MCQ Questions for Class 9 Maths Polynomials Quiz 6

State whether the statement is True or False.

Evaluate:

$(a+bc)(a-bc)(a^2+b^2c^2)$ is equal to

$a^4-b^4c^4$ .

0%
True
0%
False

Explanation

Given,

$(a+bc)(a-bc)(a^2+b^2c^2)$ .

We know,

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

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

Multiply by using direct method, we get,

$(a+bc)(a-bc)(a^2+b^2c^2)$

$= [a^2-(bc)^2](a^2+b^2c^2)$

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

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

Now, by using formula

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

$= ((a^2)^2-(b^2c^2)^2)$

$= a^4-b^4c^4$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

On evaluating

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$ we get,

$49x^2-\dfrac{4}{9}y^2$ .

0%
True
0%
False

Explanation

Given,

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$ .

We know,

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

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

Then, on multiplying, we get,

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$

$= (7x)^2-(\dfrac{2}{3}y)^2$

$= 49x^2-\dfrac{4}{9}y^2$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

$(2a+3)(2a-3)(4a^2+9)$ is equal to

$16a^4-81$ .

0%
True
0%
False

Explanation

Given,

$(2a+3)(2a-3)(4a^2+9)$ .

We know,

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

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

Multiply by using direct method, we get,

$(2a+3)(2a-3)(4a^2+9)$

$= [(2a)^2-(3)^2](4a^2+9)$

$= (4a^2-9)(4a^2+9)$

$= (4a^2+9)(4a^2-9)$ .

Now, by using formula

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

$= ((4a^2)^2-(9)^2)$

$= 16a^4-81$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

Expanding

$\left(3x-\dfrac{1}{3x}\right)^2$ , we get

$9x^2-2+\dfrac{1}{9x^2}$ .

0%
True
0%
False

Explanation

Given,

$(3x-\dfrac{1}{3x})^2$ .

We know,

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

Then,

$(3x-\dfrac { 1 }{ 3x } )^2\\ =\quad { (3x )}^{ 2 }-2(3x)(\dfrac{1}{3x})+(\dfrac { 1 }{ { 3x }} )^2 \\ =\quad 9{ x }^{ 2 }-2+\dfrac { 1 }{ { 9x }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

State whether the statement is True or False:

$(a+\dfrac{1}{2a})^2$ is equal to

$a^2+1+\dfrac{1}{4a^2}$ .

0%
True
0%
False

Explanation

Given,

$(a+\dfrac{1}{2a})^2$ .

We know,

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

Then,

$(a+\dfrac { 1 }{ 2a } )^2\\ =\quad { a }^{ 2 }+2(a)(\dfrac{1}{2a})+(\dfrac { 1 }{ 2{ a }} )^2 \\ =\quad { a }^{ 2 }+1+\dfrac { 1 }{ 4{ a }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

State whether the statement is True or False:

$(a-2b)^2$ is equal to

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

0%
True
0%
False

Explanation

Given,

$(a-2b)^2$ .

We know,

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

Then,

$(a-2b)^2\\ =\quad a^{ 2 }-2(a)(2b)+(2b)^{ 2 }\\ =\quad a^{ 2 }-4ab+4b^{ 2 }.$

That is the given statement is true.

Hence, option

$A$ is correct.

State whether the statement is True or False:

$(6-xy)(6+xy)$ is equal to

$36-x^2y^2$ .

0%
True
0%
False

Explanation

Given,

$(6-xy)(6+xy)$ .

We know, the identity

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

Here,

$a=6$ and

$b=xy$ .

Then, on applying the identity,

$(6-xy)(6+xy)\\ =(6)^2-(xy)^2=36-x^{ 2 }y^{ 2 }.$

That is, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

$\left(3x-\dfrac{1}{2y}\right)\left(3x+\dfrac{1}{2y}\right)$ is equal to

$9x^2-\dfrac{1}{4y^2}$ .

0%
True
0%
False

Explanation

Given,

$(3x-\dfrac{1}{2y})(3x+\dfrac{1}{2y})$ .

We know,

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

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

Then, on multiplying, we get,

$(3x-\dfrac{1}{2y})(3x+\dfrac{1}{2y})$

$= (3x)^2-(\dfrac{1}{2y})^2$

$= 9x^2-\dfrac{1}{4y^2}$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

$\left(2a-\dfrac{1}{a}\right)^2$ is equal to

$4a^2-4+\dfrac{1}{a^2}$ .

0%
True
0%
False

Explanation

Given,

$(2a-\dfrac{1}{a})^2$ .

We know,

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

Then,

$(2a-\dfrac { 1 }{ a } )^2\\ =\quad { (2a )}^{ 2 }-2(2a)(\dfrac{1}{a})+(\dfrac { 1 }{ { a }} )^2 \\ =\quad 4{ a }^{ 2 }-4+\dfrac { 1 }{ { a }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

State whether the statement is True or False:

$(2a+b)^2$ is equal to

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

0%
True
0%
False

Explanation

Given,

$(2a+b)^2$ .

We know,

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

Then,

$(2a+b)^2\\ =\quad { (2a )}^{ 2 }+2(2a)(b)+(b )^2 \\ =\quad 4{ a }^{ 2 }+4ab+{ { b }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

Simplify:

$\left (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2} \right)$ .

0%
$3$
0%
$5$
0%
$6$
0%
$8$

Explanation

Given,

$(\sqrt { 5 } +\sqrt { 2 } )(\sqrt { 5 } -\sqrt { 2 } )$ .

We know,

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

Applying the identity,

$(\sqrt { 5 } +\sqrt { 2 } )(\sqrt { 5 } -\sqrt { 2 } )$

$=(\sqrt { 5 })^2- (\sqrt { 2 })^2=5-2=3$ .

Hence,

$(\sqrt { 5 } +\sqrt { 2 } )(\sqrt { 5 } -\sqrt { 2 } )=3$ .

Therefore, option

$A$ is correct.

State whether the statement is True or False:

$(2x-\dfrac{1}{2x})^2$ is equal to

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

0%
True
0%
False

Explanation

Given,

$(2x-\dfrac{1}{2x})^2$ .

We know,

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

Then,

$(2x-\dfrac { 1 }{ 2x } )^2\\ =\quad { (2x )}^{ 2 }-2(2x)(\dfrac{1}{2x})+(\dfrac { 1 }{ { 2x }} )^2 \\ =\quad 4{ x }^{ 2 }-2+\dfrac { 1 }{ { 4x }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

Evaluate using expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(208)^2$

0%
45264
0%
45294
0%
43284
0%
43264

Explanation

${208}^{2} = {(200 + 8)}^{2}$

It is the form of

${(a+b)}^{2}$ , where

$a = 200, b = 8$

Applying

the formula

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

${208}^{2} = {(200 + 8)}^{2} = {200}^{2} + { 8 }^{ 2 } + 2\times 200 \times 8 = 40000 + 64 + 3200 = 43264$

State whether the statement is True or False:

The square of

$(a+\dfrac{1}{5a})$ is

$a^2+1+\dfrac{1}{4a^2}$ .

0%
True
0%
False

Explanation

Given,

$(a+\dfrac{1}{5a})^2$ .

We know,

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

Then,

$(a+\dfrac { 1 }{ 5a } )^2\\ =\quad { a }^{ 2 }+2(a)(\dfrac{1}{5a})+(\dfrac { 1 }{ 5{ a }} )^2 \\ =\quad { a }^{ 2 }+\dfrac{2}{5}+\dfrac { 1 }{ 25{ a }^{ 2 } } .$

Therefore, the statement is false.

Hence, option

$B$ is correct.

State whether the statement is True or False:

$(6-5xy)(6+5xy)$ is equal to

$36-25x^2y^2$ .

0%
True
0%
False

Explanation

Given,

$(6-5xy)(6+5xy)$ .

We know,

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

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

Multiply by using direct method, we get,

$(6-5xy)(6+5xy)$

$= 6^2-(5xy)^2$

$= 36-25x^2y^2$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

$(2x-\dfrac{3}{5})(2x+\dfrac{3}{5})$ is equal to

$4x^2-\dfrac{9}{25}$ .

0%
True
0%
False

Explanation

Given,

$(2x-\dfrac{3}{5})(2x+\dfrac{3}{5})$ .

We know,

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

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

Then, on multiplying, we get,

$(2x-\dfrac{3}{5})(2x+\dfrac{3}{5})$

$= (2x)^2-(\dfrac{3}{5})^2$

$= 4x^2-\dfrac{9}{25}$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

The square of

$(x+3y)$ is equal to

$x^2+6xy+9y^2$ .

0%
True
0%
False

Explanation

Given,

square of

$(x+3y)$ , i.e.

$(x+3y)^2$ .

We know,

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

Then,

$(x+3y)^2\\ =\quad { (x )}^{ 2 }+2(x)(3y)+(3y )^2 \\ =\quad { x }^{ 2 }+6xy+{ { 9y }^{ 2 } } .$

Therefore, the statement is true.

Hence, option

$A$ is correct.

State whether the statement is True or False:

$(4x^2-5y^2)(4x^2+5y^2)$ is equal to

$16x^4-25y^4$ .

0%
True
0%
False

Explanation

Given,

$(4x^2-5y^2)(4x^2+5y^2)$ .

We know,

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

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

Multiply by using direct method, we get,

$(4x^2-5y^2)(4x^2+5y^2)$

$= (4x^2)^2-(5y^2)^2$

$= 16x^4-25y^4$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

The square of

$(\dfrac{5a}{6b}+ \dfrac{6b}{5a} )$ is equal to

$\dfrac{25a^2}{36b^2} -2+\dfrac{36b^2}{25a^2}$ .

0%
True
0%
False

Explanation

Given, square of

$(\dfrac { 5a }{ 6b } +\dfrac { 6b }{ 5a } )$ ,

i.e.

$(\dfrac { 5a }{ 6b } +\dfrac { 6b }{ 5a } )^2$ .

Applying the formula

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

$(\dfrac { 5a }{ 6b } +\dfrac { 6b }{ 5a } )^2$

$=(\dfrac{5a}{6b})^2+2(\dfrac{5a}{6b})(\dfrac{6b}{5a})+(\dfrac{6b}{5a})^2$

$=(\dfrac{25a^2}{36b^2})+2+(\dfrac{36b^2}{25a^2})$

$=\dfrac{25a^2}{36b^2}+2+\dfrac{36b^2}{25a^2}$ .

Hence, the statement is false.

Therefore, option

$B$ is correct.

State whether the statement is True or False.

Evaluate:

$(m+3)(m-3)(m^2+9)$ is equal to

$m^4-81$ .

0%
True
0%
False

Explanation

Given,

$(m+3)(m-3)(m^2+9)$ .

We know,

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

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

Multiply by using direct method, we get,

$(m+3)(m-3)(m^2+9)$

$= [m^2-(3)^2](m^2+9)$

$= (m^2-9)(m^2+9)$

$= (m^2+9)(m^2-9)$ .

Now, by using formula

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

$= ((m^2)^2-(9)^2)$

$= m^4-81$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

The square of

$(3x+\dfrac{2}{y} )$ is equal to

$9x^2+\dfrac{12x}{y}+\dfrac{4}{y^2}$ .

0%
True
0%
False

Explanation

Given, square of

$(3x + \dfrac{2}{y})$ ,

i.e.

$(3x + \dfrac{2}{y}) ^2$ .

Applying the formula

$(a+b)^2 = { a }^{ 2 }+2ab+{ b }^{ 2 }$ , where

$a = 3x , b = \dfrac{2}{y}$ .

$(3x + \dfrac{2}{y}) ^2$

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

$=9x^{2}+12\dfrac{x}{y}+\dfrac{4}{y^2}$

$=9x^{2}+\dfrac{12x}{y}+\dfrac{4}{y^2}$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

The square of

$(8x+\dfrac{3}{2}y )$ is equal to

$64x^2+24xy+\frac{9}{4}y^2$ .

0%
True
0%
False

Explanation

Given, square of

$(8x+\dfrac { 3 }{2 }y)$ ,

i.e.

$(8x+\dfrac { 3 }{2 }y)^2$ .

We know.

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

Then,

$(8x+\dfrac { 3 }{2 }y)^2$

$=(8x)^{ 2 }+2(8x)(\dfrac{3}{2}y)+(\dfrac { 3 }{ 2 }y)^2$

$=64x^{ 2}+24xy+\dfrac { 9 }{ 4 } y^{ 2 }$ .

Here, the statement is true.

Therefore, option

$A$ is correct.

Evaluate:

$203\times 197$

0%
39991
0%
39891
0%
39981
0%
38981

Explanation

$203 \times 197 = (200 + 3) \times (200 - 3)$

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 200 , b = 3$

$203 \times 197 = (200 + 3) \times (200 - 3) = { 200 }^{ 2 }-{ 3 }^{ 2 } = 40,000 - 9 = 39991$

State whether the statement is True or False:

$(1.6x+0.7y)(1.6x-0.7y)$ is equal to

$2.56x^2-0.49y^2$ .

0%
True
0%
False

Explanation

Given,

$(1.6x+0.7y)(1.6x-0.7y)$ .

We know,

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

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

Multiply by using direct method, we get,

$(1.6x+0.7y)(1.6x-0.7y)$

$= (1.6x)^2-(0.7y)^2$

$= 2.56x^2-0.49y^2$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

Evaluate:

$20.8 \times 19.2$

0%
399.86
0%
398.16
0%
399.36
0%
398.56

Explanation

$20.8 \times 19.2 = (20+ 0.8) \times (20 - 0.8)$

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 20 , b = 0.8$

$20.8 \times 19.2 = (20+ 0.8) \times (20 - 0.8) = { 20 }^{ 2 }-{ 0.8 }^{ 2 } = 400 - 0.64 = 399.36$

State whether the statement is True or False:

The square of

$(2m^2-\dfrac{2}{3}n^2 )$ is equal to

$4m^4-\dfrac{8}{3}m^2n^2+\dfrac{4}{9}n^4$ .

0%
True
0%
False

Explanation

Given, square of

$(2m^{ 2 }-\dfrac { 2 }{ 3 } n^{ 2 })$ ,

i.e.

$(2m^{ 2 }-\dfrac { 2 }{ 3 } n^{ 2 })^{ 2 }$ .

We know.

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

Then,

$(2m^{ 2 }-\dfrac { 2 }{ 3 } n^{ 2 })^{ 2 }$

$=(2m^2)^{ 2 }+(\dfrac { 2 }{ 3 }n^2)^2 -2(2m^2)(\dfrac { 2 }{ 3 } n^{ 2 })$

$=4m^{ 4 }+\dfrac { 4 }{ 9 } n^{ 4 }-\dfrac { 8 }{ 3 } m^{ 2 }n^{ 2 }$ .

Here, the statement is true.

Therefore, option

$A$ is correct.

Find the square of:

$391$

0%
154881
0%
153881
0%
152881
0%
151881

Explanation

${391}^{2} = {(400 - 9)}^{2}$

It is the form of ${(a - b)}^{2}$ , where $a = 400, b = 9$

Applying the formula ${ (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$
${391}^{2} = {(400 - 9)}^{2} = {400}^{2} + { 9 }^{ 2 } - 2\times 400 \times 9 = 1,60,000 + 81 - 7200 = 152881$

Find the square of:

$607$

0%
368549
0%
368449
0%
368349
0%
368249

Explanation

$607$ can be written as $600+7$

$\therefore {607}^{2} = {(600 + 7)}^{2}$
It is the form of ${(a+b)}^{2}$ , where $a = 600, b = 7$
Applying the formula ${ (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab$
${607}^{2} = {(600 + 7)}^{2} = {600}^{2} + { 7 }^{ 2 } +2\times 600 \times 7 = 360000 + 49 + 8400 = 368449$

$1$ is zero of the polynomial

$f(x) =a^2x^2 - 3ax +3x - 1$ , then

$a$ is equal to

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

Explanation

$1$ is the zero of the polynomial

$f(x) =a^2x^2 - 3ax +3x - 1$
Then,

$f(x)=0$

$=>a^2(1)^2-3a(1)+3(1)-1=0$

$=>a^2-3a+3-1=0$

$=>a^2-3a+2=0$

$=>a^2-2a-a+2=0$

$=>a(a-2)-1(a-2)=0$

$=>(a-2)(a-1)=0$

$a=2$
Hence, option

$B$ is correct.

$x = 2$ and

$x = 3$ are zeros of the quadratic polynomial

$x^2 + ax + b$ , the values of

$a$ and

$b$ respectively are

0%
$5,- 6$
0%
$- 5, - 6$
0%
$-5, 6$
0%
$5, 5$

Explanation

$2$ and

$3$ are the zeros of the quadratic polynomial then

$\left( x-2 \right) \left( x-3 \right)$ are the factors of the polynomial
Thus,

$\left( x-2 \right) \left( x-3 \right)$ is the polynomial.

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

$=>{ x }^{ 2 }-5x+6$ ...(1)

Now, the given polynomial is

$\Rightarrow{ x }^{ 2 }+ax+b$ ...(2)

Comparing (1) and (2), we get

$a=-5 , b=6$

Hence, option C.

CBSE Questions for Class 9 Maths Polynomials Quiz 6 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 6 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.