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Maths 2 Marks Questions

Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 39 questions.

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Question 12 Marks
Verify : x3 + y3 = (x + y)(x2 - xy + y2)
Answer
We know that
(x + y)3 = x3 + y3 + 3xy(x + y) {Using Identity (a + b)3= a3+ b3+ 3ab (a + b)}
⇒ x3 + y3 = (x + y)3 - 3xy(x + y)
⇒ x3 + y3 = (x + y){(x + y)2 - 3xy}
⇒ x3 + y3 = (x + y)(x2 + 2xy + y2 - 3xy) {Using Identity (a + b)2= a2+ 2ab + b2 }
⇒ x3 + y3 = (x + y)(x2 - xy + y2)
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Question 22 Marks
Verify : x3 - y3 = (x - y)(x2 + xy + y2)
Answer
We know that
(x - y)3 = x3 - y3 - 3xy(x - y) {Using Identity (a – b)3= a3– b3– 3ab (a – b)}
⇒ x3 - y3 = (x - y)3 + 3xy(x - y)
⇒ x3 - y3 = (x - y){(x - y)2 + 3xy}
⇒ x3 - y3 = (x - y)(x2 - 2xy + y2 + 3xy)
⇒ x3 - y3 = (x - y)(x2 + xy+ y2)
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Question 32 Marks
Factorise : 27p3 - $\frac{1}{216}$ - $\frac{9}{2}$p2 + $\frac{1}{4}$p.
Answer
27p3 - $\frac{1}{216}$ - $\frac{9}{2}$p2 + $\frac{1}{4}$p
= $(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)\left(\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
 = $\left(3 p-\frac{1}{6}\right)^{3}$
(Using Identity (a – b)3= a3– b3– 3ab (a – b))
= $\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
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Question 42 Marks
Factorise : 8a3 + b3 + 12a2b + 6ab2
Answer
8a3 + b3 + 12a2b + 6ab2
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2a+ b)3
(Using Identity (a + b)3= a3+ b3+ 3ab (a + b))
= (2a + b)(2a + b)(2a + b)
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Question 52 Marks
Evaluate the using suitable identity: (998)3
Answer
(998)3
= (1000 - 2)3 = (1000)3 - (2)2 - 3(1000)(2)(1000 - 2)
(Using Identity (a – b)3 = a3 – b3 – 3ab (a – b))
= 1000000000 - 8 - 6000(1000 - 2)
= 1000000000 - 8 - 6000000 + 12000
= 994011992
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Question 62 Marks
Using suitable identities Evaluate : (102)3
Answer
(102)3
= (100 + 2)3 = (100)3 + (2)3 + 3(100)(2)(100 + 2)
(Using Identity (a + b)3 = a3 + b3 + 3ab (a + b))
= 1000000 + 8 + 600(100 + 2) = 1000000 + 8 + 60000 + 1200
= 1061208
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Question 72 Marks
Write the cube in expanded form : $\left(x-\frac{2}{3} y\right)^{3}$
Answer
$\left(x-\frac{2}{3} y\right)^{3}$ = $x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
(Using Identity  (a – b)3 = a3 – b3 – 3ab (a – b))
= $x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$ = $x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$
= x3 - 2x2y + $\frac{4}{3}$xy2 - $\frac{8}{27}$y3
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Question 82 Marks
Factorize:
$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
Answer
$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
We need to factorize the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
The expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$  can also be written as
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right).$
We can observe that, we can apply the identity ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$  with respect to the expression
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right)$,to get
${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$
Therefore, we conclude that after factorizing the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $ we get ${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$.
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Question 92 Marks
Evaluate the product without multiplying directly: $104 \times 96$
Answer
$104 \times 96$
$104 \times 96{\text{ can also be written as}}\left( {100 + 4} \right)\left( {100 - 4} \right).$ We can observe that, we can apply the identity $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$ with respect to the expression $\left( {100 + 4} \right)\left( {100 - 4} \right)$,to get
$\left( {100 + 4} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} - {\left( 4 \right)^2}$ $= 10000 - 16$
= 9984
Therefore, we conclude that the value of the product $104 \times 96 $ is 9984
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Question 102 Marks
Evaluate the product without multiplying directly: $95 \times 96$
Answer
$95 \times 96$
$95 \times 96{\text{ can also be written as}}\left( {100 - 5} \right)\left( {100 - 4} \right)$ We can observe that we can apply the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$

Here a = -5 and b = -4
$\left( {100 - 5} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 4} \right)} \right]\left( {100} \right) + \left( { - 5} \right) \times \left( { - 4} \right)$
 $= 10000 - 900 + 20$
= 9120 Therefore, we conclude that the value of the product $95 \times 96$ is 9120

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Question 112 Marks
What is the possible expression for the dimension of the cuboid whose volume is $12k{y^2} + 8ky - 20k$
Answer
${\text{Volume : 12}}k{y^2} + 8ky - 20k$
The expression ${\text{12}}k{y^2} + 8ky - 20k $ can also be written as $ k\left( {{\text{12}}{y^2} + 8y - 20} \right).$
$k\left( {{\text{12}}{y^2} + 8y - 20} \right) = k\left( {{\text{12}}{y^2} - 12y + 20y - 20} \right)$ $\, = k\left[ {12y\left( {y - 1} \right) + 20\left( {y - 1} \right)} \right]$
$ = k\left( {12y + 20} \right)\left( {y - 1} \right)$
$= 4k \times \left( {3y + 5} \right) \times \left( {y - 1} \right).$
Therefore, we can conclude that a possible expression for the dimension of a cuboid of volume ${\text{12}}k{y^2} + 8ky - 20k \ is \ 4k\ ,\left( {3y + 5} \right){\text{ and }}\left( {y - 1} \right)$
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Question 122 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $35{y^2} + 13y - 12$
Answer
${\text{Area : }}35{y^2} + 13y - 12$
The expression $35{y^2} + 13y - 12$ can also be written as, 35y2 + 28y - 15y - 12
= $35{y^2} + 28y - 15y - 12 = 7y\left( {5y + 4} \right) - 3\left( {5y + 4} \right)$
$= \left( {7y - 3} \right)\left( {5y + 4} \right)$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $35{y^2} + 13y - 12$ is Length = 7y - 3 and Breadth = 5y + 4
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Question 132 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $25{a^2} - 35a + 12$
Answer
${\text{Area : }}25{a^2} - 35a + 12$
The expression 25a2 - 35a + 12 can also written as 25a2 - 15a - 20a + 12
$25{a^2} - 15a - 20a + 12 = 5a\left( {5a - 3} \right) - 4\left( {5a - 3} \right)$
$\, = \left( {5a - 4} \right)\left( {5a - 3} \right).$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $25{a^2} - 35a + 12$ is ${\text{Length}} = \left( {5a - 4} \right){\text{ and Breadth}} = \left( {5a - 3} \right)$
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Question 142 Marks
Use suitable identity to find the product: (3 - 2x)(3 + 2x)
Answer
(3 - 2x)(3 + 2x) = (3)2 - (2x)2
{Using Identity a2- b2= (a- b)(a+ b)}
= 9 - 4x2
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Question 152 Marks
Without actually calculating the cube, find the value of (-12)3 + (7)3 + (5)3.
Answer
a3 + b3 + c3 = 3abc
If a + b + c = 0
Given: a = -12, b = 7, c = 5
And a + b + c = -12 + 7 + 5 = 0
Then,
(-12)3 + (7)3 + (5)3 = $ 3 \times - 12 \times 7 \times 5$
= -1260
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Question 162 Marks
Use suitable identity to find the product: (x + 8)(x - 10)
Answer
(x + 8)(x - 10)
= (x + 8){x + (-10)}
(Using Identity  (x + a)(x + b) = x2 + (a + b) x + ab)
= x2 + {8 + (-10)}x + (8)(-10)
= x2 - 2x - 80
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Question 172 Marks
Factorise :27y3 + 125z3
Answer
27y3 + 125z3
= (3y)3 + (5z)3 = (3y + 5z){(3y)2 - (3y)(5z) + (5z)2}
= (3y + 5z)(9y2 - 15yz + 25z2)
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Question 182 Marks
Find the remainder when x3 - ax2 + 6x - a  is divided by x - a.
Answer
Let p(x) = x3 - ax2 + 6x - a
x - a = 0
⇒ x = a
$\therefore$ Remainder = (a)3 - a(a)2 + 6(a) - a
= a3 - a3 + 6a - a
= 5a
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Question 192 Marks
Find the remainder when $ {x^3} + 3{x^2} + 3x + 1$ is divided by $x - \frac{1}{2}$
Answer
$x - \frac{1}{2}$
We need to find the zero of the polynomial $x - \frac{1}{2}$
$\begin{gathered} x - \frac{1}{2} = 0{\text{ }} \hfill \\ \Rightarrow {\text{ }}x = \frac{1}{2} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $x - \frac{1}{2}$ in the polynomial ${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) + 1$
$ = \frac{1}{8} + 3\left( {\frac{1}{4}} \right) + \frac{3}{2} + 1$

= ${1 \over 8} + {3 \over 4} + {3 \over 2} + 1$
$= \frac{{1 + 6 + 12 + 8}}{8}$
$\, = \frac{{27}}{8}$
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1 by \ x - \frac{1}{2}$ we will get the remainder as $\frac{{27}}{8}$

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Question 202 Marks
Verify $x = - \frac{1}{2}$ are zeroes of the polynomial $p\left( x \right) = 2x + 1$
Answer
$p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$
We need to check whether $p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$ is equal to zero or not, i.e., $p\left( {{-1 \over 2}} \right)$ is equal to 0 or not.
$p\left( { - \frac{1}{2}} \right) = 2\left( { - \frac{1}{2}} \right) + 1\,\, = - 1 + 1\,\, = 0$

Since $p\left( {{-1 \over 2}} \right)$ = 0
Therefore, $x = - \frac{1}{2}$is a zero of the polynomial $p\left( x \right) = 2x + 1$

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Question 212 Marks
Verify $x = - \frac{m}{l}$ are zeroes of the polynomial $p\left( x \right) = lx + m$
Answer
$p\left( x \right) = lx + m,\, x = - \frac{m}{l}$

We need to check whether $p\left( x \right) = lx + m{\text{ at }}x = - \frac{m}{l}$ is equal to zero or not, i.e., $p\left( {{{ - m} \over l}} \right)$ is equal to zero or not.

$p\left( { - \frac{m}{l}} \right) = l\left( { - \frac{m}{l}} \right) + m\,\, = -m + m\,\, = 0$

Therefore, $x = - \frac{m}{l}$ is a zero of the polynomial $p\left( x \right) = lx + m$

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Question 222 Marks
Verify $x = 0$ are zeroes of the polynomial $p\left( x \right) = {x^2}$
Answer
$p\left( x \right) = {x^2},\,\,\,\,x = 0$
We need to check whether $p\left( x \right) = {x^2}{\text{ at }}x = 0$ is equal to zero or not, i.e., $p\left( 0 \right)$ is equal to zero or not.
$p\left( 0 \right) = {\left( 0 \right)^2}\, = 0$
Therefore, we can conclude that $x = 0$ is a zero of the polynomial $p\left( x \right) = {x^2}$
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Question 232 Marks
Verify $x = - 1,2$ are zeroes of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$
Answer
$p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right),\, x = - 1,2$

We need to check whether $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right){\text{ at }}x = - 1,2$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( 2 \right)$ is equal to zero or not.

At $x = - 1$, $p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)\,\, = \left( 0 \right)\left( { - 3} \right)\,\, = 0$

At $x = 2$, $p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right)\,\, = \left( 3 \right)\left( 0 \right)\,\, = 0$

Therefore, $x = - 1,2$ are the zeros of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$

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Question 242 Marks
Verify $x = - 1,1$ are zeroes of the polynomial $p\left( x \right) = {x^2} - 1$
Answer
$p\left( x \right) = {x^2} - 1,\, x = - 1,1$
We need to check whether $p\left( x \right) = {x^2} - 1{\text{ at }}x = - 1,1$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( { 1} \right)$ is equal to zero or not.
At $x = - 1$
$p\left( { - 1} \right) = {\left( { - 1} \right)^2} - 1\,\, = 1 - 1\,\, = 0$
At $x = 1$
$p\left( 1 \right) = {\left( 1 \right)^2} - 1\,\, = 1 - 1\,\, = 0$
Therefore ,$x = - 1,1$ are the zeros of the polynomial $p\left( x \right) = {x^2} - 1$
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Question 252 Marks
Verify $x = \frac{4}{5}$ are zeroes of the polynomial $p\left( x \right) = 5x - \pi$
Answer
$p\left( x \right) = 5x - \pi ,\, x = \frac{4}{5}$
We need to check whether $p\left( x \right) = 5x - \pi {\text{ at }}x = \frac{4}{5}$ is equal to zero or not, i.e., $p\left( {{4 \over 5}} \right)$ is equal to zero or not.
$p\left( {\frac{4}{5}} \right) = 5\left( {\frac{4}{5}} \right) - \pi \,\, = 4 - \pi$
Therefore, $x = \frac{4}{5}$ is not a zero of the polynomial $ p\left( x \right) = 5x - \pi $
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Question 262 Marks
Verify $x = - \frac{1}{3}$ are zeroes of the polynomial $p\left( x \right) = 3x + 1$
Answer
$p\left( x \right) = 3x + 1, x = - \frac{1}{3}$
We need to check whether $p\left( x \right) = 3x + 1{\text{ at }}x = - \frac{1}{3}$ is equal to zero or not.
$p\left( { - \frac{1}{3}} \right) = 3x + 1 = 3\left( { - \frac{1}{3}} \right) + 1\, = - 1 + 1\, = 0$
Therefore, we can conclude that $x = - \frac{1}{3}$ is a zero of the polynomial $p\left( x \right) = 3x + 1$
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Question 272 Marks
Find p(0), p(1) and p(2) for the polynomial: p(x) = (x – 1)(x + 1)
Answer
p(x) = (x – 1)(x + 1)
$\therefore$ p(0) = (0 – 1)(0 + 1) = (–1)(1) = –1
p(1) = (1 – 1)(1 + 1) = (0)(2) = 0,
p(2) = (2 – 1)(2 + 1) = (1)(3) = 3
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Question 282 Marks
Find p(0), p(1) and p(2) of the polynomial: $p ( t ) = 2 + t + 2 t ^ { 2 } - t ^ { 3 }$
Answer
According to the question, 

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

$p ( 0 ) = 2 + ( 0 ) + 2 ( 0 ) ^ { 2 } - ( 0 ) ^ { 3 } = 2$

$p ( 1 ) = 2 + ( 1 ) + 2 ( 1 ) ^ { 2 } - ( 1 ) ^ { 3 } = 2 + 1 + 2 - 1 = 4$

$p ( 2 ) = 2 + ( 2 ) + 2 ( 2 ) ^ { 2 } - ( 2 ) ^ { 3 } = 4 + 8 - 8 = 4$

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Question 292 Marks
Factorise y2 – 5y + 6 by using the Factor Theorem.
Answer
Let p(y) = y2 – 5y + 6. 
The factors of 6 are 1, 2 and 3.
Now, p(2) = 22 – (5 $\times$ 2) + 6 = 0
So, y – 2 is a factor of p(y).
Also, p(3) = 32 – (5 $\times$ 3) + 6 = 0
So, y – 3 is also a factor of y2 – 5y + 6
Therefore, y2 – 5y + 6 = (y – 2)(y – 3)

This is the required factorisation.

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Question 302 Marks
Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem.
Answer
If we can find two numbers p and q such that p + q = 17 and pq = 6 $\times$ 5 = 30, then we can get the factors
So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.
So, 6x2 + 17x + 5
= 6x2 + (2 + 15)x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)

This is the required factorisation.

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Question 312 Marks
Find the value of k, if x – 1 is a factor of 4x3 + 3x2 – 4x + k.
Answer
As x – 1 is a factor of p(x) = 4x3 + 3x2 – 4x + k, therefore,

p(1) = 0
Now, p(1) = 4(1)3 + 3(1)2 – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3

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Question 322 Marks
Verify whether 2 and 0 are zeroes of the polynomial x2 – 2x.
Answer
Let p(x) = x2 – 2x
Then p(2) = 22 – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x2 – 2x.
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Question 332 Marks
Factorise : 8x3 + y3 + 27z3 – 18xyz
Answer
Here, we have
8x3 + y3 + 27z3 - 18xyz
= (2x)3 + y3 + (3z)3 – 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)2 + y2 + (3z)2 – (2x)(y) – (y)(3z) – (2x)(3z)]
= (2x + y + 3z) (4x2 + y2 + 9z2 - 2xy - 3yz - 6xz)
This is the required factorisation.
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Question 342 Marks
Factorise: 8x3 + 27y3 + 36x2y + 54xy2
Answer
The given expression can be written as
(2x)3 + (3y)3 + 3(4x2)(3y) + 3(2x)(9y2)
= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2
= (2x + 3y)3 (Using Identity (x + y)3 = x3 + y3 + 3xy (x + y))
= (2x + 3y)(2x + 3y)(2x + 3y)
This is the required factorisation.
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Question 352 Marks
Write (5p – 3q)3 in the expanded form.
Answer
Comparing the given expression with (x – y)3,
we find that x = 5p, y = 3q.
So, using Identity (x - y)3 = x3 – 3x2y + 3xy2 – y3,
we have: (5p – 3q)3 = (5p)3 – (3q)3 – 3(5p)(3q)(5p – 3q)
= 125p3 – 27q3 – 225p2q + 135pq2
This is the required expansion.
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Question 362 Marks
Write (3a + 4b)3 in the expanded form.
Answer
Comparing the given expression with (x + y)3 ,
we find that x = 3a and y = 4b.
So, using Identity (x + y)3 = x3 + y3 + 3xy (x + y),
we have: (3a + 4b)3 = (3a)3 + (4b)3 + 3(3a)(4b)(3a + 4b)
= 27a3 + 64b3 + 108a2b + 144ab2​​​​​​
This is the required expansion.
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Question 372 Marks
Factorise : 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
Answer
We have,
4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)
= [2x + (–y) + z]2
= (2x – y + z)2
= (2x – y + z)(2x – y + z)
This is the required factorisation.
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Question 382 Marks
Factorise : 49a2 + 70ab + 25b2
Answer
Here you can see that 49a2 = (7a)2, 25b2 = (5b)2, 70ab = 2(7a) (5b)
Comparing with x2 + 2xy + y2, we observe that x = 7a and y = 5b.
Using Identity (x + y)2 = x2 + 2xy + y, we get
49a2 + 70ab + 25b2 = (7a + 5b)2 = (7a + 5b) (7a + 5b)
This is the required factorisation.
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Question 392 Marks
Evaluate 105 $\times$ 106 without multiplying directly.
Answer
Firstly, write 105 as 100 + 5 and 106 as100 + 6.
$\therefore$  105 $\times$ 106 = (100 + 5) $\times$ (100 + 6)
= (100)2 + (5 + 6)100 + (5 $\times$ 6)
[$\because$ (x + a)(x + b) = x2 + (a + b)x + ab]
= 10000+ 11$\times$100 + 30
= 10000 + 1100 + 30 = 11130
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2 Marks Questions - Maths STD 9 Questions - Vidyadip