MCQ Questions for Class 9 Maths Polynomials Quiz 7

Find the square of:

$9.7$

0%
97.09
0%
94.09
0%
96.09
0%
93.09

Explanation

We can write

${9.7}^{2} = {(10 - 0.3)}^{2}$
It is the form of ${(a - b)}^{2}$ , where $a = 10, b = 0.3$
Applying the formula ${ (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$
${9.7}^{2} = {(10 - 0.3)}^{2} = {10}^{2} + { 0.3 }^{ 2 } -2\times 10 \times 0.3 = 100 + 0.09 - 6 = 94.09$

If the value of the polynomial

$x^3+2x^2-ax+1$ at

$x = 2$ is 11, then find the value of a.

0%
$a = 5$
0%
$a = 3$
0%
$a = -2$
0%
$a = -9$

Explanation

Let,

$p(x)=x^3+2x^2-ax+1$
Given

$p(2)=11$

$=>2^3+2(2)^2-a(2)+1=11$

$=>8+8-2a+1=11$

$=>-2a=11-8-8-1$

$=>-2a=-6$

$=>a=\frac{6}{2}$

$=>a=3$

Evaluate:

$\displaystyle\left(\dfrac{7}{8}{x}\, +\, \dfrac{4}{5}{y}\right )^{2}$ .

0%
$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$
0%
$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$
0%
$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$
0%
$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$

Explanation

Given, $\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$ .

We know, $(x+y)^2=x^2+2xy+y^2$ .

Then,

$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$

$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$

$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}}$ .

Therefore, option $D$ is correct.

Find the value of the polynomial

$2a^2-5 a^3+7a-3$ at

$a = 0, 2$ and

$-1$ .

0%
$p(0) = -3, \, \, \, p(2) = -21, \, \, \, p(-1) = -3$
0%
$p(0) = -2, \, \, \, p(2) = -1, \, \, \, p(-1) = -3$
0%
$p(0) = 3, \, \, \, p(2) = -13, \, \, \, p(-1) = -3$
0%
$p(0) = 2, \, \, \, p(2) = -16, \, \, \, p(-1) = -3$

Explanation

Let,

$p(a)=2a^2-5 a^3+7a-3$

$p(o)=2(0)^2-5(0)^3+7(0)-3$

$=-3$

$p(2)=2(2)^2-5(2)^3+7(2)-3$

$=8-40+14-3$

$=-21$

$p(-1)=2(-1)^2-5(-1)^3+7(-1)-3$

$=2+5-7-3$

$=-3$

Find the square of:

$2a + b$ .

0%
$4a^{2}\, -\, 4b\, +\, b^{3}$
0%
$a^{2}\, +\, 4ab\, +\, b^{3}$
0%
$a^{2}\, -\, 4b\, +\, b^{2}$
0%
$4a^{2}\, +\, 4ab\, +\, b^{2}$

Explanation

Given,

$2a + b$ .

On squaring, we get,

$(2a+b)^2$ .

We know,

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

Then,

$(2a+b)^2$

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

$=4a^2+4ab+b^2$ .

Therefore, option

$D$ is correct.

Find the square of the following number:

$998$

0%
$9,96,004$
0%
$9,15,004$
0%
$9,77,004$
0%
$9,83,004$

Explanation

$(998)^2=(1000-2)^2$

Using,

$(a-b)^2=a^2-2ab+b^2$ , we get

$(998)^2 = (1000)^2-2(1000)2+2^2$

$=1000000-4000+4$

$=996004$

Verify that the numbers given along side of the polynomial are their zeros.

$x^4+2x^3-7x^2-8x+12; -3, -2, 1, 2$

0%
Yes, $-3, -2, 1, 2$ are the zeros of given polynomial
0%
No, $-3, -2, 1, 2$ are the zeros of given polynomial
0%
Ambiguous
0%
Data insufficient

Explanation

Here the polynomial

$p(x)$ is

$x^4+2x^3-7x^2-8x+12$
Value of the polynomial

$x^4+2x^3-7x^2-8x+12$
when

$x=-3$

$P(-3)=(-3)^4+2(-3)^3-7(-3)^2-8(-3)+12$

$=81-54-63+24+12=0$
So,

$-3$ is a zero of

$p(x)$ .

On putting

$x=-2$ in the given polynomial, we have

$P(-2)=(-2)^4+2(-2)^3-7(-2)^2-8(-2)+12$

$=16-16+28+16+12=0$
So,

$-2$ is a zero of

$p(x)$ .

On putting

$x=1$ in the given polynomial, we have

$P(1)=(1)^4+2(1)^3-7(1)^2-8(1)+12$

$=1+2-7-8+12=0$
So,

$2$ is a zero of

$p(x)$ .

On putting

$x=2$ in the given polynomial, we have

$P(2)=(2)^4+2(2)^3-7(2)^2-8(2)+12$
$$=16+16-28-16+12=0$$

So,

$2$ is a zero of

$p(x)$ .
Hence,

$-3, -2, 1, 2$ are the zeros of given polynomial.

What is the zero of the binomial

$ax + b$ ?

0%
$0$
0%
$\displaystyle \frac{b}{a}$
0%
$\displaystyle \frac{-a}{b}$
0%
$\displaystyle \frac{-b}{a}$

Explanation

We will equate the given binomial with zero to get the zeros,

$ax + b = 0$

$\Rightarrow ax = - b$

$\Rightarrow x= \dfrac{-b}{a}$ .

Therefore, option

$D$ is correct.

What is the value of

$\displaystyle { ax }^{ 2 }+bx+c$ at

$\displaystyle x=\frac { -b }{ a }$ ?

0%
$a$
0%
$\displaystyle { b }^{ 2 }-4ac$
0%
$c$
0%
$0$

Explanation

Given expression is

$ax^2+bx+c$

For required value put

$x=-\cfrac{b}{a}$ in the given expression,

$\therefore$ Required Value

$=\displaystyle a\left(\dfrac { -b }{ a }\right) ^{ 2 }+b\left(\frac { -b }{ a } \right)+c$

$\displaystyle =\dfrac { a\times { b }^{ 2 } }{ { a }^{ 2 } } +\dfrac { b\times -b }{ a } +c$

$=\dfrac { { b }^{ 2 } }{ a } -\dfrac { { b }^{ 2 } }{ a } +c$

$=c$ .

Which of the number

$1, \ -1$ and

$-3$ are zeroes of the polynomial

$2x^4+9x^3+11x^2+4x-6$ ?

0%
1
0%
-1
0%
-3
0%
None of these

Explanation

Let

$f(x)=2x^4+9x^3+11x^2+4x-6$ .

$f(1)=2(1)^4+9(1)^3+11(1)^2+4(1)-6$

$=2+9+11+4-6=20\neq 0$ .

$\therefore$

$1$ is not a zero of the polynomial

$f(x)$ .

Again

$f(-1)=2(-1)^4+9(-1)^3+11(-1)^2+4(-1)-6$

$=2-9+11-4-6=-6\neq 0$ 0

$\therefore$

$-1$ is not a zero of the polynomial

$f(x)$ .

Also

$f(-3)=2(-3)^4+9(-3)^3+11(-3)^2+4(-3)-6$

$=162-243+99-12-6=0$ 0

$\therefore$

$-3$ is a zero of the polynomial

$f(x)$ .

Thus

$1$ and

$-1$ are not zeros of

$f(x)$ , whereas

$-3$ is zero of

$f(x)$ .
Therefore, option

$C$ is correct.

$p(x)=x^2-2\sqrt 2x+1$ , then

$p(2\sqrt 2)$ is equal to :

0%
$0$
0%
$1$
0%
$4\sqrt 2$
0%
$8\sqrt 2+1$

Explanation

Given,

$p(x)=x^2-2\sqrt 2x+1$

$\therefore p(2\sqrt2)=(2\sqrt2)^2-2\sqrt 2(2\sqrt2)+1$

$=8-8+1=1$
Option B is correct.

State true or false:

A polynomial cannot have more than one zero.

0%
True
0%
False

Explanation

The given statement is false,"a polynomial can have more than one zero".
E.g.

$\space p(x)=x^2-5x+6$ has two zeros.
As,

$x^2-5x+6=0\Rightarrow x^2-2x-3x+6=0$

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

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

$\Rightarrow x=2,3$

That is the given statement is wrong.
Therefore, option

$B$ is correct.

Which out of the following options is a trinomial, having degree 7?

0%
$x^7 + 6x - 8$
0%
$5x^2 + 12x^{-7} + x$
0%
$y^3 - 4x^2 + 12xy$
0%
$y^7 - y^6 + 5y^4 + 6 - 8y^5 + \sqrt y$

Explanation

The polynomial

$x^7+6x-8$ has three terms. The first one is

$x^7$ , the second is

$6x$ , and the third is

$-8$ .

The exponent of the first term is

$7$ .

The exponent of the second term is

$1$ because

$6x = 6x^1$ .

The exponent of the third term is

$0$ because

$-8 = -8x^0$ .

Since the highest exponent is

$7$ , therefore, the degree of

$x^7+6x-8$ is

$7$ .

The degree of trinomial

$\displaystyle ax^{5}-bx^{4}+c$ is

0%
$9$
0%
$5$
0%
$4$
0%
$1$

Explanation

The polynomial

$ax^5-bx^4+c$ has three terms. The first one is

$ax^5$ , the second is

$-bx$ , and the third is

$c$ .

The exponent of the first term is

$5$ .

The exponent of the second term is

$1$ because

$-bx = -bx^1$ .

The exponent of the third term is

$0$ because

$c = cx^0$ .

Since the highest exponent is

$5$ , therefore, the degree of

$ax^5-bx^4+c$ is

$5$ .

The value of

$25x^2\, +\, 16y^2\, +\, 40xy$ at

$x = 1$ and

$y = -1$ is :

0%
$81$
0%
$- 49$
0%
$1$
0%
None of these

Explanation

Substitute

$x=1$ and

$y=-1$ in the given polynomial.

$\therefore 25x^2 + 16y^2 + 40xy = 25(1)^2 + 16(-1)^2 + 40(1)(-1)$

$25(1) + 16(1) + 40(-1)$

$25 + 16 - 40$

$1$

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$ =______

0%
$\displaystyle \left ( 2a+2b \right )$
0%
$\displaystyle \left ( 2a-2b \right )$
0%
$\displaystyle 4ab$
0%
$\displaystyle -4ab$

Explanation

Given,

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$ .

We know,

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

Then,

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$
=

$\displaystyle \left \{ (a+b)+\left ( b-a \right ) \right \}\left \{ (a+b)-\left ( b-a \right ) \right \}$

$=(a+b+b-a)(a+b-b+a)$

$= 2b \times 2a = 4ab$ .

Therefore, option

$C$ is correct.

The value of

$(501)^2\, -\, (500)^2$ is :

0%
$1$
0%
$101$
0%
$1,001$
0%
None of these

Explanation

Given,

$(501)^{2}- (500)^{2}$ .

We know,

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

Then,

$(501)^{2}- (500)^{2}$

$=(501+500)(501-500)$

$=(1001)(1)$

$= 1001$ .

Therefore, option

$C$ is correct.

Simplify:

$\displaystyle \left ( p-q \right )^{2}+4pq$ .

0%
$\displaystyle p^{2}-q^{2}$
0%
$\displaystyle \left ( p+q \right )^{2}$
0%
$\displaystyle \left ( 2p-q \right )^{2}$
0%
$\displaystyle \left ( 2p-2q \right )^{2}$

Explanation

Given,

$(p-q)^2+4pq$ .

We know,

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

and

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

Therefore,

$(p-q)^2+4pq$

$=p^2-2pq+q^2$ + 4

$pq$ [Since,

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

$=p^2 + 2pq +q^2$

$=(p+q)^2$ . [Since,

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

Therefore option

$B$ is correct.

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$ is equal to:

0%
$\displaystyle x^{2}-64$
0%
$\displaystyle x^{4}-64$
0%
$\displaystyle x^{4}-256$
0%
$\displaystyle x^{2}-256$

Explanation

Given,

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$ .

We know,

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

Then,

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$
=

$\displaystyle \left ( x^{2}-4^{2} \right )\left ( x^{2}+16 \right )$
=

$\displaystyle \left ( x^{2}-16 \right )\left ( x^{2}+16 \right )=\left ( x^{2} \right )^{2}-\left ( 16 \right )^{2}$
=

$\displaystyle x^{4}-256$ .

Therefore, option

$C$ is correct.

$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )$ =_____

0%
$\displaystyle m^{2}x^{2}+2mxny-n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}-2mxny-n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}+2mxny+n^{2}y^{2}$

Explanation

$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )=\left ( mx-ny \right )^{2}$
=

$\displaystyle \left ( mx \right )^{2}-2mx\times ny+\left ( ny \right )^{2}$
=

$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$

$a^{2} + 4a + 4$ =

0%
$(a + 2)^{2}$
0%
$(a + 1)^{2}$
0%
$(a - 2)^{2}$
0%
$(a - 1)^{2}$

Explanation

$a^{2} + 4a + 4 = a^{2} + 2(a) (2)+ (2)^{2}$

$=(a + 2)^{2}$

Find the missing term in the following problem:

$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, =\, \displaystyle \frac{9x^2}{16}\, +\, ..........\, +\, \displaystyle \frac{16y^2}{9}$ .

0%
$2xy$
0%
$- 2xy$
0%
$12xy$
0%
$- 12xy$

Explanation

Given, $\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\,$ .

We know, $(a-b)^2=a^2-2ab+b^2$ .

Then,

$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\,$

$=\left( \displaystyle \frac{3x}{4} \right )^2\, -\, 2\, \left(\displaystyle \frac{3x}{4} \right )\left(\displaystyle \frac{4y}{3} \right )\, +\, \left(\displaystyle \frac{4y}{3}\right)^2$

$=\, \displaystyle \frac{9x^2}{16}\, -\, 2xy\, +\, \displaystyle \frac{16y^2}{9}$

$=\, \displaystyle \frac{9x^2}{16}\, +\, (-\, 2xy)\, +\, \displaystyle \frac{16y^2}{9}$ .

Hence, the missing term is $-2xy$ .

Therefore, option $B$ is correct.

Simplify:

$(7m -8n)^2 + (7m + 8n)^2$ .

0%
$49m^2+164n^2$
0%
$98m^2+128n^2$
0%
$49m^2+64n^2$
0%
$128m^2+64n^2$

Explanation

Given,

$(7m -8n)^2 + (7m + 8n)^2$ .

We know,

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

and

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

Then,

$(7m -8n)^2 + (7m + 8n)^2$

$= ((7m)^2-2(7m)(8n) + (8n)^2)+((7m)^2 + 2(7m)(8n) + (8n)^2)$

$= (49m^2-112mn + 64n^2)+(49m^2 + 112mn + 64n^2)$

$= 49m^2-112mn + 64n^2+49m^2 + 112m + 64n^2$

$= 98m^2 + 128n^2$ .

Therefore, option

$B$ is correct.

Simplify:

$3x(4x -5) + 3$ and find its value for

$x =$

$\displaystyle \frac{1}{2}$

0%
$-\dfrac{3}{2}$
0%
$-\dfrac{1}{2}$
0%
$-\dfrac{3}{7}$
0%
$-\dfrac{3}{8}$

Explanation

We have,

$3x(4x -5)+ 3 =3x \times 4x-3x \times 5+3$

$=12x^2-15x + 3$
When

$x=\displaystyle \frac{1}{2}$ , then

$3x (4x-5) +3$

$=\displaystyle 3 \times \frac{1}{2} \left( 4 \times \frac{1}{2} - 5\right) + 3$

$= \displaystyle \frac{3}{2} \times -3 + 3 = -\frac{9}{2} + 3 = -\frac{3}{2}$

The value of the polynomial

$x^2+5$ , at

$x=3$ is

0%
$12$
0%
$13$
0%
$14$
0%
$15$

Explanation

Let

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

Therefore, at

$x=3$ ,

$f(x) = f(3)$ .

Hence,

$f(3) = 3^2+5 = 9+5 = 14$

The value of

$(x + 3y)^2 + (x - 3y)^2$ is:

0%
$2x^2 + 18 y^2$
0%
$2x^2 - 18 y^2$
0%
$2x^2 + 18y^2 - 2xy$
0%
None of these

Explanation

We know,

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

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

Then,

$(x + 3y)^2 + (x - 3y)^2=[(x)^2+(3y)^2+2\times x\times3y]+[(x)^2+(3y)^2-2\times x\times3y]$

$=x^2+9y^2+6xy+x^2+9y^2-6xy$

$=2x^2+18y^2$ .

Hence, Option

$A$ is correct.

$f(x) = -3x - 5$ , then the value of

$f(2)$ is

0%
$-11$
0%
$-1$
0%
$1$
0%
$11$
0%
$5$

Explanation

Given,

$f(x)=-3x-5$

Then

$F(2)=-3(2)-5$

$=-6-5$

$=-11$

$\displaystyle x = \frac{4}{3}$ is a root of the polynomial

$f(x)= 6x^3-11x^2+ kx-20$ , then find the value of

$k$ .

0%
$17$
0%
$19$
0%
$22$
0%
$23$

Explanation

$f(x) = 6x^3-11x^2+ kx -20$

$\displaystyle f \left ( \frac{4}{3} \right ) = 6 \left ( \frac{4}{3} \right )^3 - 11 \left ( \frac{4}{3} \right )^2 + k \left ( \frac{4}{3} \right ) - 20 = 0$

$\Rightarrow \displaystyle 6 \left ( \frac{64}{27}\right) - 11 \left ( \frac{16}{9} \right ) + \frac{4k}{3} - 20 = 0$

$\Rightarrow 128-176+12k -180 = 0$

$\Rightarrow 12k+128 -356=0$

$\Rightarrow 12k = 228$

$\Rightarrow k =19.$

Identify the degree of the given equation:

$\displaystyle { x }^{ 2 }+3x-5={ x }^{ 2 }+9x-23$

0%
Zero
0%
One
0%
Two
0%
Three

Explanation

$x^2+3x-5=x^2+9x-23$

We need to bring the given equation to it standard form.

$\therefore -5+23=9x-3x$ .....Transpose

$\therefore 18 = 6x$

$\therefore x=3$

Here, the highest power of variable

$x$ is

$1$ .

Hence, the degree of the given equation is

$1$ .

Using

$(x + a) (x + b) = x^2+(a+b)x+ab$ , find

$103 \times 104$ .

0%
$14982$
0%
$11992$
0%
$10712$
0%
$11482$

Explanation

Given,

$103 \times 104$

$=(100+3)(100+4)$ .

Using,

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

Then,

$103 \times 104$

$=(100+3)(100+4)$

$=(100)^2+(3+4)(100) +(3)(4)$

$=(100)^2+(7)(100) +12$

$= 10000 +700 + 12$

$= 10712$ .

Therefore, option

$C$ is correct.

CBSE Questions for Class 9 Maths Polynomials Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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