MCQ Questions for Class 9 Maths Polynomials Quiz 11

Write whether the following statement is True or False ? Justify your answer.
Zero of polynomial is always

$0$

0%
True
0%
False

If one zero of the polynomial

$p(x)={x}^{3}-6{x}^{2}+11x-6$ is

$3$ , find the other two zeroes.

0%
$0$ and $2$
0%
$2$ and $-2$
0%
$1$ and $2$
0%
$2$ and $-3$

The number of rational roots of

$(2x+3)(2x+5)(x-1)(x-2)=30$ is

0%
$4$
0%
$3$
0%
$1$
0%
$5$

Using identities, evaluate:

$71^2$ .

0%
$5041$
0%
$3041$
0%
$6041$
0%
$2041$

Explanation

Given,

$71^2$ ,

i.e.

$71^2$

$=(70+1)^2$ .

We know,

$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$ .

Then,

$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$

Hence,

$71^2=5041$ .

Therefore, option

$A$ is correct.

Find the zeros of the polynomial

$3\sqrt2 x^2+ 13x + 6 \sqrt2$ and verify the relationship between the zeros.

0%
$x=\dfrac{\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$
0%
$x={2\sqrt2}\text{ and } \dfrac{-3}{\sqrt{2}}$
0%
$x=\dfrac{-2\sqrt2}{3}\text{ and } {-3}{\sqrt{2}}$
0%
$x=\dfrac{2\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$

Explanation

Let us first factorize the given equation

$3\sqrt 2x^2+13x+6\sqrt2$ as shown below:

$\Rightarrow 3\sqrt { 2 } x^{ 2 }+13x+6\sqrt { 2 } \\ =3\sqrt { 2 } x^{ 2 }+9x+4x+6\sqrt { 2 } \\ =3x(\sqrt { 2 } x+3)+2\sqrt { 2 } (\sqrt { 2 } x+3)\\ =(3x+2\sqrt { 2 } )(\sqrt { 2 } x+3)$

Therefore, the zeroes of the given polynomial are:

$(3x+2\sqrt { 2 } )=0\\ \Rightarrow x=-\dfrac { 2\sqrt { 2 } }{ 3 }=\alpha\ \\ (\sqrt { 2 } x+3)=0\\ \Rightarrow x=-\dfrac { 3 }{ \sqrt { 2 } }=\beta$

Now, the sum and product of the roots is as follows:

$\alpha +\beta =-\dfrac { 2\sqrt { 2 } }{ 3 } -\dfrac { 3 }{ \sqrt { 2 } } =-\left( \dfrac { 2\sqrt { 2 } }{ 3 } +\dfrac { 3 }{ \sqrt { 2 } } \right) =\dfrac { -4-9 }{ 3\sqrt { 2 } } =-\dfrac { 13 }{ 3\sqrt { 2 } } =-\dfrac { b }{ a } \\ \alpha \beta =\left( -\dfrac { 2\sqrt { 2 } }{ 3 } \right) \left( -\dfrac { 3 }{ \sqrt { 2 } } \right) =2=\dfrac { 6\sqrt { 2 } }{ 3\sqrt { 2 } } =\dfrac { c }{ a }$

Hence verified.

The value of

${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$ , corrected to three decimal places is:

0%
$2.001$
0%
$2.004$
0%
$2.006$
0%
$1.995$

Explanation

Given,

$(1.02)^2+(0.98)^2$ .

We know,

$(a+b)^2= a^2+2ab+b^2$ .

Then,

$(1.02)^2+(0.98)^2$

$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$ )

$=(1.02+0.98)^2-1.9992$

$=(2)^2-1.9992$

$=4-1.9992$

$=2.0008$ .

When corrected to three decimal digit,

$2.0008$

$=2.001$ .

Therefore, option

$A$ is correct.

Which of the following is the zeroes of the polynomial

$x^2 + 4x + 4 =$ ?

0%
$2$
0%
$-2$
0%
$4$
0%
$-4$

Explanation

Now,

$x^2 + 4x + 4$

$=(x+2)^2$ .

$x^2+4x+4=0$ then

$x=-2,-2$ .

So the the zeroes of

$x^2+4x+4$ is

$-2$ .

The zeroes of the polynomial

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

0%
$0$
0%
$0,2,3$
0%
$0,2,-3$
0%
$None\ of\ these$

Explanation

$P\left(x\right)=x\left(x-2\right) \left(x+3\right)$

$\therefore \$ To find zeroes,

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

$\therefore \left(x\right) \left(x-2\right) \left(x+3\right)=0$

$\therefore x=0$ OR

$x-2=0$ OR

$x+3=0$

$\therefore x=0$ OR

$x=2$ OR

$x=-3$

$\therefore$ value of

$x$ is

$0, 2, -3$

Degree of the polynomial

$13 + 11x + 12x^3 + 3x^2$ is

0%
$2$
0%
$1$
0%
$3$
0%
$4$

Explanation

The degree of an individual term of a polynomial is the exponent of its variable

$13+11x+12{x}^{3}+3{x}^{2}=12{x}^{3}+3{x}^{2}+11x+13$

The highest exponent of

$x$ is

$3$

$\therefore\,$ Degree

$=3$

The roots of

$x^{3}-4x^{2}-8x+8$ are :

0%
$1,2\pm \sqrt{3}$
0%
$-2,3\pm \sqrt{5}$
0%
$-2,2\pm \sqrt{3}$
0%
$1,3\pm \sqrt{5}$

Explanation

$\begin{array}{l} { x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ { { suppose } }, \\ y={ x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ and\, \, \, replace\, y\, with\, \, 0. \\ { x^{ 3 } }-4{ x^{ 2 } }-8x+8=0 \\ Now,\, factorizing\, the\, equ: \\ (x+2)\, ({ x^{ 2 } }-6x+4)=0 \\ \Rightarrow x+2=0 \\ \therefore \, \, \, x=-2 \\ and,\, \\ \, \, \, { x^{ 2 } }-6x+4=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { u\sin g\, quadratic{ { } }formula=-\frac { { b\pm \sqrt { { b^{ 2 } }-4(ac) } } }{ { 2a } } } \right. \\ \, \, \, \, substitute\, the\, values\& \, find\, the\, value\, of\, x: \\ \, \, \, \, \, \, \, a=1,\, b=-6,\, c=4 \\ \, \Rightarrow x=\, \, \frac { { -6\pm \sqrt { { { (-6) }^{ 2 } }-4(1)\, (4) } } }{ { 2\, .1 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { { 6\pm \sqrt { 36-16 } } }{ { 2\, } } =\frac { { 6\pm \sqrt { 20 } } }{ 2 } =\frac { { 6\pm 2\sqrt { 5 } } }{ { 2\, } } =3\pm \sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, x=3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ so,that\, we\, can\, say\, the\, value\, of\, \, x=-2,\, 3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, or\, \, x=-2,\, 3\pm \sqrt { 5 } \, \\ And\, the\, \, correct\, option\, is\, B. \end{array}$

Factorise :

${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$

0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$

Explanation

$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$

Here we will use the following identity,

$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$

and

$\left(m-n\right)^2=m^2+n^2-2mn$

Using this we have :

$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$

$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$

$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$

Hence, the answer is

$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$

If one zero of a quadratic polynomial

$x^2+3x+k$ is

$2$ . Find the value of

$k$ .

0%
$2$
0%
$-10$
0%
$-5$
0%
$9$

Explanation

$2$ is a zero of

$x^2+3x+k$ .

Thus,

$(2)^2+3(2)+k=0$

$k+10=0$

$k=-10$

$xy=20$ and

$(x+y)^{2}=70$ , then

$x^{2}+y^{2}$ is equal to

0%
$25$
0%
$35$
0%
$30$
0%
$50$

Explanation

$(x+y)^2=x^2+y^2+2xy$

It is given that

$(x+y)^2=70$ and

$xy=20$

$\Rightarrow 70=x^2+y^2+(2\times 20)$

$x^2+y^2=70-40$

$=30$

The zeroes of the polynomial

${ p }({ x })={ x }({ x }-1)({ x }-2)$ are

0%
$0$
0%
$0 , - 1 , - 2$
0%
$0,1 , - 2$
0%
$0,1,2$

Explanation

$\begin{array}{l} We\, have \\ p\left( x \right) =x\left( { x-1 } \right) \left( { x-2 } \right) \\ for\, the\, \, zero\, polynomial \\ p\left( x \right) =x \\ \therefore x\left( { x-1 } \right) \left( { x-2 } \right) =0 \\ Here\, we\, get \\ x=0\, \, \, or\, \, \left( { x-1 } \right) =0\, \, and\, \left( { x-2 } \right) =0 \\ \therefore x=0\, \, ,\, \, \, x=1\, \, and\, x=2 \\ So,\, \left( { 0,1,2 } \right) \, are\, the\, zero\, polynomila\, of\, p\left( x \right) \, \\ Hence,\, option\, D\, is\, the\, required\, asnwer. \end{array}$

$(a-3)x^{2}+(2a+b)x+(b-c)$ is a zero polynomial, then

0%
$b+2a=0$
0%
$ab+c=0$
0%
$a=3$
0%
$b-c=0$

The zero of the polynomial

$P\left( x \right) =\surd 5x-5$ is ......

0%
$\surd 5$
0%
$-\surd 5$
0%
$-5$
0%
$4$

Explanation

Given,

$p(x)=\sqrt{5}x-5$ .

Here, we find zero of a polynomial when

$p(x)=0$ .

Then,

$\sqrt{5}x-5=0$

$\implies$

$\sqrt{5}x=5$

$\implies$

$x=\dfrac{5}{\sqrt{5}}$

$\implies$

$x=\dfrac{5}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}$

$\implies$

$x=\dfrac{5\sqrt{5}}{5}=\sqrt{5}$ .

Therefore, option

$A$ is correct.

The degree of the polynomial

$\frac { 4 }{ 5 } { x }^{ 2 }-\frac { 7 }{ 5 } x+\frac { 2 }{ 3 } { x }^{ 3 }+6$ is :

0%
2
0%
1
0%
3
0%
0

The value of

$(a +b)^{2} + (a-b)^{2}$ is

0%
$2a + 2b$
0%
$2a - 2b$
0%
$2a^{2} + 2b^{2}$
0%
$2a^{2} - 2b^{2}$

Explanation

$\textbf{Step-1: Apply the relevant formula & simplify.}$

$\text{We have,}$

$(a + b)^{2} + (a - b)^{2}$

$\text{Now,}$

$(a+b)^2 = a^2+2ab+b^2$

$......(i)$

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

$......(ii)$

$\textbf{Step-2: Add the above expressions to get the result}$

$(a + b)^{2} + (a - b)^{2}$

$a^2+2ab+b^2 + a^2-2ab+b^2$

$=2a^2 +2b^2$

$\textbf{Hence, option - C is the answer}$

Write whether the following statement is True or False. Justify your answer.
A polynomial cannot have more than one zero

0%
True
0%
False

Write whether the following statement is True or False. Justify your answer.
Zero of a polynomial is always

$0$

0%
True
0%
False

Write whether the following statement is True or False. Justify your answer.
The degree of the sum of two polynomials each of degree

$5$ is always

$5$ .

0%
True
0%
False

Explanation

The given statement is False. For example, consider the two polynomial

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

$x^5 + x^4 + 2x^3 + 3.$ The degree of each of these polynomial is

$5$ . Their sum is

$x^4 + 2x^3 + 3x^2 + 7.$ The degree of this polynomial is

$4$ not

$5$ .

Verify whether the following is true or false.

$\dfrac{-4}{5}$ is a zero of

$4 - 5y.$

0%
True
0%
False

Explanation

A zero of a polynomial

$p(x)$ is a number

$c$ such that

$p(c) = 0$

Let

$p(y) = 4 - 5y$

$\therefore p \left ( -\dfrac{4}{5} \right ) = 4 - 5 \left ( \dfrac{-4}{5} \right ) = 4 + 4 = 8 \neq 0$

Hence,

$-\dfrac{4}{5}$ is not a zero of

$4 - 5y.$

So, given statement is false.

Verify whether the following is true or false.

$0$ and

$2$ are the zeroes of

$t^2 - 2t.$

0%
True
0%
False

Explanation

A zero of a polynomial

$p(x)$ is a number

$c$ such that

$p(c) = 0$

Let

$p(t) = t^2 - 2t$

$\therefore p(0) = (0)^2 - 2(0) = 0$

And

$p(2) = (2)^2 - 2(2) = 4 - 4 = 0$

Hence,

$0$ and

$2$ are zeroes of the polynomial

$p(t) = t^2 - 2t.$

So, given statement is true.

Verify whether the following is true or false.

$-\dfrac{1}{3}$ is a zero of

$3x + 1.$

0%
True
0%
False

Explanation

A zero of a polynomial

$p(x)$ is a number

$c$ such that

$p(c) = 0$

Let

$p(x) = 3x + 1$

$\therefore p\left ( -\dfrac{1}{3} \right ) = 3 \left ( -\dfrac{1}{3} \right ) + 1 = -1 + 1 = 0$

Hence,

$-\dfrac{1}{3}$ is a zero of

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

So, given statement is true.

Verify whether the following is true or false.

$-3$ is a zero of

$x - 3$

0%
True
0%
False

Explanation

A zero of a polynomial

$p(x)$ is a number

$c$ such that

$p(c) = 0$

Let

$p(x) = x - 3$

$\therefore p(-3) = -3 - 3 = -6 \neq 0$

Hence,

$-3$ is not a zero of

$x - 3$ .

So, given statement is false.

Verify whether the following is true or false.

$-3$ is a zero of

$y^2 + y - 6.$

0%
True
0%
False

Explanation

A zero of a polynomial

$p(x)$ is a number

$c$ such that

$p(c) = 0$

Let

$p(y) = y^2 + y - 6$

$\therefore p(-3) = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0$

Hence,

$-3$ is a zero of the polynomial

$y^2 + y - 6.$

So, given statement is true.

Find the zeroes of the polynomial

$4-\dfrac{1}{2}x^2$

0%
2
0%
$2\sqrt{2}$
0%
0
0%
4

Explanation

$4-\dfrac{1}{2}x^2=0$

$\dfrac{1}{2}x^2=4$

$x^2=8$

$\therefore x=\pm 2\sqrt{2}$

$\therefore$ The zeroes of the given polynomial

$2\sqrt{2}$ .

Which of the following are the zeroes of the quadratic polynomial

$9-4x^2$ ?

0%
4
0%
9
0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$

Explanation

$9-4x^2=0$

$4x^2=9$

$x^2=\dfrac{9}{4}$

$\therefore x=\pm \dfrac{3}{2}$

$\therefore$ The zeroes of the given polynomial are

$\dfrac{3}{2}$ and

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

The zeroes of the polynomials,

$t^2 15$ are.

0%
$\pm \sqrt{15}$
0%
$\pm \sqrt{5}$
0%
$\pm \sqrt{3}$
0%
$\pm 3$

Explanation

$t^2 – 15=0$

$t^2=15$

$t=\pm \sqrt{15}$

The degree of a quadratic polynomial is:

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

The highest power of the variable in a polynomial in one variable is called the degree of the polynomial.

Quadratic polynomial is a polynomial in which highest order of the variable is

$2$ .

For example,

$2x^2+5x+1=0$ is a quadratic equation and it has a degree

$2$ .

Thus, degree of quadratic polynomial is

$2$ .

Therefore, option

$C$ is correct.

CBSE Questions for Class 9 Maths Polynomials Quiz 11 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 11 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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