{2} - 12x + 19$ so that the result is exactly divisibly by $x^{2} + x - 6$ ?

1649690_1668926_ans_589f6913792e42aca037401989790ea7.jpeg

Find the cubic polynomial with the sum of zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as $-4,\dfrac { 1 }{ 2 } ,\dfrac { -1 }{ 3 }$

Let the zeroes be

$\alpha, \beta$ and

$\gamma$ .
Then, we have

$\alpha + \beta + \gamma = - 4$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{1}{2}$

$\alpha \beta \gamma = \dfrac{1}{3}$
Now , required cubic polynomial

$= x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha)x - \alpha \beta \gamma$

$= x^3 - (-4)x^2 + \left ( \dfrac{1}{2} \right )x - \left ( \dfrac{1}{3} \right )$

$= \dfrac{6x^3 + 24x^2 + 3x - 2}{6}$

So,

$6x^3 + 24x^2 + 3x - 2$ is the required cubic polynomial which satisfy the given conditions.

Find the zeros of the following polynomial by factorization method and verify the relations between the zeroes and the coefficient of the polynomials :
${ 4x }^{ 2 }-3x-1$

Let

$f(x) = 4x^2 - 3x - 1$

Splitting the middle term, we get

$=4x^2 - 4x + 1x - 1$

$=4x(x-1)+1(x-1)$

$=(x-1)(4x+1)$

For

$f(x) = 0$ , we have

$4x^2 - 3x - 1 = 0$

$(x-1)(4x+1)=0$

Either

$x - 1 = 0 \Rightarrow x = 1$

$4x + 1 = 0$

$\Rightarrow 4x = -1$

$\Rightarrow x = \dfrac{-1}{4}$

$\therefore$ The zeroes of

$f(x)$ are

$1$ and

$\dfrac{-1}{4}$ .

Verification:

$\alpha =1, \beta = \dfrac{-1}{4}$

$a = 4, b= -3$ and

$c= -1$

$\therefore \alpha + \beta = \dfrac{-b}{a}$

$\Rightarrow 1 - \dfrac{1}{4} = \dfrac{-(-3)}{4}$

$\Rightarrow \dfrac{3}{4} = \dfrac{3}{4}$

$\Rightarrow LHS = RHS$

Hence, verified

$\alpha . \beta = \dfrac{c}{a}$

$\Rightarrow 1 \times \left(\dfrac{-1}{4}\right) = \dfrac{-1}{4}$

$\Rightarrow \dfrac{-1}{4} = \dfrac{-1}{4}$

$\Rightarrow LHS = RHS$

Hence, verified

What must be added to $3x^{3} + x^{2} - 22x + 9$ so that the result is exactly divisible by $3x^{2} + 7x - 6$ ?

1650393_1668929_ans_56747b78d1fd4ded97d5a29ee7b4a66c.jpeg

Divide each of the following and find the quotient and remainder:
$30x^{4} + 11x^{3} - 82x^{2} - 12x + 48$ by $3x^{2} + 2x - 4$ .

Given:

$30x^4 + 11x^3 - 82x^2 - 12x + 48\ \text{divided by}\ 3x^2 + 2x - 4$

To find:

Quotient and remainder

Solution:

Consider,

$10x^2 - 3x -12$

$3x^2 + 2x - 4\ \overline{ )30x^4 + 11x^3 - 82x^2 - 12x + 48}$

$\underline{30x^4 + 20x^3 - 40x^2 }$

$-9x^3 - 42x^2 + 6x$

$\underline{-9x^3 - 6x^2 + 12x}$

$- 36x^2 + 0x + 48$

$\underline {-36x^2 - 12x + 48}$

$12x$

From the above long division,

$\text{Quotient}= 10x^2 - 3x^2 - 12$

$\text{Remainder}= 12x$

Verify division algorithm i.e, $Dividend = Divisor \times Quotient + Remainder$ , in each of the following. Also, write the quotient and remainder:
$4y^{3} + 8y + 8y^{2} + 7$ by $2y^{2} - y + 1$ .

$\dfrac{4y^3 + 8y + 8y^2 + 7}{2y^2 - y + 1} \Rightarrow \dfrac{4y^3 + 8y^2 + 8y + 7}{2y^2 - y + 1}$

ref. image

$Q = 2y + 5 , R = 11 y+ 2$

Divided = (Divisor

$\times$ Quotient ) + Remainder

$= ((2y^2 - y + 1) \times (2y + 5)) + 11y + 2$

$= 4y^3 - 2y^2 + 2y + 10y^2 - 5y + 5 + 11y + 2$

$= 4y^3 + 8y^2 + 8y + 7$

1506154_1672547_ans_cfff26a74c244778b8d713e417a75531.png

Divide each of the following and find the quotient and remainder:
$9x^{4} - 4x^{2} + 4$ by $3x^{2} - 4x + 2$ .

Given:

$9x^4 - 4x^2 + 4\ \text{divided by}\ 3x^2 - 4x + 2$

To find:

Quotient and remainder

Solution:

Consider,

$3x^2 + 4x + 2$

$3x^2 - 4x + 2\ \overline{ )9x^4 - 4x^2 + 4}$

$\underline{9x^4 + 6x^2 -12x^3}$

$12x^3 - 10x^2 + 4$

$\underline{12x^3 - 16x^2 + 8x}$

$6x^2 - 8x + 4$

$\underline {6x^2 - 8x + 4}$

$0$

From the above long division,

$\text{Quotient}= 3x^2 +4x +2$

$\text{Remainder}= 0$

Divide the polynomial $f(x) = 3x^2 - x^3 - 3x + 5$ by the polynomial $g(x) = x - 1 - x^2$ and verify the division algorithm.

Given data :- Given Polynomial ,

$f(x) = 3x^2 - x^3 - 3x + 5$

$g(x) = x - 1 - x^2$

Solution :-

Ref. image

Now, to verify the division algorithm.

$f(x) = g(x) . p(x) + r(x)$

$= (-x^2 + x - 1) (x - 2) + 3 ... (p(x) = (x - 2)$ &

$= -x^3 + 2x^2 + x^2 - 2x + 2 + 3 - x$

$= -x^3 + 3x^2 - 3x + 5$

Hence its verified.

1501228_1700960_ans_4f735c31d3644c3f9b604d8bfed5606c.png

Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:
$10x^{2} - 7x + 8, 5x - 3$ .

Given that:

$10x^2 - 7x + 8 \ \text{is divided by}\ 5x - 3$

To find:

Quotient and Remainder

Solution:

$2x - \dfrac{1}{5}$

$5x - 3\ \overline{)10x^2 - 7x + 8}$

$\underline{10x^2 - 6x}$

$-x + 8$

$\underline{-x + \dfrac{3}{5}}\\$

$\dfrac{37}{5}$

From the above long division,

$\text{Quotient}\ = 2x - \dfrac{1}{5}\\$

$\text{Remainder}\ = \dfrac{37}{5}$

Write the quotient and remainder:
Divide $14x^{2} + 13x - 15$ by $7x - 4$

$\dfrac{14x^2+13x-15}{7x-4}$

Quotient

$=2x+3$

Remainder

$=-3$

1506627_1672542_ans_e1170b7773fe40d59f54ce0cd43d08b0.png

Verify divison algorithm for the polynomial $f(x) = 8 +20 x +x^2 - 6x^3$ by $g(x) = 2 +5x - 3x^2$

1961104_1714059_ans_f92a8c1e37a440eb9fb7091e4d9f5901.jpg

Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:
$x^{4} - x^{3} + 5x, x - 1$ .

Given that:

$x^4 - x^3 + 5x \ \text{is divided by} \ x - 1$

To find:

Quotient and Remainder

Solution:

$x^3 + 5$

$x - 1\ \overline{)x^4 - x^3 + 5x}$

$\underline{x^4 - x^3}$

$5x$

$\underline{5x - 5}$

$5$

From the above long division,

$\text{Quotient}\ = x^3 + 5$

$\text{Remainder}\ = 5$

Verify division algorithm i.e, $Dividend = Divisor \times Quotient + Remainder$ , in each of the following. Also, write the quotient and remainder:
$15y^{4} - 16y^{3} + 9y^{2} - \dfrac {10}{3} y + 6$ by $3y - 2$ .

$\dfrac{15y^4 - 16y^3 + 9y^2 - \dfrac{10}{3} y + 6}{3y - 2}$

Ref. image

$Q = 5y^3 - 2y^2 + \dfrac{5}{3} y, \, R= 6$

Verification:

$Divi$

$= (Divis \times Q) + R$

$= ((5y^3 - 2y^2 + \dfrac{5}{3} y) (3y - 2)) + 6$

$= 15y^4 - 6y^3 + 5y^2 - 10y^3 + 4y^2 + \dfrac{10}{3} y + 6$

$= 15 y^4 - 16 y^3 + 9y^2 - \dfrac{10}{3} y + 6$

1505624_1672546_ans_1c44113f758a4712a08a5873605b95cc.png

State division algorithm for polynomials.

Division algorithm for polynomials states that, suppose

$f(x)$ and

$g(x)$ are the two polynomials, where

$g(x)≠0,$ we can write:

$f(x) = q(x) g(x) + r(x)$

which is same as the

$Dividend = Divisor \times Quotient + Remainder$ and where

$r(x)$ is the remainder polynomial and is equal to

$0$ and degree

$r(x) <$ degree

$g(x).$

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.
$6y$ .

$6y$ is a linear polynomial because it has a degree of

$1$

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.
$-7+x$ .

$-7+x$ is a linear polynomial because it has a degree of

$1$

Identify constant, linear, quadratic, cubic polynomial from the following.
$-z^3$ .

$-z^3$ is a cubic polynomial because it has a degree of

$3$

What must be added to $f(x) = 4x^4 + 2x^3 - 2x^2 + x - 1$ so that the resulting polynomial is divisible by $g(x) = x^2 + 2x - 3?$

By division algorithm, we have

$f(x) = g(x) \times q(x) + r(x)$

$\Rightarrow f(x) - r(x) = g(x) \times q(x)$

Clearly,

$RHS$ is divisible

$g(x)$ . Therefore

$LHS$ is also divisible by

$g(x)$ .

Thus, if we add

$-r(x)$ to

$f(x)$ , then the resulting polynomial is divisible by

$g(x)$

Let us now find the remainder

$r(x)$ when

$f(x)$ is divided by

$g(x)$

$\therefore r(x) = -61x + 65$

and

$-r(x) = 61x -65$

Hence we should add

$-r(x) = 61x - 65$ to

$f(x)$ so that the resulting polynomial is divisible by

$g(x)$ .

Identify constant, linear, quadratic, cubic and quartic polynimial from the following.
$1+x+x^2$ .

$1+x+x^2$ is a quadratic polynomial because it has a degree of

$2$

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.
$-6x^2$ .

$-6x^2$ is a quadratic polynomial because it has a degree of

$2$

Identify constant, linear, quadratic, cubic polynomial from the following.
$1-y-y^3$ .

$1-y-y^3$ is a cubic polynomial because it has a degree of

$3$

The graphs of $y=p(x)$ for some polynomials (for question 1 to 4) are given below. Find the number of Zero in each case.

1. There is no zero as the graph does not intersect the x-axis.
2. The number of zeros in four as the graph intersects the x-axis at four points.
3. The number of zeros is three as the graph intersects the x-axis at three points.
4. The number of zeros is three as the graph intersects the x-axis at three points.

If $\alpha , \beta , \gamma$ be Zero of polynomial $6x^3 +3x^2 -5x +1$ then find the value of $\alpha^{-1} , \beta^{-1} , \gamma^{-1}$

$P(x) = 6x^3 +3x^2 - 5x + 1$ so

$a=6,b = 3,c= -5, d= 1$

$\because \alpha, \beta$ and

$\gamma$ are zeros of the polynomial

$p(x)$

$\therefore \alpha +\beta + \gamma= \frac{-b}{a}= \frac{-3}{6} = \frac{-1}{2}$

$\alpha \beta + \alpha \gamma +\beta \gamma = \frac{c}{a} = \frac{-5}{6}$ and

$\alpha \beta \gamma=\frac{-d}{a} = \frac{-1}{6}$
Now

$\alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{1}{\alpha } = \frac{1}{\beta }=\frac{1}{\gamma}=\frac{\beta \gamma + \alpha \gamma+\alpha \beta }{\beta \gamma}= \frac{-5/6}{-1/6}=5$

Verify the division algorithm for the polynomials $p(x)=2x^4-6x^3+2x^2-x+2$ and $g(x)=x+2$ .

1677182_1715931_ans_bce5e8c0bbbc40a39308d27adeea7be8.jpg

Find the values of $a$ for which all the roots of the equation $x^{4}$ $-4 x^{3}-8 x^{2}+a=0$ are real

Given,

$x^4-4x^3-8x^2+a=0$

Let

$f(x)=x^{4}-4 x^{3}-8 x^{2}$

$\Rightarrow f^{\prime}(x)=4 x^{3}-12 x^{2}-16 x=4 x\left(x^{2}-3 x-4\right)=4 x(x-4)(x+1)$ Hence, graph of

$y=f(x)$

Now, graph of

$y=f(x)+a$ moves up or down depending on values of

$a$ . It is clear that if equation

$f(x)+a=0$ has four real roots,

then

$0 \leq a<3$ .

1578229_1746235_ans_b02bedb3e3d645b5bcf756fdce452258.png

If the graph of a polynomial intersects the $x-axis$ at exactly two points,is it necessarily a quadratic polynomial?

No, the polynomial is not necessarily to be a quadratic polynomial.

For ex.

$x^4 - 1$ is a polynomial intersecting the

$x - axis$ at exactly two points.

State whether the following expression is polynomials or not? Justify your answer.
$\dfrac{(x - 2)(x - 4)}{x}$

$\dfrac{(x - 2)(x - 4)}{x} = \dfrac{x^2 - 6x + 8}{x} = x - 6 + \dfrac{8}{x} = x - 6 + 8x^{-1}$
Here, the exponent of variable x in the third term, i.e., in

$8x^{-1},$ is

$-1,$ which is not a whole number. So, this algebraic expression is not a polynomial.

State whether the following expression are polynomials or not? Justify your answer.
$\dfrac{1}{x + 1}$

$\dfrac{1}{x + 1} = (x + 1)^{-1}$ which cannot be reduced to an expression in which the exponent of the variable x has only whole numbers in each of its terms. So, this algebraic expression is not a polynomial.

State whether the following expression is polynomials or not? Justify your answer.
$\dfrac{1}{5x^{-2}} + 5x + 7$

$\dfrac{1}{5x^{-2}} + 5x + 7 = \dfrac{1}{5}x^2 + 5x + 7$
In each term of this expression, the exponent of the variable x is a whole number. Hence, it is a polynomial.

State whether the following expression is polynomials or not? Justify your answer.
$1 - \sqrt{5x}$

$1 - \sqrt{5x} = 1 - \sqrt{5} \ x^{\frac{1}{2}}$
Here, the exponent of the second term, i.e.,

$x^{\frac{1}{2}}, \dfrac{1}{2},$ which is not a whole number. Hence, the given algebraic expression is not a polynomial.

State whether the following expression are polynomials or not? Justify your answer.
$\dfrac{1}{2x}$

$\dfrac{1}{2x} = \dfrac{1}{2}x^{-1}$
Here, the exponent of the variable x is

$-1,$ which is not a whole number so, this algebraic expression is not a polynomial.

State whether the following expression is polynomials or not? Justify your answer.
$\sqrt {3}\ x^2 - 2x$

$\sqrt {3} \ x^2 - 2x$
In each term of this expression, the exponent of the variable x is a whole number. Hence, it is a polynomial.

Classify the following as a constant, linear quadratic and cubic polynomials:
$5t - \sqrt{7}$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a linear polynomial

Classify the following polynomial as polynomials in one variable, two variable etc.
$x^2 + x + 1$

$x^2 + x + 1$ is a polynomial in one variable.As

$x$ is the only variable present here.

Classify the following as a constant, linear quadratic and cubic polynomials:
$2 - x^2 + x^3$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a Cubic polynomial.

Classify the following polynomial as polynomials in one variable, two variable etc.
$x^2 - 2xy + y^2 + 1$

$x^2 - 2xy + y^2 + 1$ is a polynomial in two variable(x,y).

Classify the following as a constant, linear quadratic and cubic polynomials:
$4 - 5y^2$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a quadratic polynomial

Classify the following as a constant, linear quadratic and cubic polynomials:
$3x^3$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a cubic polynomial

Classify the following as a constant, linear quadratic and cubic polynomials:
$1 + x + x^3$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a cubic polynomial.

Classify the following as a constant, linear quadratic and cubic polynomials:
$2 + x$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

$1$ is called a linear polynomial.
(c) A polynomial of degree

$2$ is called a quadratic polynomial.
(d) A polynomial of degree

$3$ is called a cubic polynomial.
Hence, given polynomial is a linear polynomial.

Classify the following as a constant, linear quadratic and cubic polynomials:
$y^3 - y$

We know that
(a) a polynomial in which exponent of the variable is zero is called a constant polynomial.
(b) a polynomial of degree

11 is called a cubic polynomial.
Hence, given polynomial is a cubic polynomial.

Classify the following as a constant into linear quadratic and cubic polynomials:
$\sqrt{2}x - 1$

We know that

(a) a polynomial in which exponent of the variable is zero, is called a constant polynomial.

(b) a polynomial of degree

$1$ is called a linear polynomial.

$2$ is called a quadratic polynomial.

(d) A polynomial of degree

$3$ is called a cubic polynomial.

Here,

$\sqrt2x-1$ has maximum degree

$1$ , so it is a linear polynomial.

Classify the following as a constant into linear, quadratic and cubic polynomials:
$t^2$

We know that

(a) a polynomial in which exponent of the variable is zero, is called a constant polynomial.

(b) a polynomial of degree

$1$ is called a linear polynomial.

$2$ is called a quadratic polynomial.

(d) A polynomial of degree

$3$ is called a cubic polynomial.

Here

$t^2$ has degree

$2$ , so it is a quadratic polynomial

Which of the following is quadratic polynomial?
i) $2-\dfrac{1}{3}x^2$
ii) $x+\dfrac{1}{\sqrt{x}}$
iii) $x+\dfrac{1}{x}$
iv) $x^2+3\sqrt{x^{+2}}$

i) On solving the equations,

$2-\dfrac{1}{3}x^2$

Re-writing in the format of

$ax^2+bx+c=0$

$(-\dfrac{1}{3})x^2+x(0)+2=0$

$\because a=-\dfrac{1}{3}b=0$ &

$c=2$

So, following the ideal pattern of a quadratic polynomial

$2-\dfrac{1}{3}x^2$ is a quadratic polynomial.

ii) On solving the equations,

$x+\dfrac{1}{\sqrt{x}}$

$\because$ it can't re-written in the format of

$ax^2+bx+c=0$

So,following the ideal pattern of a quadratic polynomial ,

$x+\dfrac{1}{\sqrt{x}}$ is not a quadratic polynomial.

iii) On solving the equations,

$x+\dfrac{1}{x}$

$\because$ it can't be re-written in the format of

$ax^2+bx+C=0$

So,following the ideal pattern of a quadratic polynomial.

iv) On solving the equations,

$x^2+3\sqrt{x}+2$

$\because$ it can't be re-written in the format of

$ax^2+bx+c=0$

So,following the ideal pattern of a quadratic polynomial,

$x^2+3\sqrt{x}+2$ is not a quadratic polynomial.

Which of the following is a quadratic polynomial?
$2x^2+1$

On solving the equations,

$2x^2+1=0$

Re-writing in the format of

$ax^2+bx+c=0$

$2x^2+(0)x+1=0$

$\because a=2\ \ \ b=0$ and

$c=1$

So,following the ideal pattern of a quadratic polynomial .

Fill in the blanks:
$ax^n+bx+c$ is a quadratic polynomial if n=........

$ax^n+bx+c$ is a quadratic polynomial if

$n=2$

and also

$a \neq 0$ , as it will make the polynomial a monomial.

Which of the following is a quadratic polynomial?
$x^2+\dfrac{1}{\sqrt{x}}$

On solving the equations,

$x^2+\dfrac{1}{\sqrt{x}}$

$\because$ it can't be re-written in the format of

$ax^2+bc+c=0$

So,following the ideal pattern of quadratic polynomial,

$x^2+\dfrac{1}{\sqrt{x}}$ is not a quadratic polynomial.

Which of the following is a quadratic polynomial?
$3\sqrt{x^2+1}+x$

On solving the equations,

$3\sqrt{(x^2+1)}+x=0$

$\because$ it can\t be written in the format of

$ax^2+bx+c=0$

So, following the ideal pattern of a quadratic polynomial,

$3\sqrt{(x^2+1)}+x=0$ is not a quadratic polynomial.

Which of the following is a polynomial?
$\dfrac{\sqrt{3}}{2}+x^2$

On solving the equations,

$\dfrac{\sqrt{3}}{2}+x^2=0$

$\because$ After simplifying the equation, one of the term has a positive (2) exponent.

So,following the ideal pattern of a polynomial,

$\dfrac{\sqrt{3}}{2}+x^2=0$ is a polynomial.

Fill in the blanks:
The value of the quadratic polynomial $x^2-5x+4$ for $x=1$ is..............

The value of the quadratic polynomial

$x^2-5x+4$ for

$x=1$ is 10

Putting the value of x=-1. in

$x^2-5x+4$

$(1)^2-5(1)+4$

$1-5+4$

$0$

$\therefore$ The value is 0.

Fill in the blanks: $x^2+x+3$ is a ................. polynomial.

$x^2+x+3$ is a Quadratic polynomial.

because it is in the form of

$ax^2+bx^2+c=0$

Which of the following is a quadratic polynomial?
$\sqrt{x^2+1}+\dfrac{1}{\sqrt{x}}$

On solving the equations,

$\sqrt{x^2+1}+\dfrac{1}{\sqrt{x}}$

$\because$ it can\t be written in the format of

$ax^2+bx+c=0$

So, following the ideal pattern of a quadratic polynomial,

$\sqrt{x^2+1}+\dfrac{1}{\sqrt{x}}$ is not a quadratic polynomial.

Apply Division Algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $p(x)$ by $g(x)$ as given below:
$p(x) = 6x^3 + 13x^2 + x - 2, g(x) = 2x + 1$

$p(x) = 6x^3 + 13x^2 + x - 2, g(x) = 2x + 1$
Dividend

$= 6x^3 + 13x^2 + x - 2$
Divisor

$= 2x + 1$
Here, dividend and divisor both are in the standard form.
Now, on dividing

$p(x)$ by

$g(x)$ we get the following division process
Quotient

$= 3x^2 + 5x - 2$
Remainder

$= 0$

1788842_1815370_ans_1ecb3d4faedd470f89b8ab8e3663ee8f.png

Apply Division Algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $p(x)$ by $g(x)$ as given below:
$p(x) = x^3 - 3x^2 + 4x + 2, g(x) = x^2+2x+1$

$p(x) = x^3 - 3x^2 + 4x + 2, g(x)$

$= x^2+2x+1$
Dividend

$= x^3 - 3x^2 + 4x + 2$
Divisor

$= x^2+2x+1$
Here, dividend and divisor both are in the standard form.
Now, on dividing

$p(x)$ by

$g(x)$ we get the following division process
Quotient

$= x - 5$
Remainder

$= 13x + 7$
1788874_1815363_ans_e1842d6b6d734a9bb84bea418129d318.png

Apply Division Algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $p(x)$ by $g(x)$ as given below:
$p(x) = 2x^2 + 3x + 1, g(x) = x + 2$

$p(x) = 2x^2 + 3x + 1, g(x) = x + 2$
Dividend

$= 2x^2 + 3x + 1$
Divisor

$= x + 2$
Apply the division algorithm we get,

$q(x)= 2x - 1$

$r(x)= 3$

1788893_1815356_ans_b9e43faeec56497fa8180c06ee759e68.png

Apply Division Algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $p(x)$ by $g(x)$ as given below:
$p(x) = x^4 - 1, g(x) = x + 1$

$p(x) = x^4 - 1, g(x) = x + 1$
Dividend

$= x^4 - 1$
Divisor

$= x + 1$
Here, dividend and divisor both are in the standard form.
Now, on dividing

$p(x)$ by

$g(x)$ we get the following division process

Quotient

$=x^3 - x^2 + x - 1$ ;

Remainder

$= 0$

1788886_1815360_ans_246602af360b479180b6749b8f647395.png

Examine, seeing the graph of the polynomial given below, whether it is a linear or quadratic polynomial or neither linear nor quadratic polynomial:

The given graph have a straight line but it doesn't intersect at

$\mathrm{x}-$ axis and the shape of the graph is also not a parabola.
So, it is not a graph of a quadratic polynomial.
Therefore, it is not a graph of linear polynomial or quadratic polynomial.

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial: $x^2 - 3x + 4, 2x^4 - 11x^3 + 29x^2 - 30x + 29$

Let us divide

$2x^4 - 11x^3 + 29x^2 - 30x + 29$ by

$x^2 - 3x + 4$
The division process is
Here, the remainder is

$58x - 115$ ,
therefore,

$x^2 - 3x + 4$ is not a factor of

$2x^4 - 11x^3 + 29x^2 - 30x + 29$
1788824_1815387_ans_e17a313bd63b40d0abdd35a419545d64.png

Apply Division Algorithm to find the quotient $q(x)$ and remainder $r(x)$ on dividing $p(x)$ by $g(x)$ as given below:
$p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2$

$p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2$
Dividend

$= x^3 - 3x^2 + 5x - 3$
Divisor

$= x^2 - 2$
Apply the division algorithm we get,

$q(x)= x - 3$

$r(x)= 7x - 9$
1788892_1815359_ans_4c2174969cdf4f00b3244ddd2c83a43b.png

Examine, seeing the graph of the polynomial given below, whether it is a linear or quadratic polynomial or neither linear nor quadratic polynomial:

In general, we know that for a linear polynomial ax

$+b$ ,

$\mathrm{a} \neq \mathrm{0},$ the graph of

$\mathrm{y}=\mathrm{ax}+\mathrm{b}$ is a straight line which intersects the

$\mathrm{x}-$ axis at exactly one point.
And here, we can see that the graph of

$y=p(x)$ is a straight line and intersects the

$\mathrm{x}-$ axis at exactly one point.
Therefore, the given graph is of a Linear Polynomial.

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial: $x - 2, x^3 + 3x^2 - 12x + 4$

Let us divide

$x^3 + 3x^2 - 12x + 4$ by

$x - 2$
The division process is
Here, the remainder is

$0$ ,
therefore,

$x - 2$ is a factor of

$x^3 + 3x^2 - 12x + 4$
1788832_1815377_ans_f91c3831c71e43c886784ac08c53bef5.png

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial: $x^2 - 4x + 3, x^3 - x^3 - 3x^4 - x + 3$

Let us divide

$x^3 + 3x^2 - x + 3$ by

$x^2 - 4x + 3$
The division process is
Here, the remainder is

$0$ ,
therefore,

$x^2 - 4x + 3$ is a factor of

$x^3 - 3x^2 - x + 3$
1788818_1815406_ans_3f9110fcf20342e184e533d43008b163.png

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial: $t^2 - 5t + 6, t^2 + 11t - 6$

Let us divide

$t^2 + 11t - 6$ by

$t^2 - 5t + 6$
The division process is
Here, the remainder is

$16t - 12$ ,
Therefore,

$t^2 - 5t + 6$ is not a factor of

$t^2 + 11t - 6$
1788807_1815421_ans_252aea205eda4a77aa1e388d08089c55.png

Verify that the numbers given alongside of the cubic polynomial are its zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
$x^3 - 6x^2 + 11x - 6; 1, 2, 3$

Let

$p(x) = x^3 - 6x^2 + 11x - 6$

Then,

$p (1) = (1)^3 - 6(1)^2 + 11(1) - 6$

$= 1 - 6 + 11 - 6$

$= 0$

$p(2) = (2)^3 - 6(2)^2 + 11(2) - 6$

$= 8 - 24 + 22 - 6$

$= 0$

$p(3) = (3)^3 - 6(3)^2 + 11(3) - 6$

$= 27 - 54 + 33 - 6$

$= 0$

Hence,

$1, 2$ and

$3$ are the zeroes of the given polynomial

$x^3 - 6x^2 + 11x - 6$ .

Now, Let

$\alpha = 1, \beta = 2$ and

$\gamma = 3$

Then,

$\alpha + \beta + \gamma = 1 + 2 + 3 \\= 6$

Also,

$\alpha+\beta+\gamma = - \dfrac{Coefficient of x^2}{Coefficient of x^3} \\= - \dfrac{-6}{1} \\= 6$

Solving further,

$\alpha \beta + \beta \gamma + \gamma \alpha = (1)(2) + (2) (3) + (3) (1)$

$= 2 + 6 + 3$

$= 11$

Also,

$\alpha \beta+\beta \gamma+\alpha \gamma= \dfrac{Coefficient of x}{Coefficient of x^3}\\ = \dfrac{11}{1} \\= 11$

And

$\alpha \beta \gamma = 1 \times 2 \times 3$

$= 6$

$\alpha \beta \gamma= - \dfrac{Constant term}{Coefficient of x^3} \\= - \dfrac{-6}{1} \\= 6$

Thus, the relationship between the zeroes and the coefficients is verified.

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as the numbers are: $\dfrac{2}{5}, \dfrac{1}{10}, \dfrac{1}{2}$

Let the zeroes be

$\alpha, \beta$ and

$\gamma$ and the polynomial be

$x^3+\dfrac{b}{a}x^2+\dfrac{c}{a}x+\dfrac{d}{a}$ .
Then, we have

$\alpha + \beta + \gamma = \dfrac{2}{5}=-\dfrac{b}{a}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{1}{10} = \dfrac{c}{a}$

$\alpha \beta \gamma = \dfrac{1}{2}=\dfrac{-d}{a}$

Now, required cubic polynomial is

$= x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha)x - \alpha \beta \gamma$

$= x^3 - \left ( \dfrac{2}{5} \right )x^2 + \left ( \dfrac{1}{10} \right )x - \left ( \dfrac{1}{2} \right )$

$= \dfrac{10x^3 - 4x^2 + x - 5}{10}$

So,

$10x^3 - 4x^2 + x - 5$ is the required cubic polynomial which satisfy the given conditions.

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial: $t - 1, t^3 + t^2 - 2t + 1$

Let us divide

$t^3 + t^2 - 2t + 1$ by

$t - 1$
The division process is
Here, the remainder is

$1$ ,
therefore,

$t - 1$ is not a factor of

$t^3 + t^2 - 2t + 1$
1788812_1815409_ans_2ec7eaa0760e43d894abc4407f54dc61.png

Verify that, $3, - 1, - \frac{1}{3}$ are the zeroes of the cubic polynomial $p(x) = 3x^2 - 5x^2 - 11x - 3$ and then verify the relationship between the zeroes and the coefficients.

Let $p(x) = 3x^3 - 5x^2 - 11x - 3$

(i) $p (-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3$

$= -3 - 5 + 11 - 3$

$= 0$

(ii) $p(-\dfrac{1}{3}) = 3(- \dfrac{1}{3})^3 - 5(- \dfrac{1}{3})^2 - 11(-\dfrac{1}{3}) - 3$

$= (- \dfrac{1}{9}) - (\dfrac{5}{9}) + (\dfrac{11}{3}) - 3$

$= (\dfrac{-1 - 5 + 33 - 27}{9})$

$= 0$

(iii) $p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3$

$= 81 - 45 - 33 - 3 = 0$

Hence, we verified that $3, -1$ and $- \dfrac{1}{3}$ are the zeroes of the given polynomial.

Let us verify the relationship between the zeroes of coefficients.

Let $\alpha = 3, \beta = -1$ and $\gamma = - \dfrac{1}{3}$

Verification:

(i) $\alpha + \beta + \gamma = 3 + (-1) + (-\dfrac{1}{3}) = (\dfrac{5}{3})$

$= - \dfrac{Coefficient of x^2}{Coefficient of x^3} = - \dfrac{5}{3}$

(ii) $\alpha \beta + \beta \gamma + \gamma \alpha$

$= (3)(-1) + (-1) (- \dfrac{1}{3}) + (- \dfrac{1}{3}) (3) = (- \dfrac{11}{3})$

$= \dfrac{Coefficient of x}{Coefficient of x^3} = \dfrac{-11}{3} = (- \dfrac{11}{3})$

And

(iii) $\alpha \beta \gamma = 3 \times -1 \times (- \dfrac{1}{3}) = 1$

$= - \dfrac{Constant term}{Coefficient of x^3} = - \dfrac{-3}{3} = 1$

Thus, the relationship between the zeroes and the coefficients is verified.

Use division algorithm to show that the square any positive integer is of the form or $3p$ or $3p+1$ .

Let

$'a'$ be any positive integer and

$b=3$ ,
Then, by division algorithm

$a=3q+r$ ,
for some integer

$q\geqslant 0$ and

$0\leqslant r\leqslant 3$ .
So,

$a=3q,3q+1,3q+2$ .
Then, the square of positive integer,

$a=3q$

$a^{2}=(3q)^{2}$

$=9q^{2}$

$=3(3q^{2})$

$=3p$
(Where

$p=3q^{2}$ )

$a=3q+1$

$a^2=(3q+1)^{2}$

$=9q^{2}+6q+1$

$=3(3q^{2}+2q)+1$

$=3p+1$

(Where

$p=3q^{2}+2q$ )

$a=3q+2$

$a^2=(3q+2)^{2}$

$=9q^{2}+12q+4$

$=3(3q^{2}+4q+1)+1$

$=3p+1$

(Where

$p=3q^{2}+4q+1$ )

Since

$p$ is some positive integer.

$\therefore$ The square of any positive integer is of the form

$3p$ or

$3p+1$ .

Verify that 1,-1 and + 3 are the zeroes of the cubic polynomial $x^{3} - 3x^{2}- x + 3$ and also verify the relationship between zeroes and the coefficient.

$P (x)= x^3- 3x^{2} - x + 3$

$P (1) = 1^{3} - 3 (1)^{2} - 1 + 3$

$= 1-3 -1 + 3 = 4-4 =0$

$P(-1) = (-1)^{3}-3 (-1)^{2}- (-1) +3 =-1-3+1+3=0$

$P(3) =3^{3} - 3(3)^{2} - (3) + 3$

$=27 -27 - 3 +3 =0$

$\therefore \alpha=1,\beta=-1$ and

$\gamma=3$ are the zeroes of the polynomial

$ax^{3} + bx^{2} + cx + d=0$ i.e,

$x^3- 3x^{2} - x + 3=0$

for cubic polynomial having

$\alpha,\beta,\gamma$ as zeroes

$\alpha+\beta+\gamma=\dfrac{-b}{a}$

$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac ca$

$\alpha\beta\gamma=\dfrac{-d}{a}$

$\alpha+\beta+\gamma=1-1+3=3$ and

$\dfrac{-b}{a}=\dfrac{-(-3)}{1}=3$ [verified]

$\alpha\beta+\beta\gamma+\gamma\alpha=1(-1)+(-1)3+3(1)=-1$ and

$\dfrac ca=\dfrac{-1}{1}=1$ [verified]

$\alpha\beta\gamma=1(-1)(3)=-3$ and

$\dfrac{-d}{a}=\dfrac{-3}{1}=-3$ [verified]

State which of the following statement are true and which are false ? Given reasons for your choice
$\dfrac{1}{x^{2}-5x + 6}$ is a quadratic polynomial

False ,
because it is not a polynomial at all.

Show that one and only one out of $n,n+2$ or $n+4$ is divided by $3$ , where $n$ is any positive integer.

Let

$'n'$ be any positive integer and

$b=3$ .
Then, by division algorithm

$n=3q+r$
for some integer

$q\geqslant 0$ and

$0\leqslant r\leqslant 3$
If

$r=0$ , then

$n=3q$ (divisible by

$3$ )

$n+2=3q+2$ (not divisible by

$3$ )

$n+4=3q+4$ (not divisible by

$3$ )
If

$r=1$ ,

$n=3q+1$ (not divisible by

$3$ )

$n+2=3q+1+2=3q+3=3(q+1)$ (divisible by

$3$ )

$n+4=3q+1+4=3q+5$ (not divisible by

$3$ )
If

$r=2$

$n=3q+2$ (not divisible by

$3$ )

$n+2=3q+2+2=3q+4$ (not divisible by

$3$ )

$n+4=3q+2+4=3q+6=3(q+2)$ (divisible by

$3$ )

$\therefore$ Since, one and only one out of

$n,n+2$ or

$n+4$ is divisible by

$3$ .

Use division algorithm to show that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$ .

Let

$'a'$ be any positive integer and

$b=3$ ,
Then by division algorithm

$a=3q+r$ ,
For some integer

$q\geqslant 0$ and

$0\leqslant r\leqslant 3$ .
So,

$a=3q,3q+1$ or

$3q+2$
The cube of the positive integer,

$a=3q\Rightarrow a^{3}=(3q)^{3}=27q^{3}=9(3q^{3})$

$a^{3}=9m$ , where

$m=3q^{3}$ such that

$m$ is an integer.

$a=3q+1\Rightarrow a^{3}=(3q+1)^{3}$

$a^{3}=27q^{3}+27q^{2}+9q+1$

$=9(3q^{3}+3q^{2}+q)+1$

$=9m+1$
Where,

$m=3q^{3}+3q^{2}+q$ such that

$m$ is an integer.

$a=3q+2\Rightarrow a^{3}=(3q+2)^{3}$

$a^{3}=27q^{3}+54q^{2}+36q+8$

$=9(3q^{3}+6q^{2}+4q)+8$

$=9m+8$
Where,

$m=3q^{3}+6q^{2}+4q$ such that

$'m'$ is an integer.

$\therefore$ The cube of any positive integer is in the form

$9m,9m+1$ or

$9m+2$ .

Draw the graphs of
$6 - x - x^{2}$
and find the zeroes in each case.

$6 - x - x^{2}$
When

$x=-3$

$y=6-(-3)-(-3)^2=6+3-9=0$
When

$x=-2$

$y=6-(-2)-(-2)^2=6+2-4=4$
When

$x=-1$

$y=6-(-1)-(-1)^2=6+1-1=6$
When

$x=0$

$y=6-0-0^2=6$
When

$x=1$

$y=6-1-1^2=4$
When

$x=2$

$y=6-2-2^2=4-4=0$

List of

$x$ and

$y$

$x \,\,\,\,\ -3 \,\,\,\,\,\ -2\,\,\,\,\,\ -1 \,\,\,\,\,\ 0 \,\,\,\,\,\ 1 \,\,\,\,\,\,\,\ 2$

$y \,\,\,\,\,\,\,\,\,\, 0\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\ 6 \,\,\,\,\, 4 \,\,\,\,\,\,\,\,\ 0$

$(x, y)= (-3, 0), (-2, 4), ((-1, 6), (0,6), (1, 4), (2, 0).$

$\therefore$ graph

$6 - x - x^{2}$ cuts

$X -$ axis at the points

$- 3$ and

$-2$

$\therefore$ The zeroes of quadratic polynomial are

$- 3$ and

$2.$

1760730_1835280_ans_7cd5d040d7604b1eb99aafe764b9e93c.PNG

Draw the graph of
$2x$ and find the point of intersection on X - axis. Is the zeroes of the polynomial?

Let

$y = 2x$
When

$x=0$

$y=2\times 0=0$
When

$x=-1$

$y=2\times-1=-2$
When

$x=1$

$y=2\times1=2$

List of

$x$ and

$y$

$x \,\,\,\,\,\,\,\ 0 \,\,\,\,\,\,\ -1 \,\,\,\,\,\,\ 1$

$y \,\,\,\,\,\,\,\, 0 \,\,\,\,\,\,\ -2 \,\,\,\,\,\,\ 2$

$(x, y)= (0,0), (-1, -2), ((1, 2)$

$\therefore$ The graph

$y = 2x$ passes through the origin

$(0, 0)$ .

$\therefore$ The zeroes of the polynomial is zero.
1760631_1835274_ans_dbae57872b5849b79e98489ceb3ccf0b.PNG

Draw the graph of $2x + 5$ and find the point of intersection on X - axis. Is the zeroes of the polynomial?

Let

$y = 2x + 5$
when

$x=0$

$y=2(0)+5=5$
when

$x=-1$

$y=2(-1)+5=3$
when

$x=1$

$y=2(1)+5=7$
when

$x=2.5$

$y=2(2.5)+5=10$
when

$x=-2.5$

$y=2(-2.5)+5=0$

List of

$x$ and

$y$

$x \,\,\,\,\,\,\,0\,\,\,\,\,\,\,-1\,\,\,\,\,\,\,\ 1\,\,\,\,\,\,\, 2.5 \,\,\,\,\,\,\,\,- 2.5$

$y \,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\ 3\,\,\,\,\,\,\,\,\,7\,\,\,\,\,\,\,10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,0$

$(x,y)= (0, 0), (-1, 3), ((1, 7), (2,5, 10), (-2.5, 0)$

$\therefore$ The graph

$y = 2 x + 5$ cuts

$X -$ axis at the point

$(-2.5, 0)$

$\therefore$ the zeroes of the polynomial is

$- 2.5$

1761594_1835269_ans_961131de749d4f61b99eaeb7342d0027.PNG

General form of a quadratic polynomial and cubic polynomial in variable x.

The general form of a quadratic polynomial in variable

$x$ is

$f(x) = ax^{2} + bx + c (a \neq 0)$
The general form of a cubic polynomial in variable

$x$ is

$f(x) = ax^{3} + bx^{2} +cx + d (a \neq 0)$

Draw the graph of $2x - 5$ and find the point of intersection on X - axis. Is the zeroes of the polynomial?

Let

$y = 2x - 5$
When

$x=0$

$y=2\times 0-5=-5$
When

$x=-1$

$y=2\times-1-5=-7$
When

$x=1$

$y=2\times1-5=-3$
When

$x=2.5$

$y=2\times2.5-5=0$
When

$x=-2.5$

$y=2\times(-2.5)-5=-10$

List of

$x$ and

$y$

$x \,\,\,\,\,\,0\,\,\,\,\,\,\ -1\,\,\,\,\,\ 1,\,\,\,\,\ 2.5\,\,\,\,\,\,\,-2.5$

$y\,\,\,\,\,\,5 \,\,\,\,\,\,\,\,\ -7\,\,\,\ -3\,\,\,\,\,\,\ 0 \,\,\,\,\,\,\,\,-10$

$(x, y )= (0, -5), (-1, -7), ((1, -3), (2.5, 0), (-2.5, 0)$

$\therefore$ The graph

$y = 2x - 5$ cuts

$X -$ axis at he point

$(2.5, 0)$

$\therefore$ The zeroes of the polynomial is

$2.5$

1760675_1835271_ans_09502304789d4a39b690956165af707d.PNG

Write three quadratic polynomial that gave 2 zeroes each.

The quadratic polynomial that have two zeries

$x^{2} - 3x - 4=(x+1)(x-4),$ it has 2 zeroes at - 1 and 4

ii.

$-x^{2} - 2x + 3=(1-x)(x+3),$ it has 2 zeroes at 1 and -3

iii.

$x^{2} - x - 2=(x-2)(x+3),$ it has 2 zeroes at 2 and -3

Verify that the numbers given alongside that cubic polynomials below are its zeroes . Also verify the relationship between the zeroes and coefficient in each case.
$2x^{3} + x^{2} - 5x + 2 ,\left ( \dfrac{1}{2},1 .-2 \right )$

$2x^{3} + x^{2} - 5x + 2 ,\left ( \dfrac{1}{2},1 .-2 \right )$

$2x^{3} + x^{2} - 5x + 2$

Compare with this

$ax^{3} + bx^{2} + cx + d$

$x = \dfrac{1}{2}$

$p \left ( \dfrac{1}{2} \right ) = 2 \left ( \dfrac{1}{2} \right )^{3} +\left ( \dfrac{1}{2} \right )^{2} - 5 \left ( \dfrac{1}{2} \right ) + 2$

$= \dfrac{1}{4} +\dfrac{1}{4}- \dfrac{5}{2} +2$

$= \dfrac{1 + 1 - 10 + 8}{4} = \dfrac{0}{4} = 0$

$x = 1$

$p (1) = 2(1)^{3} + (1)^{2} -5 (1) + 2$

$= 2+1-5+2=0$

$x=2$

$P(-2)=2(-2)^3+(-2)^2-5(-2)+2=- 16 + 4 + 10 +2= 0$

$\therefore \alpha=\dfrac{1}{2},\beta=1$ and

$\gamma=-2$ are zeroes of

$2x^{3} + x^{2} - 5x + 2$

$\therefore$ a = 2,b = 1 , c = -5 , d= 2

$\therefore \alpha = \dfrac{1}{2} , \beta = 1 , and \gamma = -2$

$\alpha +\beta +\gamma = \dfrac{1}{2} + 1 +(-2)$

$=\dfrac{1 + 2 -4}{2} = -\dfrac{1}{2} = -\dfrac{b}{a}$

$\alpha\beta+\beta\gamma+\gamma\alpha=(1/2)(1)+(1)(-2)+(-2)(1/2)=\dfrac{-5}{2}=\dfrac{c}{a}$

$\alpha\beta\gamma=(1/2)(1)(-2)=-1=\dfrac{-2}{2}=\dfrac{-d}{a}$

Write one quadratic polynomial that has one zero.

The quadratic polynomial

$y = x^{2} - 6x + 9=(x-3)^2$ has only one zero i.e, 3

Draw the graphs of
$y = x^{2} - x - 6$
and find the zeroes in each case.

$y = x^{2} - x - 6$
When

$x=-3$

$y=-3^2-(-3)-6=9+3-6=6$
When

$x=-2$

$y=-2^2-(-2)-6=4-2-6=0$
When

$x=-1$

$y=-1^2-(-1)-6=1+1-6=-4$
When

$x=0$

$y=0^2-0-6=6$
When

$x=1$

$y=1^2-1-6=-6$
When

$x=2$

$y=2^2-2-6=4-8=-4$
When

$x=3$

$y=3^2-3-6=9-9=0$

List of

$x$ and

$y$

$x \,\,\,\ -3\,\,\,\,\ -2 \,\,\,-1 \,\,\,\,\,\,\, 0 \,\,\,\,\,\,\,\,\,\,\,1 \,\,\,\,\,\ 2\,\,\,\,\, 3$

$y \,\,\,\,\,\,\,\,\,\,6 \,\,\,\,\,\,\,\,\,\,\,0\,\,\ -4 \,\,\ -6 \,\,\,\, -6\,\, -4 \,\,\, 0$

$(x, y)= (-3, 6), (-2, 0), ((-1, -4), (0, -6), (1, -6), (2, -4).(3,0)$

$\therefore$ The graph

$y = x^{2} - x - 6$ cuts

$X-$ axis at the points

$3$ and

$- 2$

$\therefore$ The zeroes of quadratic polynomial are

$-2$ and

$3.$

1760793_1835279_ans_147461bd3ddb4de893b33624589e5a1a.PNG

Use division algorithm to show that any odd positive integer is of the form $6q+1$ or $6q+3$ or $6q+5$ , where $q$ is some integer.

Let

$'a'$ be any positive integer and

$b=6$ ,
Then, by division algorithm

$a=6q+r$ ,
for some integer

$q\geqslant 0$ and

$0\leqslant r\leqslant 6$ .
So,

$a=6q,6q+2,6q+3,6q+4$ or

$6q+5$ .
Since

$a$ is odd,

$'a'$ can not be

$6q=2(3q)$
or

$6q+2=2(3q+1)$ or

$6q+4=2(3q+2)$ .
(Since they are divisible by

$2$ ).

$\therefore$ Any odd integer is of the form

$6q+1$ or

$6q+3$ or

$6q+5$ .

If the zeroes of the polynomial $x^{3} - 3x^{2}+ x+1$ are $a - b , a, a + b$ , find $a$ and $b.$

Polynomial

$p (x) = x^{3} - 3x^{2} + x + 1$

$a -b ,a,a +b$ are zeroes of the polynomial

$p (x) = x^{3} - 3x^{2} + x + 1$

$\therefore$ Sum of its zeroes

$= \left ( a - b \right ) + a + \left ( a - b \right )$

$= -\dfrac{-3}{1}$

$\Rightarrow 3a = 3\Rightarrow a = 1$

Sum of the product of its zeroes

$= a \left ( a - b \right ) + a \left ( a + b \right ) + \left ( a - b \right )\left ( a + b \right ) = \dfrac{1}{1}$

$\Rightarrow a^{2} - ab + a^{2} + ab + a^{2} - b^{2} = 1$

$\Rightarrow 3 a^{2} - b^{2} = 1 \Rightarrow 3 \left ( 1 \right )^{2} - b^{2} = 1$

$\Rightarrow b^{2} = 3-1 = 2\Rightarrow b = \pm \sqrt{2}$

Since

$a = 1$ and

$b = \pm \sqrt{2}$

How will you verify if a quadratic polynomial has only one zero?

If the graph of a given polynomial touches X - axis at exactly one point, then it has only one zero.

Also we know that if a given polynomial

$f(x)=ax^2+bx+c$ has only one zero when

$D=b^2-4ac=0$

If $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial find the values of the exoressions given below:
$\alpha +\beta +\gamma , \alpha \beta + \beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$ .

$x^{3} + + 4x^{2} - 5x - 2$
The expression is in the form

$ax^{3} + bx^{2} + cx + d,$ then

$a = 1, b = 4, c = -5, d = -2$

$\therefore \alpha + \beta + \gamma = \dfrac{-b}{a} = \dfrac{-4}{1} = -4$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} = \dfrac{-5}{1} = -5$

$\alpha \beta \gamma = \dfrac{-d}{a} = \dfrac{-(-2)}{1} = 2$

Given examples of polynomials p (x) , g (x) , q (x) , q (x) and r (x) , Which satisfy the division algirithm and deg p ( x ) = deg (x)

deg p (x) = deg q (x)
Let

$p (x) = 12 x^{2}+4x+12$
deg of p ( x) = 2

$q (x) = 3x ^{2}+x +3$ deg of q (x) = 2Using division algorithm

$12x^{2} +4x+12$ - - - +
...............
0Since , remainder is zero

$3x^{2} +x +$ is a factor of $$12x^{2+4x12$$

If $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial $4x^{3} + 8x^{2} - 6x - 2$ , find the values of the expressions given below:
$$\alpha +\beta + \gamma , \alpha \beta + \beta \gamma + \gamma \alpha $and$ \alpha \beta \gamma $$.

$4x^{3} ++ 8x^{2} - 6x - 2$

The expression in in the form

$ax^{3} + bx^{2} + cx + d$ ,
then

$a = 4, b = 8, c = -6, d = -2$

$\therefore \alpha + \beta + \gamma = \dfrac{-b}{a} = \dfrac{-8}{4} = -2$

$\therefore \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} = \dfrac{-6}{4} = \dfrac{-3}{2}$

$\alpha \beta \gamma = \dfrac{-d}{a} = \dfrac{-(-2)}4{} = \dfrac{1}{2}$

If $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomials find the values of the expressions given below: $\alpha +\beta +\gamma , \alpha \beta + \beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$ .

$x^{3} + 5x^{2} + 4$
The expression is in the form

$ax^{3} + bx^{2} + cx + d$
then

$a = 1, b = 5, c = 0, d = 4$

$\therefore \alpha + \beta +\gamma = \dfrac{-b}{a} = \dfrac{-5}1{} = -5$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} = \dfrac{0}{1} = 0$

$\alpha \beta \gamma = \dfrac{-d}{a} = \dfrac{-4}{1} = -4$

If $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial find the values of the expressions given below:
$\alpha + \beta + \gamma , \alpha \beta + \beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$ .

$x^{3} + 3x^{2} - x - 2$

The expression is in the from

$ax^{3} + bx^{2} + cx + d$ ,
then

$a = 1, b = 3, c = -1, d = -2$

$\therefore \alpha + \beta + \gamma =\dfrac{-b}{a} = \dfrac{-3}{1} = -3$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} = \dfrac{-1}{1} = -1$

$\alpha \beta \gamma = \dfrac{-d}{a} = \dfrac{-(-2)}1{} = 2$

Given examples of polynomials p (x) , g (x) , q (x) , q (x) and r (x) , Which satisfy the division algirithm and deg r (x ) = 0

Let

$P (x)= x^{4}+4x^{3}+3x^{2}+2x+2if r (x )=1, degree of r (x)= 0Dividide 3x^{4}+9x^{3}+3x^{2}+6x+2by 3 3) 3x^{4}+9x^{3}+3x^{2}+6x +2 (x^{3}4+3x^{4}$ -...............

6x +2

-..........

$\therefore g(x )=x^{4}+3x^{3}x^{2}+3x$

$\partial degree of r = 0$

Given examples of polynomials p (x) , g (x) , q (x) , q (x) and r (x) , Which satisfy the division algirithm and
deg q (x) = deg r (x)

deg q (x) = deg r (x)Let p(x) =

$2x^{5}+4x^{4}+6x^{3}+10x^{2}+4q(x)=2x^{2}+2x +2$
Using division algorithm

$2x^{3}+2x^{2}+2x$ - - -

.................

$\therefore deg of q (x)= deg of r (x)$

Classify the following polynomial as linear, quadratic or cubic polynomial:
$2x^{2} + 3x + 1$

The given polynomial is

$2x^{2} + 3x+ 1$

The degree of the polynomial

$2x^{2} + 3x+ 1$ is

$2$

So, the polynomial

$2x^{2} + 3x + 1$ is a quadratic polynomial .

Classify the following polynomial as linear, quadratic and cubic polynomial .
$3r^{3}$

The degree of the polynomial

$3r^{3}$ is

$3$
So, the polynomial

$3r^{3}$ is a cubic polynomial

Classify the following polynomial as linear, quadratic and cubic polynomial .
$m^{3} + 7m^{2} + \dfrac{5}{2}m - \sqrt{7}$

The degree of the polynomial

$m^{3} + 7m^{2} + \dfrac{5}{2}m - \sqrt{7}$ is 3
So, the polynomial

$m^{3} + 7m^{2} + \dfrac{5}{2}m - \sqrt{7}$ is a cubic polynomial

Classify the following polynomial as linear, quadratic and cubic polynomial .
$\sqrt{2}y - \dfrac{1}{2}$

The degree of the polynomial

$\sqrt{2}y - \dfrac{1}{2}$ is 1
So, the polynomial

$\sqrt{2}y - \dfrac{1}{2}$ is a liner polynomial

Classify the following polynomial as linear, quadratic and cubic polynomial .
$5p$

The degree of the polynomial

$5p$ is 1
So, the polynomial

$5p$ is a liner polynomial

Classify the following as linear , quadratic and cubic polynomial :
$x^{2} + x$

$x^{2} + x$ is a quadratic polynomial as its degree is

$2$

Using division algorithm, find quotient and remainder dividing $f(x)$ by $g(x)$ .
$f(x)=x^3-3x^2+5x+3,g(x)=x^2-2$

given

$f(x)=x^3-3x^2+5x-3$
and

$g(x)=x^2-2$
Dividing

$f(x)$ and

$g(x)$

$\quad \quad \quad \quad \quad \quad \quad x-3\\ \left. x^{ 2 }-2 \right) \overline { x^{ 3 }-3x^{ 2 }+5x-3 } \\ \quad \quad \quad \quad \quad x^{ 3 }\quad \quad \quad \quad -2x\\ \quad \quad \quad \quad \_ -\_ \_ \_ \_ \_ +\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad -3x^{ 2 }+7x-3\\ \quad \quad \quad \quad \quad \quad -3x^{ 2 }\quad \quad +6\\ \quad \quad \quad \quad \quad \_ +\_ \_ \_ \_ \_ -\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 7x-9$
Quotient

$q(x)=x-3$
Remainder

$r(x)=7x-9$
By Euclid division algorithm

$f(x)=g(x).q(x)+r(x)$

$=(x^2-2)(x-3)+7x-9$

$=x^3-x^2-2x+6+7x-9$

$=x^3-3x^2+5x-3$

$=f(x)$
Thus division algorithm is verified.

If $a-b,a, a+b$ are zero of polynomial $x^3 -3x^2 +x+1$ , then find $a$ and $b$ .

Given polynomial

$=x^3-3x^2 +x+1$
Comparing it with

$Ax^3+Bx^2+Cx+D$

$A=1, B=-3, C=1$ and

$D=1$
Sum of zeroes

$=-\dfrac {B}{A}=-\left(\dfrac {-3}{1}\right)=3$
But zeros are

$a-b, a$ and

$a+b$
Thus

$a-b+a+a+b=3$

$\Rightarrow \ 3a=3\Rightarrow a=\dfrac {3}{3} =1$
Product of zeroes

$=-\dfrac {D}{A}=-\left(\dfrac {1}{1}\right)=-1$
But product of zeros

$=(a-b)(a) (a+b)=a(a^2 -b^2)$
Then

$a(a^2 -b^2)=-1$

$\therefore 1(1^2 -b^2)=-1$

$\Rightarrow 1-b^2=-1$

$\Rightarrow b^2=2\Rightarrow b=\pm \sqrt 2$
Hence

$a=1$ and

$b=\pm \sqrt 2$

Classify the following as linear , quadratic and cubic polynomial :
$x - x^{3}$

$x - x^{3}$ is a cubic polynomial as its degree is

$3$

State whether the following expression is a polynomials in one variable or not ? State reasons for your answers.
$y^{2} + 2\sqrt{3}$

$y^{2} + 2\sqrt{3}$ is a polynomial in variable because degree of the variable y is a whole number i.e.

$2$ .

Classify the following as linear, quadratic or cubic polynomial:
$7x^{3}$

Given expression is

$7x^{3}$ .

We can see that its degree is

$3$ .

And we know that a polynomial of degree

$3$ is called a cubic polynomial.

Hence,

$7x^3$ is a cubic polynomial.

Using division algorithm, find quotient and remainder dividing $f(x)$ by $g(x)$ .
$f(x)=3x^3+x^2+2x+5,g(x)=1+2x+x^2$

Given:

$f(x)=3x^3+x^2+2x+5$
and

$g(x)=1+2x+x^2$
or

$g(x)=x^2+2x+1$
Dividing

$f(x)$ by

$g(x)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad 3x-5\\ \left. x^{ 2 }+2x+1 \right) \overline { 3x^{ 2 }+x^{ 2 }+2x+5 } \\ \quad \quad \quad \quad \quad \quad \quad \quad 3x^{ 3 }+6x^{ 2 }+3x\\ \quad \quad \quad \quad \quad \quad \quad \quad -\_ \_ \_ -\_ \_ \_ -\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -5x^{ 2 }-x+5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -5x^{ 2 }-10x-5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \_ +\_ \_ +\_ \_ \_ +\_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 9x+10$
Quotient

$q(x)=3x-5$
Remainder

$r(x)=9x+10$
Here, Quotient

$\times$ Divisor

$+$ Remainder

$(3x-5)(1+2x+x^2)+9x+10$

$=3x+6x^2+3x^3-5-10x-5x^2+9x+10$

$=3x^3+x^2-7x-5+9x+10$

$=3x^3+x^2+2x+5$

$=$ devidend
Thus division algorithm is verified.

Using division algorithm, find quotient and remainder dividing $f(x)$ by $g(x)$ .
$f(x)=x^3-6x^2+11x-6,g(x)=x+2$

Given:

$f(x)=x^3-6x+11x-6$
and

$g(x)=x+2$
Dividing

$f(x)$ and

$g(x)$

$\quad \quad \quad \quad \quad x^{ 2 }-8x+27\\ \left. x+2 \right) \overline { x^{ 3 }-6x^{ 2 }+11x-6 } \\ \quad \quad \quad \quad \quad x^{ 3 }+2x^2\\ \quad \quad \quad \quad \_ -\_ \_ -\_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad -8x^{ 2 }+11x-6\\ \quad \quad \quad \quad \quad \quad -8x^{ 2 }-16x\\ \quad \quad \quad \quad \quad \_ +\_ \_ \_ +\_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 27x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 27x+54\\ \quad \quad \quad \quad \quad \_ \_ \_ \_ -\_ \_ \_ \_ -\_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -60$
Quotient

$q(x)=x^2-8x+27$
Remainder

$r(x)=-60$
Here, Divisor

$\times$ Quotient

$+$ Remainder

$(x+2)(x^2-8x+27)-60$

$=x^3-6x^2+11-6$

$=devidend$
Here, division algorithm is verified.

Using division algorithm, find quotient and remainder dividing $f(x)$ by $g(x)$ .
$f(x)=9x^4-4x^2+4,g(x)=3x^2+x-1$

Given

$f(x)=9x^4-4x^2+4$
and

$g(x)=3x^2+x-1$
Dividing

$f(x)$ by

$g(x)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad 3x^{ 2 }-x\\ \left. 3x^{ 2 }+x-1 \right) \overline { 9x^{ 4 }-4x^{ 2 }+4 } \\ \quad \quad \quad \quad \quad \,\, 9x^{ 4 }-3x^{ 3 } -3x^{ 2 }\\ \quad \quad \quad \quad \_ -\_ \_ +\_ \_ \_ \_ \_ +\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad -3x^{ 3 }-x^{ 2 }+4\\ \quad \quad \quad \quad \quad \quad \quad \quad -3x^{ 3 }-x^{ 2 }\quad +x\\ \quad \quad \quad \quad \quad \quad \quad \quad \_ +\_ \_ \_ +\_ \_ \_ \_ -\_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -x+4$
Thus quotient

$q(x)=3x^2-x$
Remainder

$f(x)=-x+4$
Here:

$f(x).q(x)+f(x)$

$=(3x^2+x-1)(3x^2-x)+(-x)+4$

$=9x^4+3x^3-3x^2-3x^3-x^2+x-x+4$

$=9x^4-4x^2+4$

$=f(x)$
Thus, division algorithm is verified.

Divide $x^3 -3x^2 +3x-5$ by $x-1-x^2$ and test division algorithm.

Let

$f(x)=x^3-3x^2+3x-5$
and

$g(x)=x-1-x^2=-x^2+x-1$
Now divide

$f(x)$ by

$g(x)$

$\left. -x^{ 2 }+x-1 \right) x^{ 3 }-3x^{ 2 }+3x-5\left( -x+2 \right. \\ \quad \quad \quad \quad \quad x^{ 3 }-x^{ 2 }+x\\ \quad \quad \quad \quad \quad -\quad +\quad -\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad -2x^{ 2 }+2x-5\\ \quad \quad \quad \quad \quad \quad -2x^{ 2 }+2x-5\\ \quad \quad \quad \quad \quad \quad +\quad -\quad +\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -3$

Thus quotient

$q(x)=-x+2$
and remainder

$r(x) =-3$
Using divisional algorithm
Dividend = divisor

$\times$ Quotient + remainder

$\Rightarrow x^3 -3x^2+3x-5$

$=(-x^2+x-1)(-x+2)-3$

$\Rightarrow x^3-3x^2+3x-5$

$=(x^3-2x^2-x^2 +2x+x-2)-3$

$\Rightarrow x^3-3x^2+3x-5=x^3-3x^2+3x-2-3$

$\Rightarrow x^3 -3x^2+3x-5=x^3 -3x^2+3x-5$

$L.H.S =R.H.S$
Hence, division algorithm is verified.

State whether the following expression is a polynomials in one variable or not ? State reasons for your answers.
$3x^{2} - 5x + 13$

$3x^{2} - 5x + 13$ is a polynomial in one variable, because degree of the variable

$x$ is a whole number.

State whether the following expression is a polynomials in one variable or not ? State reasons for your answers.
$2\sqrt{x} + \sqrt{3}x$

$2\sqrt{x} + \sqrt{3}x$ is not a polynomial because degree of

$x$ in first term is

$\dfrac{1}{2}.$

State whether the following expression is a polynomials in one variable or not ? State reasons for your answers.
$y + \dfrac{3}{y}$

$y + \dfrac{3}{y}$ is not a polynomial because degree of the variable y in second term is

$-1.$

If $1$ and $2$ are two zeros of the polynomial $x^4-3x^3-2x^2+12x-8$ , then the remaining of two zeros are _____.

1985779_1946154_ans_61aae9419cc548068391dedbb4d996d1.jpeg

If the polynomial $x^2-2x+k$ is a factor of $x^4- 6x^3 + 16x^2 -26x + 10-a$ , then find the value of $k$ and $a.$

After dividing polynomial

$x^4- 6x^3 + 16x^2 -26x + 10 - x - a$ by

$x^2-2x+k$ we got remainder,

$(-10+2k)x+(10-a-8k+k^2).$

Which will be equal to zero if,

$-10 + 2k = 0\text{ and } 10-a-8k+k^2 = 0$

For

$-10 + 2k = 0$

$\Rightarrow 2k = 10$

$\Rightarrow k = 5$

For

$10-a-8k+k^2 = 0$

$\Rightarrow 10 - a - 8 \times 5+25 = 0$

$\Rightarrow -5 - a = 0$

$\Rightarrow a = -5$

So,

$k = 5, a = -5$

Divide $p(x)$ by $g(x)$ in the following case and verify division algorithm
$p(x) = x^{2} + 4x + 4; g(x) = x + 2$

The division of

$p(x)=x^2+4x+4$ by

$g(x)=x+2$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$x^2+4x+4$ , the divisor is

$x+2$ , the quotient is

$x+2$ and the remainder is

$0$ , therefore,

$x^{ 2 }+4x+4=[(x+2)(x+2)]+0\\ \Rightarrow x^{ 2 }+4x+4=[x(x+2)+2(x+2)]\\ \Rightarrow x^{ 2 }+4x+4=x^{ 2 }+2x+2x+4\\ \Rightarrow x^{ 2 }+4x+4=x^{ 2 }+4x+4$

Hence, the division algorithm is verified.

Give 4 different reasons why the graph cannot be the graph of the polynomial $p$ given by
$p(x)={x}^{4}-{x}^{2}+1$

1 - The given polynomial has degree

$4$ and positive leading coefficient and the graph should rise on the left and right sides.

2 -

$p(0) = 1$ , graph shows negative value.

3 - The equation

$x^ 4 - x ^2 + 1 = 0$ has no solution which suggests that the polynomial

$p(x) = x^ 4 - x ^2 + 1$ has no zeros. The graph shows x intercepts.

4 - The graph has a zero of multiplicity

$2$ , a zero of multiplicity

$3$ and and a zero of multiplicity

$1$ . So the degree of the graphed polynomial should be at least

$6$ . The given polynomial has degree

$4$ .

Divide $p(x)$ by $g(x)$ in the following cases and verify division algorithm.
$p(x) = x^{4} - 4x^{2} + 12x + 9; g(x) = x^{2} + 2x - 3$

The division of

$p(x)=x^4-4x^2+12x+9$ by

$g(x)=x^2+2x-3$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$x^4-4x^2+12x+9$ , the divisor is

$x^2+2x-3$ , the quotient is

$x^2-2x+3$ and the remainder is

$18$ , therefore,

$x^{ 4 }-4x^{ 2 }+12x+9=[(x^{ 2 }+2x-3)(x^{ 2 }-2x+3)]+18\\ \Rightarrow x^{ 4 }-4x^{ 2 }+12x+9=[x^{ 2 }(x^{ 2 }-2x+3)+2x(x^{ 2 }-2x+3)-3(x^{ 2 }-2x+3)]+18\\ \Rightarrow x^{ 4 }-4x^{ 2 }+12x+9=(x^{ 4 }-2x^{ 3 }+3x^{ 2 }+2x^{ 3 }-4x^{ 2 }+6x-3x^{ 2 }+6x-9)+18\\ \Rightarrow x^{ 4 }-4x^{ 2 }+12x+9=x^{ 4 }-2x^{ 3 }+2x^{ 3 }+3x^{ 2 }-4x^{ 2 }-3x^{ 2 }+6x+6x-9+18\\ \Rightarrow x^{ 4 }-4x^{ 2 }+12x+9=x^{ 4 }-4x^{ 2 }+12x+9$

Hence, the division algorithm is verified.

Divide $p(x)$ by $g(x)$ in the following case and verify division algorithm.
$p(x) = 2x^{2} - 9x + 9; g(x) = x - 3$

The division of

$p(x)=2x^2-9x+9$ by

$g(x)=x-3$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$2x^2-9x+9$ , the divisor is

$x-3$ , the quotient is

$2x-3$ and the remainder is

$0$ , therefore,

$2x^{ 2 }-9x+9=[(x-3)(2x-3)]+0\\ \Rightarrow 2x^{ 2 }-9x+9=[x(2x-3)-3(2x-3)]\\ \Rightarrow 2x^{ 2 }-9x+9=x^{ 2 }-3x-6x+9\\ \Rightarrow 2x^{ 2 }-9x+9=2x^{ 2 }-9x+9$

Hence, the division algorithm is verified.

Let $P(x)$ be a cubic polynomial with zeroes $\alpha,\beta, \gamma$ if $\displaystyle \frac{P\left(\dfrac{1}{2}\right)+P\left(-\dfrac{1}{2}\right)}{P(0)}=100$ find $\sqrt{\displaystyle \frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma a}}$ .

Let

$P(x) = ax^3 + bx^2 + cx + d$
So

$P\left(\dfrac{1}{2}\right) + P\left(-\dfrac{1}{2}\right) = 2b \left(\dfrac{1}{4}\right) + 2d$
and

$P(0) = d$

$\Rightarrow \dfrac{2\dfrac{b}{4}+2d}{d}=100$

$\Rightarrow \dfrac{2b}{4d}+2=100 \Rightarrow \dfrac{b}{d}=196$
also

$\alpha+\beta+\gamma=-\dfrac{b}{a},\alpha\beta\gamma=-\dfrac{d}{a}$
Hence

$\sqrt{\displaystyle \dfrac{1}{\alpha\beta}+\dfrac{1}{\beta\gamma}+\dfrac{1}{\gamma a}} = \sqrt{\dfrac{\alpha+\beta+\gamma}{\alpha\beta\gamma }}=\sqrt{196}=14$

Divide $p(x)$ by $g(x)$ in the following case and verify division algorithm.
$p(x) = x^{3} + 4x^{2} - 5x + 6; g(x) = x + 1$

The division of

$p(x)=x^3+4x^2-5x+6$ by

$g(x)=x+1$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$x^3+4x^2-5x+6$ , the divisor is

$x+1$ , the quotient is

$x^2+3x-8$ and the remainder is

$14$ , therefore,

To check:

$x^{ 3 }+4x^{ 2 }-5x+6=[(x+1)(x^{ 2 }+3x-8)]+14\\ \text{R.H.S}\Rightarrow [x(x^{ 2 }+3x-8)+1(x^{ 2 }+3x-8)]+14\\ =x^{ 3 }+3x^{ 2 }-8x+x^{ 2 }+3x-8+14\\ =x^{ 3 }+3x^{ 2 }+x^{ 2 }-8x+3x-8+14\\ =x^{ 3 }+4x^{ 2 }-5x+6=\text{L.H.S}$

Hence, the division algorithm is verified.

Divide $p(x)$ by $g(x)$ in the following case and verify division algorithm.
$p(x) = x^{4} - 3x^{2} - 4; g(x) = x + 2$

The division of

$p(x)=x^4-3x^2-4$ by

$g(x)=x+2$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$x^4-3x^2-4$ , the divisor is

$x+2$ , the quotient is

$x^3-2x^2+x-2$ and the remainder is

$0$ , therefore,

$x^{ 4 }-3x^{ 2 }-4=[(x+2)(x^{ 3 }-2x^{ 2 }+x-2)]+0\\ \Rightarrow x^{ 4 }-3x^{ 2 }-4=[x(x^{ 3 }-2x^{ 2 }+x-2)+2(x^{ 3 }-2x^{ 2 }+x-2)]+0\\ \Rightarrow x^{ 4 }-3x^{ 2 }-4=x^{ 4 }-2x^{ 3 }+x^{ 2 }-2x+2x^{ 3 }-4x^{ 2 }+2x-4+0\\ \Rightarrow x^{ 4 }-3x^{ 2 }-4=x^{ 4 }-2x^{ 3 }+2x^{ 3 }+x^{ 2 }-4x^{ 2 }-2x+2x-4\\ \Rightarrow x^{ 4 }-3x^{ 2 }-4=x^{ 4 }-3x^{ 2 }-4$

Hence, the division algorithm is verified.

Find the divisor $g(x)$ , when the polynomial $p(x) = 4x^{3} + 2x^{2} - 10x + 2$ is divided by $g(x)$ and the quotient and remainder obtained are $(2x^{2} + 4x + 1)$ and $5$ respectively

We know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here it is given that the dividend is

$p(x)=4x^3+2x^2-10x+2$ , the divisor is

$g(x)$ , the quotient is

$2x^2+4x+1$ and the remainder is

$5$ , therefore,

$4x^{ 3 }+2x^{ 2 }-10x+2=[g(x)(2x^{ 2 }+4x+1)]+5\\ \Rightarrow g(x)=\cfrac { 4x^{ 3 }+2x^{ 2 }-10x+2 }{ 2x^{ 2 }+4x+1 }$

The division is shown above.

Hence, from the above division, we get that the divisor is

$g(x)=2x-3$

Divide $p(x)$ by $g(x)$ in the following case and verify division algorithm.
$p(x) = x^{3} - 1; g(x) = x - 1$

The division of

$p(x)=x^3-1$ by

$g(x)=x-1$ is as shown above:

Now we know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here, the dividend is

$x^3-1$ , the divisor is

$x-1$ , the quotient is

$x^2+x+1$ and the remainder is

$0$ , therefore,

$x^{ 3 }-1=[(x-1)(x^{ 2 }+x+1)]+0\\ \Rightarrow x^{ 3 }-1=[x(x^{ 2 }+x+1)-1(x^{ 2 }+x+1)]+0\\ \Rightarrow x^{ 3 }-1=x^{ 3 }+x^{ 2 }+x-x^{ 2 }-x-1+0\\ \Rightarrow x^{ 3 }-1=x^{ 3 }+x^{ 2 }-x^{ 2 }+x-x-1\\ \Rightarrow x^{ 3 }-1=x^{ 3 }-1$

Hence, the division algorithm is verified.

Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$ .

Let us first divide the given polynomial

$x^4+x^3+8x^2+ax+b$ by

$(x^2+1)$ as shown in the above image:

From the division, we observe that the quotient is

$x^2+x+7$ and the remainder is

$(a-1)x+(b-7)$ .

Since it is given that

$x^4+x^3+8x^2+ax+b$ is exactly divisible by

$x^2+1$ , therefore, the remainder must be equal to

$0$ that is:

$(a-1)x+(b-7)=0\\ \Rightarrow (a-1)x+(b-7)=0\cdot x+0\\ \Rightarrow (a-1)=0,\quad (b-7)=0\quad \quad \quad \quad (By\quad comparing\quad coefficients)\\ \Rightarrow a=1,\quad b=7$

Hence,

$a=1$ and

$b=7$ .

1237902_1086130_ans_56917c69fff5451faee27e18d3c239d9.jpg

What must be subtracted from $(x^{3} + 5x^{2} + 5x + 8)$ so that the resulting polynomial is exactly divisible by $(x^{2} + 3x - 2)$ ?

It is given that the dividend is

$p(x)=x^3+5x^2+5x+8$ , the divisor is

$g(x)=x^2+3x-2$ , therefore, the division is as shown above.

Therefore, from the division, we get that the remainder is

$x+12$ .

Hence,

$x+12$ must be subtracted from

$x^3+5x^2+5x+8$ so that the resulting polynomial is exactly divisible by

$x^2+3x-2$ .

What should be added to $(x^{4} - 1)$ so that it is exactly divisible by $(x^{2} + 2x + 1)$ ?

It is given that the dividend is

$p(x)=x^4-1$ , the divisor is

$g(x)=x^2+2x+1$ , therefore, the division is as shown above.

Therefore, from the division, we get that the remainder is

$r(x)=(-4x-4)$ that is

$-r(x)=4x+4$ .

Hence,

$4x+4$ should be added to

$x^4-1$ so that the resulting polynomial is exactly divisible by

$x^2+2x+1$ .

The product of two polynomials is $6x^{3} + 29x^{2} + 44x + 21$ and one of the polynomials is $3x + 7$ . Find the other polynomial.

Let

$A$ be the other polynomial. It is given that one of the polynomial is

$3x+7$ and the product of the polynomials is

$6x^3+29x^2+44x+21$ that is:

$(3x+7)\times A=6x^{ 3 }+29x^{ 2 }+44x+21\\ A=\dfrac { 6x^{ 3 }+29x^{ 2 }+44x+21 }{ 3x+7 }$

We now divide

$6x^3+29x^2+44x+21$ by

$(3x+7)$ as shown below:

From the division, we observe that the quotient is

$2x^2+5x+3$ and the remainder is

$0$ .

Hence, the other polynomial is

$2x^2+5x+3$ .

Rewrite the terms in proper order and then divide
$a+a^{2}+8$ by $a+4$

The given polynomial

$a+a^2+8$ can be rewritten as

$a^2+a+8$ .

We now divide

$a^2+a+8$ by

$(a+4)$ as shown in the above image:

From the division, we observe that the quotient is

$a-3$ and the remainder is

$20$ .

Hence, the quotient is

$a-3$ and the remainder is

$20$ .

1248964_1129550_ans_40cee2d4ffe745b68c967163f9876e5c.png

If polynomial $p(x)$ is divided by $x^{2} + 3x + 5$ , the quotient polynomial and the remainder polynomials are $2x^{2} + x + 1$ and $x = 3$ respectively. Find $P(x)$ .

We know that the division algorithm states that:

Dividend

$=$ Divisor

$\times$ Quotient

$+$ Remainder

It is given that the divisor is

$x^2+3x+5$ , the quotient is

$2x^2+x+2$ and the remainder is

$3$ . And let the dividend be

$p(x)$ . Therefore, using division algorithm we have:

$p(x)=[(x^{ 2 }+3x+5)(2x^{ 2 }+x+1)]+3\\ =[x^{ 2 }(2x^{ 2 }+x+1)+3x(2x^{ 2 }+x+1)+5(2x^{ 2 }+x+1)]+3\\ =2x^{ 4 }+x^{ 3 }+x^{ 2 }+6x^{ 3 }+3x^{ 2 }+3x+10x^{ 2 }+5x+5+3\\ =2x^{ 4 }+7x^{ 3 }+14x^{ 2 }+8x+8$

Hence, the dividend

$p(x)$ is

$2x^{ 4 }+7x^{ 3 }+14x^{ 2 }+8x+8$ .

On dividing the polynomial $p(x) = x^{3} - 3x^{2} + x + 2$ by a polynomial $g(x)$ , the quotient and remainder were $(x - 2)$ and $(-2x + 4)$ respectively. Find $g(x)$

We know that the division algorithm states that:

Dividend

$=($ Divisor

$\times$ quotient

$)+$ Remainder

Here it is given that the dividend is

$p(x)=x^3-3x^2+x+2$ , the divisor is

$g(x)$ , the quotient is

$x-2$ and the remainder is

$-2x+4$ , therefore,

$x^{ 3 }-3x^{ 2 }+x+2=[g(x)(x-2)]+(-2x+4)\\ \Rightarrow g(x)=\cfrac { x^{ 3 }-3x^{ 2 }+x+2 }{ x-2 }$

The division is shown above.

Hence, from the above division, we get that the divisor is

$g(x)=x^2-x+1$

Divide using the long division method and check the answer.
$6x^{2}+5x-4$ by $2x-1$

$\left. 2x-1 \right) 6{ x }^{ 2 }+5x-4\left( 3x+4 \right. \\ \quad \quad \quad 6{ x }^{ 2 }-3x\\ \quad \quad \ \ \ -\quad \quad +\quad \\ -------------\\ \quad \quad \quad 8x-4\\ \quad \quad \quad 8x-4\\ \quad \quad \ \ -\quad +\\ ------------\\ \quad \quad \quad X\quad$

Divide $32x^{2}+200-160x$ by $\left(8x-20\right)$

$\left. 8x-20 \right) 32{ x }^{ 2 }+200-160x\left( 4x-10 \right. \\ \quad \quad \quad \ \ \ 32{ x }^{ 2 }\quad \quad -80x\\ \quad \quad \quad \quad -\quad \quad \quad +\quad \\ -----------------\\ \quad \quad \quad \quad 200-80x\\ \quad \quad \quad \quad 200-80x\\ \quad \quad \quad \quad -\quad \quad +\\ \quad \quad --------------\\ \quad \quad \quad \quad \quad X\quad$

Divide using the long division method and check the answer.
$x^{2}+5x+10$ by $x+3$

We divide

$x^2+5x+10$ by

$(x+3)$ as shown in the above image:

From the division, we observe that the quotient is

$x+2$ and the remainder is

$4$ .

We know that the division algorithm states that:

Dividend

$=$ Divisor

$\times$ Quotient

$+$ Remainder

We apply the division algorithm to check the answer as follows:

$Dividend=Divisor\times Quotient+Remainder\\ x^{ 2 }+5x+10=[(x+3)(x+2)]+4\\ x^{ 2 }+5x+10=(x^{ 2 }+2x+3x+6)+4\\ x^{ 2 }+5x+10=x^{ 2 }+5x+6+4\\ x^{ 2 }+5x+10=x^{ 2 }+5x+10\\ Hence\quad verified.$

1248438_1127725_ans_9b43953a68ef4954b2d816f8ec68316c.png

Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:
$3x^{2} + 4x + 5, x - 2$ .

2039812_1672556_ans_93f02d20ef164374a96ac7a47d801aa3.png

Check whether ${ x }^{ 2 }+3x+1$ is a factor of
$3{ x }^{ 4}+{ 5x }^{ 3 }-{ 7x }^{ 2 }+2x+2?$

1294257_1512701_ans_b90d09d10fb14d3c9af82e407401154f.jpg

Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:
$5y^{3} - 6y^{2} + 6y - 1, 5y - 1$ .

2039813_1672558_ans_f7a8acec77994e0e8518abd5a86f3555.png

Sl.	$p(x)$	$g(x)$	$q(x)$	$r(x)$
i	?	$x - 2$	$x^{2} - x + 1$	$4$
ii	?	$x + 3$	$2x^{2} + x + 5$	$3x + 1$
iii	?	$2x + 1$	$x^{3} + 3x^{2} - x + 1$	$0$
iv	?	$x - 1$	$x^{3} - x^{2} - x - 1$	$2x - 4$
v	?	$x^{2} + 2x + 1$	$x^{4} - 2x^{2} + 5x - 7$	$4x + 12$

Polynomials - Class 10 Maths - Extra Questions

Can (x2−1)(x^2 -1) be the quotient on division of x6+2x3+x−1x^6 + 2x^3 + x- 1 by a polynomial in xx of degree 5?

Look at the graphs given. It is the graph of y=p(x)y = p(x), where p(x) is a polynomial. Find the number of zeros of p(x).p(x).

Let f(x)=x3f(x)=x^3. The graph of the polynomial is shown in the figure. Find the number of zeros of polynomial f(x).

Look at the graph given. It is the graph of y=p(x)y = p(x), where p(x) is a polynomial. Find the number of zeros of p(x).p(x).

Look at the graph given. It is the graph of y=p(x)y = p(x), where p(x) is a polynomial. Find the number of zeros of p(x)p(x).

Look at the graph given. It is the graph of y = p(x), where p(x) is a polynomial. Find the number of zeros of p(x).

The graph of y = p(x) is given in fig below, for some polynomial p(x). Find the number of zeros of p(x).

The graph of y=p(x)y = p(x) is given in figure for some polynomial p(x)p(x). The number of zeros of p(x)p(x) are

Let f(x)=x3f(x)=x^3The graph of the polynomial is shown in fig.The co-ordinates of the points, at which the graph intersects the x-axis is (0,0)If true then enter 11 and if false then enter 00

The graph of y=p(x)y = p(x) is given in the figure for some polynomial p(x)p(x). The number of zeros of p(x)p(x) is:

Construct a graph and represent the zeroes of the cubic polynomial for the equation: y=x3−6x2+11x−6y=x^3-6x^2+11x-6

On dividing 3x3+x2+2x+53x^{3} + x^{2} + 2x + 5 by a polynomial g(x)g(x), the quotient and remainder are (3x−5)(3x - 5) and (9x+10)(9x + 10) respectively. Find g(x)g(x)

A polynomial p(x)p(x) is divided by (2x−1)(2x - 1). The quotient and remainder obtained are (7x2+x+5)(7x^{2} + x + 5) and 44 respectively. Find p(x)p(x)

What must be subtracted from 6x^{4} + 13x^{3} + 13x^{2} + 30x + 206x^{4} + 13x^{3} + 13x^{2} + 30x + 20, so that the resulting polynomial is exactly divisible by 3x^{2} + 2x + 53x^{2} + 2x + 5?

Which type of polynomial, the given expression x^{3} + x + 4x^{3} + x + 4 is?

Based on degree, which type of polynomial, the given expression 5x^{2} + x - 75x^{2} + x - 7 is?

Which type of polynomial, the given expression x - 1x - 1 is?

Which type of polynomial, the given expression 3p3p is?

Which type of polynomial, the given expression x - x^{3}x - x^{3} is?

Classify the following expression as linear, quadratic, or cubic polynomials.\pi r^{2}\pi r^{2}

Draw the graph of y = x^2 - x - 6y = x^2 - x - 6 and find zeroes. Justify the answer.

Divide the polynomials p(x)p(x) by the polynomials q(x)q(x). Find the quotient and remainder. p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,\,\,\,\,\,\,\,\,\,\, q\left( x \right) = {x^2} - 2p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,\,\,\,\,\,\,\,\,\,\, q\left( x \right) = {x^2} - 2

Classify the following as linear, quadratic and cubic polynomials5{x}^{2}+x-75{x}^{2}+x-7

Find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by \dfrac{-8}{3},\dfrac{4}{3}\dfrac{-8}{3},\dfrac{4}{3}

Classify the following as linear, quadratic and cubic polynomialsx-1x-1

Classify the following as linear, quadratic and cubic polynomials{x}^{2}+x+4{x}^{2}+x+4

What is meant by division algorithm give example?

Classify whether x-x^3x-x^3 is linear, quadratic, or a cubic polynomial.

Divide the polynomial p(x)p(x) by the polynomial g(x)g(x) and find the quotient and remainder in each of the following.p(x)=x^{4}-3x^{2}+4x-3\quad g(x)=x^{2}-2p(x)=x^{4}-3x^{2}+4x-3\quad g(x)=x^{2}-2

Find the remainder when x^4 + 5x^2 + 6 x^4 + 5x^2 + 6 is divided by x^2 + 1x^2 + 1.

Divide (y^{3} - 3y^{2} + 5y - 1)(y^{3} - 3y^{2} + 5y - 1) by (y - 1)(y - 1) to find quotient and remainder.

Divide the polynomial p(x)p(x) by the polynomial g(x)g(x) and find the quotient and remainder in each of the following.p(x)=x^{4}-5x+6\quad g(x)=2-x^{2}p(x)=x^{4}-5x+6\quad g(x)=2-x^{2}

Divide 6x^2-31x+476x^2-31x+47 by 2x-52x-5 and verify the division algorithm.

Classify the following as linear, quadratic and cubic polynomials3p3p

Divide and also verify the result {x^2} + 3x + 2{x^2} + 3x + 2 by \left( {x + 2} \right)\left( {x + 2} \right)

Give an example of a quadratic binomial.

Divide and write the quotient and remainder.(a) (y^2+10y+24) \div (y+4)(y^2+10y+24) \div (y+4)(b) (p^2+7p-5) \div (p+3)(p^2+7p-5) \div (p+3)

Divide and write the quotient and remainder.(y^2+10y+24) \div (y+4)(y^2+10y+24) \div (y+4)

Divide:2{x^3} - {x^2} - 3x + 1\,\,\,\ by\,\,\ (x + 1)2{x^3} - {x^2} - 3x + 1\,\,\,\ by\,\,\ (x + 1)

Find the quotient and remainder.(a) \dfrac{{{a^2} + 8a - 9}}{{a - 7}}\dfrac{{{a^2} + 8a - 9}}{{a - 7}}(b) \dfrac{{{x^2} + 6x - 11}}{{x - 3}}\dfrac{{{x^2} + 6x - 11}}{{x - 3}}

Fill in the blacks.A polynomial of degree 11 is called a _____ polynomial

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.-p-p.

If the graph of a polynomial intersects the x - asis x - asis at only one point, can it be a quadratic polynomial ?

Examine, seeing the graph of the polynomial given below, whether they are a linear or quadratic polynomial or neither linear nor quadratic polynomial:

Classify the following as linear , quadratic and cubic polynomial : 1 + x 1 + x

Classify the following as linear , quadratic and cubic polynomial : 3t

Classify the following as linear , quadratic and cubic polynomial : y + y^{2} + 4 y + y^{2} + 4

Classify the following polynomial as linear, quadratic and cubic polynomial .a^{2}a^{2}

Classify the following as linear , quadratic and cubic polynomial : r^{2} r^{2}

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:3x^2 + 4x- 43x^2 + 4x- 4.

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:4x^2 + 5 \sqrt{2}x -34x^2 + 5 \sqrt{2}x -3

If one root of the equation x^{3}-13x^{2}+15x+189= 0x^{3}-13x^{2}+15x+189= 0 exceeds other by 22 then find its smallest root.

If \alpha\alpha and \beta\beta are the zeros of the polynomial x^2 -8x + kx^2 -8x + k such that \alpha^2+\beta^2=40\alpha^2+\beta^2=40, then the value of kk is

Prove that there do not exist natural numbers xx and y,y, with x > 1,x > 1, such that \displaystyle \frac {x^7-1}{x-1}=y^5+1\displaystyle \frac {x^7-1}{x-1}=y^5+1.

Verify that -5, \displaystyle \frac{1}{2}, \frac{3}{4}-5, \displaystyle \frac{1}{2}, \frac{3}{4} are zeros of cubic polynomial 8x^3 + 30x^2 -47x + 158x^3 + 30x^2 -47x + 15. Also verify the relationship between the zeros and the coefficients.

3x^{2}+2x-1=03x^{2}+2x-1=0 is a equation of degree _____

Classify the following as linear, quadratic and cubic polynomials:(i) x^2+xx^2+x(ii) x-x^3x-x^3(Iii) y+y^2+4y+y^2+4(iv) 1+x1+x(v) 3t3t(vi) r^2r^2(vii) 7x^37x^3

The graphs of y = p(x)y = p(x) are given in the figure, for some polynomials p(x)p(x). Find the number of zeroes of p(x)p(x), in each case.

Give the geometric representations of 2x+9=02x+9=0 as an equation(i) in one variable(ii) in two variable

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.(i) x^2 - 2x - 8x^2 - 2x - 8 (ii) 4s^2 - 4s + 14s^2 - 4s + 1 (iii) 6x^2 - 3 - 7x6x^2 - 3 - 7x (iv) 4u^2 + 8u4u^2 + 8u (v) t^2 - 15t^2 - 15 (vi) 3x^2 - x - 43x^2 - x - 4

If two zeroes of the polynomial x^4- 6x^3 - 26x^2 + 138x -35x^4- 6x^3 - 26x^2 + 138x -35 are 2\pm \sqrt{3}2\pm \sqrt{3}, find the other zeroes.

On dividing x^3- 3x^2 + x + 2x^3- 3x^2 + x + 2 by a polynomial g(x), the quotient and remainder were ( x-2 )( x-2 ) and (-2x + 4)(-2x + 4), respectively. Find g(x).g(x).

Give examples of polynomials p(x), g(x), q(x)p(x), g(x), q(x) and r(x)r(x), which satisfy the division algorithm and(i) deg p(x) = p(x) = deg q(x)q(x) (ii) deg q(x) = q(x) = deg r(x)r(x) (iii) deg r(x) = 0r(x) = 0

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:(i) 2x^3+x^2-5x+2; \dfrac{1}{2}, 1, -22x^3+x^2-5x+2; \dfrac{1}{2}, 1, -2 (ii) x^3-4x^2+5x-2; 2, 1, 1x^3-4x^2+5x-2; 2, 1, 1

Sketch the graph and identify zeroes of a linear polynomial: \dfrac{x + 10}{4}\dfrac{x + 10}{4}.

Construct a graph which represent the zeroes of the quadratic polynomial for an equationy=x^2-10xy=x^2-10x.

Graphically represent the zeros of the given linear polynomial: y=x-12y=x-12

Graphically represent the zeroes of the quadratic polynomial for an equation y=2x^2-11x+5y=2x^2-11x+5 in graph.

What are the zeroes of the quadratic polynomial from the graph?

Find the zeroes of the quadratic polynomial from the graph.

Represent the zeroes of the quadratic polynomial for an equation y=x^2+x-9 y=x^2+x-9 in graph.

Sketch a graph and represent the zeroes of the quadratic polynomial for the equation y = x^2+2x-3y = x^2+2x-3.

Graphically represent the zeroes of the cubic polynomial: y=x^3+4x^2+x-6y=x^3+4x^2+x-6

Using division algorithm, divide the polynomial 2x^3+2x^2-5x+12x^3+2x^2-5x+1 by (x-1)(x-1).

Graphically represent the zeroes of the cubic polynomial: y=x^3-3x+2y=x^3-3x+2

y=x^3-3x^2-6x+8y=x^3-3x^2-6x+8, Draw a graphical representation of the polynomial.

Quadratic polynomial: y=x^2-4x+4y=x^2-4x+4, draw a graph and represent the zeroes of the polynomial.

Sketch a graphical representation of the zeros of the given polynomial: y=x^3-12x^2+44x-48y=x^3-12x^2+44x-48

Divide: 2x^4+x^3+8x^2+14x+52x^4+x^3+8x^2+14x+5 by (2x+1)(2x+1) (use division algorithm method).

Can $(x^2 -1)$ be the quotient on division of $x^6 + 2x^3 + x- 1$ by a polynomial in $x$ of degree 5?

Look at the graphs given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x).$

Let $f(x)=x^3$ . The graph of the polynomial is shown in the figure. Find the number of zeros of polynomial f(x).

Look at the graph given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x).$

Look at the graph given. It is the graph of $y = p(x)$ , where p(x) is a polynomial. Find the number of zeros of $p(x)$ .

The graph of $y = p(x)$ is given in figure for some polynomial $p(x)$ . The number of zeros of $p(x)$ are

Let $f(x)=x^3$
The graph of the polynomial is shown in fig.
The co-ordinates of the points, at which the graph intersects the x-axis is (0,0)
If true then enter $1$ and if false then enter $0$

The graph of $y = p(x)$ is given in the figure for some polynomial $p(x)$ . The number of zeros of $p(x)$ is:

Construct a graph and represent the zeroes of the cubic polynomial for the equation: $y=x^3-6x^2+11x-6$

On dividing $3x^{3} + x^{2} + 2x + 5$ by a polynomial $g(x)$ , the quotient and remainder are $(3x - 5)$ and $(9x + 10)$ respectively. Find $g(x)$

A polynomial $p(x)$ is divided by $(2x - 1)$ . The quotient and remainder obtained are $(7x^{2} + x + 5)$ and $4$ respectively. Find $p(x)$

What must be subtracted from $6x^{4} + 13x^{3} + 13x^{2} + 30x + 20$ , so that the resulting polynomial is exactly divisible by $3x^{2} + 2x + 5$ ?

Which type of polynomial, the given expression $x^{3} + x + 4$ is?

Based on degree, which type of polynomial, the given expression $5x^{2} + x - 7$ is?

Which type of polynomial, the given expression $x - 1$ is?

Which type of polynomial, the given expression $3p$ is?

Which type of polynomial, the given expression $x - x^{3}$ is?

Classify the following expression as linear, quadratic, or cubic polynomials.
$\pi r^{2}$

Draw the graph of $y = x^2 - x - 6$ and find zeroes. Justify the answer.

Divide the polynomials $p(x)$ by the polynomials $q(x)$ . Find the quotient and remainder. $p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,\,\,\,\,\,\,\,\,\,\, q\left( x \right) = {x^2} - 2$

Classify the following as linear, quadratic and cubic polynomials
$5{x}^{2}+x-7$

Find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by
$\dfrac{-8}{3},\dfrac{4}{3}$

Classify the following as linear, quadratic and cubic polynomials
$x-1$

Classify the following as linear, quadratic and cubic polynomials
${x}^{2}+x+4$

Classify whether $x-x^3$ is linear, quadratic, or a cubic polynomial.

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following.
$p(x)=x^{4}-3x^{2}+4x-3\quad g(x)=x^{2}-2$

Find the remainder when $x^4 + 5x^2 + 6$ is divided by $x^2 + 1$ .

Divide $(y^{3} - 3y^{2} + 5y - 1)$ by $(y - 1)$ to find quotient and remainder.

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following.
$p(x)=x^{4}-5x+6\quad g(x)=2-x^{2}$

Divide $6x^2-31x+47$ by $2x-5$ and verify the division algorithm.

Classify the following as linear, quadratic and cubic polynomials
$3p$

Divide and also verify the result ${x^2} + 3x + 2$ by $\left( {x + 2} \right)$

Divide and write the quotient and remainder.
(a) $(y^2+10y+24) \div (y+4)$
(b) $(p^2+7p-5) \div (p+3)$

Divide and write the quotient and remainder.
$(y^2+10y+24) \div (y+4)$

Divide:
$2{x^3} - {x^2} - 3x + 1\,\,\,\ by\,\,\ (x + 1)$

Find the quotient and remainder.

(a) $\dfrac{{{a^2} + 8a - 9}}{{a - 7}}$

(b) $\dfrac{{{x^2} + 6x - 11}}{{x - 3}}$

Fill in the blacks.
A polynomial of degree $1$ is called a _____ polynomial

Identify constant, linear, quadratic, cubic and quartic polynomial from the following.
$-p$ .

If the graph of a polynomial intersects the $x - asis$ at only one point, can it be a quadratic polynomial ?

Classify the following as linear , quadratic and cubic polynomial :
$1 + x$

Classify the following as linear , quadratic and cubic polynomial :
3t

Classify the following as linear , quadratic and cubic polynomial :
$y + y^{2} + 4$

Classify the following polynomial as linear, quadratic and cubic polynomial .
$a^{2}$

Classify the following as linear , quadratic and cubic polynomial :
$r^{2}$

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$3x^2 + 4x- 4$ .

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$4x^2 + 5 \sqrt{2}x -3$

If one root of the equation $x^{3}-13x^{2}+15x+189= 0$ exceeds other by $2$ then find its smallest root.

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2 -8x + k$ such that $\alpha^2+\beta^2=40$ , then the value of $k$ is

Prove that there do not exist natural numbers $x$ and $y,$ with $x > 1,$ such that $\displaystyle \frac {x^7-1}{x-1}=y^5+1$ .

Verify that $-5, \displaystyle \frac{1}{2}, \frac{3}{4}$ are zeros of cubic polynomial $8x^3 + 30x^2 -47x + 15$ . Also verify the relationship between the zeros and the coefficients.

$3x^{2}+2x-1=0$ is a equation of degree _____

Classify the following as linear, quadratic and cubic polynomials:
(i) $x^2+x$
(ii) $x-x^3$
(Iii) $y+y^2+4$
(iv) $1+x$
(v) $3t$
(vi) $r^2$
(vii) $7x^3$

The graphs of $y = p(x)$ are given in the figure, for some polynomials $p(x)$ . Find the number of zeroes of $p(x)$ , in each case.

Give the geometric representations of $2x+9=0$ as an equation
(i) in one variable
(ii) in two variable

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$ (ii) $4s^2 - 4s + 1$ (v) $t^2 - 15$ (vi) $3x^2 - x - 4$

If two zeroes of the polynomial $x^4- 6x^3 - 26x^2 + 138x -35$ are $2\pm \sqrt{3}$ , find the other zeroes.

On dividing $x^3- 3x^2 + x + 2$ by a polynomial g(x), the quotient and remainder were $( x-2 )$ and $(-2x + 4)$ , respectively. Find $g(x).$

Give examples of polynomials $p(x), g(x), q(x)$ and $r(x)$ , which satisfy the division algorithm and
(i) deg $p(x) =$ deg $q(x)$ (ii) deg $q(x) =$ deg $r(x)$ (iii) deg $r(x) = 0$

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) $2x^3+x^2-5x+2; \dfrac{1}{2}, 1, -2$

(ii) $x^3-4x^2+5x-2; 2, 1, 1$

Sketch the graph and identify zeroes of a linear polynomial: $\dfrac{x + 10}{4}$ .

Construct a graph which represent the zeroes of the quadratic polynomial for an equation $y=x^2-10x$ .

Graphically represent the zeros of the given linear polynomial: $y=x-12$

Graphically represent the zeroes of the quadratic polynomial for an equation $y=2x^2-11x+5$ in graph.

Represent the zeroes of the quadratic polynomial for an equation $y=x^2+x-9$ in graph.

Sketch a graph and represent the zeroes of the quadratic polynomial for the equation $y = x^2+2x-3$ .

Graphically represent the zeroes of the cubic polynomial: $y=x^3+4x^2+x-6$

Using division algorithm, divide the polynomial $2x^3+2x^2-5x+1$ by $(x-1)$ .

Graphically represent the zeroes of the cubic polynomial: $y=x^3-3x+2$

$y=x^3-3x^2-6x+8$ , Draw a graphical representation of the polynomial.

Quadratic polynomial: $y=x^2-4x+4$ , draw a graph and represent the zeroes of the polynomial.

Sketch a graphical representation of the zeros of the given polynomial: $y=x^3-12x^2+44x-48$

Divide: $2x^4+x^3+8x^2+14x+5$ by $(2x+1)$ (use division algorithm method).

Use synthetic division to find the quotient Q and Remainder R when dividing
$\dfrac12{x}^{3}-\dfrac13{x}^{2}-\dfrac32x+\dfrac13$ by $x-\dfrac12$

Use synthetic division to find the quotient Q and Remainder R when dividing
$7{x}^{5}-2{x}^{4}-5{x}^{3}+{x}^{2}-3x+5$ by $x-2$

Use synthetic division to find the quotient Q and Remainder R when dividing
$4{x}^{3}+{x}^{2}-2x+3$ by $x+2$

$\dfrac{3x^4+x^3+6x^2-16x-6}{3x+1}$ = ? (Use division algorithm method).

Given $f(x) =$ $4x^4+6x^3+6x^2+6x+2$ , $g(x) =$ $2x+2$ . Divide $\dfrac{f(x)}{g(x)}$ (Use division algorithm method)

Find the quotient and remainder on dividing $p(x) = x^{3} - 6x^{2} + 15x - 8$ by $g(x) = x - 2$

Divide $(5x + x^{2} + 14 + 2x^{3})$ by $(x + 2)$

What must be added to the polynomial $p(x) = x^{4} + 2x^{3} - 2x^{2} + x - 1$ so that the resulting polynomial is exactly divisible by $x^{2} + 2x - 3$