NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Class : 9th
Subject : Maths
Chapter : 2
Chapter Name : Polynomials

Exercise 2.1

Q1 Which of the following expressions are polynomials in one variable and which are not? State
reasons for your answer
(i) 4x2 − 3x + 7
(ii) y 2 + √2
(iii) 3√t + t√2
(iv) y + 2y
(v) x2 + y 3 + t

Answer. (i) 4x2 − 3x + 7
Yes, this expression is a polynomial in one variable x.
(ii) y 2 + √2
Yes, this expression is a polynomial in one variable x.
(iii) 3√t + t√2
No It can be observed that the exponent of variable t in term 3√t is 12 which is not a whole number
.Therefore this expression is not a polynomial.
(iv) y + 2y No It can be observed that the exponent of variable t in term 2y which is not a whole
is −1,
number .Therefore this expression is not a polynomial.
(v) x2 + y 3 + t No It can be observed that this expression is a polynomial in 3 variables x,y and and
t .Therefore , this expression is not a polynomial .

Page : 32 , Block Name : Exercise 2.1

Q2 Write the coef�cients of x2 in each of the following:

(i) 2 + x2 + x
(ii) 2 − x2 + x3
(iii) 2 x2 + x
π

(iv) √2x − 1

Answer. (i) 2 + x2 + x
In the above expression the coef�cient of x2 is 1.

(ii) 2 − x2 + x3
In the above expression the coef�cient of x2 is −1

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(iii) 2 x2 + x
π

In the above expression the coef�cient of x2 is 2
π

(iv) √2x − 1 0x2 + √2x − 1 or 0x2 + √2x − 1
In the above expression the coef�cient of x2 is 0

Page : 32 , Block Name : Exercise 2.1

Q3 Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer. Degree of a polynomial is the highest power of the variable in the polynomial.
Binomial has two terms in it. Therefore, binomial of degree 35 can be written as x35 + x34
Monomial has only one term in it. Therefore, monomial of degree 100 can be written as X 100

Page : 32 , Block Name : Exercise 2.1

Q4 Write the degree of each of the following polynomials
(i) 5x3 + 4x2 + 7x
(ii) 4 − y 2
(iii) 5t − √7
(iv) 3

Answer. Degree of a polynomial is the highest power of the variable in the polynomial.
(i) 5x3 + 4x2 + 7x
This is a polynomial in a variable x and the highest power of variable x is 3 . Therefore, the degree
of this polynomial is 3.
(ii) 4 − y 2 This is a polynomial in variable y and the highest power of variable y is 2. Therefore, the
degree of this polynomial is 2 .
(iii) 5t − √7 This is a polynomial in variable t and the highest power of variable t is 1. Therefore,
the degree of this polynomial is 1.
(iv) 3 This is a constant polynomial. Degree of a constant polynomial is always 0.

Page : 32 , Block Name : Exercise 2.1

Q5 Classify the following as linear, quadratic and cubic polynomials:
(i) x2 + x
(ii) x − x3
(iii) y + y 2 + 4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3

Answer. Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and
3 respectively.

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(i) x2 + x is a quadratic polynomial as its degree is 2 .
(ii) x − x3 is a cubic polynomial as its degree is 3 .
(iii) y + y 2 + 4 is a quadratic polynomial as its degree is 1.
(iv) 1 + x is a linear polynomial as its degree is 1.
(v) 3t is a linear polynomial as its degree is 1.
(vi) r2 is a quadratic polynomial as its degree is 2.
(vii) 7x3 s a cubic polynomial as its degree is 3 .

Page : 32 , Block Name : Exercise 2.1

Exercise 2.2

Q1 Find the value of the polynomial 5x − 4x2 + 3 at

(i) x = 0
(ii) x = −1
(iii) x = 2

Answer. (i) p(x) = 5x − 4x2 + 3
p(0) = 5(0) − 4(0)2 + 3
=3

(ii) p(x) = 5x − 4x2 + 3
p(−1) = 5(−1) − 4(−1)2 + 3
= −5 − 4(1) + 3 = −6

(iii) p(x) = 5x − 4x2 + 3
p(2) = 5(2) − 4(2)2 + 3
= 10 − 16 + 3 = −3

Page : 34 , Block Name : Exercise 2.2

Q2 Find p(0), p(1) and p(2) for each of the following polynomials;

(i) p(y) = y 2 − y + 1
(ii) p(t) = 2 + t + 2t2 − t3
(iii) p(x) = x3
(iv) p(x) = (x − 1)(x + 1)

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Answer. (i) p(y) = y 2 − y + 1
p(0) = (0)2 − (0) + 1 = 1
p(1) = (1)2 − (1) + 1 = 1
p(2) = (2)2 − (2) + 1 = 3

(ii) p(t) = 2 + t + 2t2 − t3
p(0) = (0)2 − (0) + 1 = 1
p(1) = (1)2 − (1) + 1 = 1
p(2) = (2)2 − (2) + 1 = 3

(iii) p(x) = x3
p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8

(iv) p(x) = (x − 1)(x + 1)
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

Page : 34 , Block Name : Exercise 2.2

Q3 Verify whether the following are zeroes of the polynomial, indicated against them.
(i) P (x) = 3x + 1, x = − 13
(ii) p(x) = 5x − π, x = 45
(iii) p(x) = x2 − 1, x = 1, −1
(iv) p(x) = (x + 1)(x − 2), x = −1, 2
(v) p(x) = x2 , x = 0
(vi) p(x) = [x + m, x = − l
m

(vii) P (x) = 3x2 − 1, x = − 1 , 2
√3 √3
(viii) p(x) = 2x + 1, x = 12

−1
Answer. (i) If x =3
is a zero of given polynomial p(x) = 3x + 1 should be 0
Here p ( 3 ) = 3 ( 3 ) + 1 = −1 + 1 = 0
−1 −1

−1
Therefore x = 3 is a zero of the given polynomial .

(ii) If x = 45 is a zero of polynomial p(x) = 5x − π then p ( 45 ) should be 0.

Here p ( 45 ) = 5 ( 45 ) − π = 4 − π

As p ( 45 ) ≠ 0
Therefore x = 45 is a zero of given polynomial .

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(iii) If x = 1 and x = −1 are zeroes of polynomial p(x) = x2 − 1 then p(1) and p(−1) should be
0.
Here p(1) = (1)2 − 1 = 0 and p(−1) = (−1)2 − 1 = 0
p(−1) = (−1)2 − 1 = 0

(iv) If x = −1 and x = 2 is a zero of polynomial p(x) = (x + 1)(x − 2) then p(−1)P (2) should
be 0.
Here p(−1) = (−1 + 1)(−1 − 2) = 0(−3) = 0 and p(2) = (2 + 1)(2 − 2) = 3(0) = 0
Therefore x = −1 and x = 2 are zeroes of the given polynomial

(v) If x = 0x = 2 is a zero of polynomial p(x) = x2 then p(0) should be 0.
Here p(0) = (0)2 = 0
Hence x = 0 is a zero of the given polynomial .

is a zero of polynomial p(x) = |x + m then p ( l ) should be 0.
−m −m
(vi) If x = l

Here, p ( ) = l ( l ) + m = −m + m = 0
−m −m
l
m
Therefore x = − l is a zero of the given polynomial .

(vii) If x =
−1
and x = 2 are zeroes of polynomial p(x) = 3x2 − 1 then p ( −1 ) and p ( 2 )
√3 √3 √3 √3
should be 0.
2
Here p ( ) = 3( −1 ) − 1 = 3 ( 13 ) − 1 = 1 − 1 = 0
−1
√3 √3
2
And p ( 2 ) = 3( 2 ) − 1 = 3 ( 43 ) − 1 = 4 − 1 = 3
√3 √3
−1
Hence ,x = is a zero of the given polynomial .However, x = 2 is not a zero of the given
√3 √3
polynomial .

(viii) If x = 12 is a zero of polynomial p(x) = 2x + 1 then p ( 12 ) should be 0.

Here p ( 12 ) = 2 ( 12 ) + 1 = 1 + 1 = 2

As p ( 12 ) ≠ 0
Therefore x = 12 is not a zero of polynomial

Page : 35 , Block Name : Exercise 2.2

Q4 Find the zero of the polynomial in each of the following cases

(i) p(x) = x + 5
(ii) p(x) = x − 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x − 2
(v) p(x) = 3x

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(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c are real numbers.

Answer. (i) Zero of a polynomial is that value of the variable at which the value of the polynomial is
obtained as 0.
p(x) = x + 5
p(x) = 0
x+5=0
x = −5
Therefore for x = 5 the value of the polynomial n is 0 and hence x = 5 is a zero of the given
polynomial .

(ii) p(x) = x − 5
p(x) = 0
x-5=0
x=5
Therefore for x = 5 the value of the polynomial n is 0 and hence x = 5 is a zero of the given
polynomial .

(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = −5
x = − 52
5 −5
Therefore for x = − 2 the value of the polynomial n is 0 and hence x = 2
is a zero of the given
polynomial .

(iv) p(x) = 3x − 2
p(x) = 0
3x − 2 = 0
x = 23
Therefore for x = 23 the value of the polynomial n is 0 and hence x = 23 is a zero of the given
polynomial .

(v) p(x) = 3x
p(x) = 0
ax = 0
x=0
Therefore for x = 0 the value of the polynomial n is 0 and hence x = 0 is a zero of the given
polynomial .

(vi) p(x) = ax
p(x) = 0
ax = 0
x=0
Therefore for x = 0 the value of the polynomial n is 0 and hence x = 0 is a zero of the given
polynomial .

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(vii) p(x) = cx + d
p(x) = 0
cx + d = 0
−d
x= c
−d −d
Therefore for x = c the value of the polynomial n is 0 and hence x = c is a zero of the given
polynomial .

Page : 35 , Block Name : Exercise 2.2

Exercise 2.3

Q1 Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1
(ii) x = 12
(iii) x
(iv) x + π
(v) 5 + 2x

Answer.

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(ii)

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Page : 40 , Block Name : Exercise 2.3

Q2 Find the remainder when x3 + 3x2 + 3x + 1 is divided by x – a.

Answer.

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Page : 40 , Block Name : Exercise 2.3

Q3 Check whether 7 + 3x is a factor of 3x3 + 7x

Answer.

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Page : 40 , Block Name : Exercise 2.3

Exercise 2.4

Q1 Determine which of the following polynomials has (x + 1) a factor :

(i) x3 + x2 + x + 1
(ii) x4 + x2 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 − x2 − (2 + √2)x + √2

Answer. (i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1 then p(−1) must be zero , otherwise
(x + 1) is not of p(x) .
p(x) = x3 + x2 + x + 1
p(−1) = (−1)3 + (−1)2 + (−1) + 1
= −1 + 1 − 1 − 1 = 0

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Hence x + 1 is not a factor of this polynomial .

(ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1 then p(−1) must be zero , otherwise
(x + 1) is not of p(x) .
p(x) = x4 + x3 + x2 + x + 1
p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1
=1−1+1−1+1=1
Therefore x + 1 is not a factor of this polynomial .

(iii) If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1 then p(−1) must be 0
otherwise (x + 1) is not a factor of this polynomial .
p(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1
=1−3+3−1+1=1
As p(−1) ≠ 0
Therefore, x+1 is not a factor of this polynomial .

(iv) If x + 1 is a factor of polynomial p(x) = x3 − x2 − (2 + √2)x + √2 then p(−1) must be 0
otherwise (x + 1) is not a factor of this polynomial .
p(−1) = (−1)3 − (−1)2 − (2 + √2)(−1) + √2
= −1 − 1 + 2 + √2 + √2
= 2√2
As p(−1) ≠ 0
Therefore, x+1 is not a factor of this polynomial .

Page : 43 , Block Name : Exercise 2.4

Q2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following
cases:

(i) p(x) = 2x3 + x2 − 2x − 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) P (x) = x3 − 4x2 + x + 6, g(x) = x − 3

Answer. If g(x) = x + 1 is a factor of the given polynomial p(x) then p(-1) must be zero .
p(x) = 2x3 + x2 − 2x − 1
p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1
= 2(−1) + 1 + 2 − 1 = 0
Hence g(x) = x + 1 is a factor of the polynomial .

(ii) If g(x) = x + 2 is a factor of the given polynomial . p(x) then p(−2) must be 0.
p(x) = x3 + 3x2 + 3x + 1
p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1
= −8 + 12 − 6 + 1
= −1
As p(−2) ≠ 0
Hence g(x) = x + 2 is not a factor of the given polynomial.

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(iii) If g(x) = x − 3 is a factor of the given polynomial . p(x) then p(3) must be 0.
p(x) = x3 − 4x2 + x + 6
p(3) = (3)3 − 4(3)2 + 3 + 6
= 27 − 36 + 9 = 0
Hence g(x) = x − 3 is not a factor of the given polynomial.

Page : 43 , Block Name : Exercise 2.4

Q3 Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x2 + x + k
(ii) p(x) = 2x2 + kx + √2
(iii) p(x) = kx2 − √2x + 1
(iv) p(x) = kx2 − 3x + k

Answer. (i) p(x) = x2 + x + k
p(1) = 0
⇒ (1)2 + 1 + k = 0
⇒2+k=0
⇒ k = −2
Therefore the value of k is -2

(ii) p(x) = 2x2 + kx + √2
p(1) = 0
⇒ 2(1)2 + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = −2 − √2 = −(2 + √2)
Therefore ,the value of k is −(2 + √2)

(iii) p(x) = kx2 − √2x + 1
p(1) = 0
⇒ k(1)2 − √2(1) + 1 = 0
⇒ k − √2 + 1 = 0
⇒ k = √2 − 1
Therefore the value of k is √2 − 1

(iv) p(x) = kx2 − 3x + k
⇒ p(1) = 0
⇒ k(1)2 − 3(1) + k = 0
⇒k−3+k=0
⇒ 2k − 3 = 0
⇒ k = 32
3
Therefore the value of k is 2

Page : 44 , Block Name : Exercise 2.4

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Q4 Factorise :

(i) 12x2 − 7x + 1
(ii) 2x2 + 7x + 3
(iii) p(x) = kx2 − √2x + 1
(iv) 3x2 − x = 4

Answer. (i) 12x2 − 7x + 1
We can �nd two numbers such that pq = 12 × 1 = 12 and p + q = −7 .They are p = 4 and
q = −3
Here 12x2 − 7x + 1 = 12x2 − 4x − 3x + 1
= 4x(3x − 1) − 1(3x − 1)
= (3x − 1)(4x − 1)

(ii) 2x2 + 7x + 3
We can �nd two numbers such that pq = 2 × 3 = 6 and p + q = 7
They are p = 6 and q = 1 .
Here 2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)

(iii) 6x2 + 5x − 6
We can �nd two numbers such that pq = −36 and p + q = 5
They are p = 9 and q = −4
Here,
6x2 + 5x − 6 = 6x2 + 9x − 4x − 6
= 3x(2x + 3) − 2(2x + 3)
= (2x + 3)(3x − 2)

(iv) 3x2 − x − 4
We can �nd two numbers such that pq = 3 × (−4) = −12
And p + q = −1
They are p = −4 and q = 3
Here,
3x2 − x − 4 = 3x2 − 4x + 3x − 4
= x(3x − 4) + 1(3x − 4)
= (3x − 4)(x + 1)

Page : 44 , Block Name : Exercise 2.4

Q5 Factorise :

(i) x3 − 2x2 − x + 2
(ii) x2 − 3x2 − 9x − 5
(iii) x3 + 13x2 + 32x + 20

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(iv) 2y 3 + y 2 − 2y − 1

Answer. (i) Let p(x) = x3 − 2x2 − x + 2
All the factors of 2 have to be considered .These are ±1, ±2
By trial method
p(2) = (2)3 − 2(2)2 − 2 + 2
=8−8−2+2=0
Therefore (x − 2) is a factor of polynomial p(x)
Let us �nd out the quotient on dividing x3 − 2x2 − x + 2 by p(x)
By long division ,

It is known that
Dividend = Divisor * Quotient + Remainder
∴ x3 − 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0
= (x + 1) [x2 − 2x − x + 2]
= (x + 1)[x(x − 2) − 1(x − 2)]
= (x + 1)(x − 1)(x − 2)
= (x − 2)(x − 1)(x + 1)

(ii) Let p(x) = x3 − 3x2 − 9x − 5
All the factors of 5 have to considered .These are ±1, ±5
By Trial method
p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5
= −1 − 3 + 9 − 5 = 0
Therefore, x + 1 is a factor of this polynomial .
Let us �nd the quotient on dividing x3 + 3x2 − 9x − 5 by x + 1
By long division,

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It is known that
Dividend = Divisor * Quotient + Remainder
∴ x3 − 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0
= (x + 1) (x2 − 5x + x − 5)
= (x + 1)[(x(x − 5) + 1(x − 5)]
= (x + 1)(x − 5)(x + 1)
= (x − 5)(x + 1)(x + 1)

(iii) Let p(x) = x3 + 13x2 + 32x + 20
All the factors of 20 have to be considered Some of them are ±1
±2, ±4, ±5 … …
By Trial Method
p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20
= −1 + 13 − 32 + 20
= 33 − 33 = 0
As p(−1) is zero , therefore x + 1 is a factor of his polynomial p(x)
Let us �nd the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1)
By long division ,

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It is known that
Dividend = Divisor * Quotient + Remainder
x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
= (x + 1) (x2 + 10x + 2x + 20)
= (x + 1)[x(x + 10) + 2(x + 10)]
= (x + 1)(x + 10)(x + 2)
= (x + 1)(x + 2)(x + 10)

(iv) Let p(y) = 2y 3 + y 2 − 2y − 1
By Trial Method
p(1) = 2(1)3 + (1)2 − 2(1) − 1
=2+1−2−1=0
Therefore y − 1 is a factor of this polynomial .
Let us �nd the quotient on dividing 2y 3 + y 2 − 2y − 1 by y − 1

Page : 44 , Block Name : Exercise 2.4

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Exercise 2.5

Q1 Use suitable identities to �nd the following products

(i) (x + 4)(x + 10)
(ii) (x + 8)(x − 10)
(iii) (3x + 4)(3x − 5)
(iv) (y 2 + 2 ) (y 2 − 2 )
3 3

(v) (3 − 2x)(3 + 2x)

Answer. (i) By using the identity (x + a)(x + b) = x2 + (a + b)x + ab
(x + 4)(x + 10) = x2 + (4 + 10)x + 4 × 10
= x2 + 14x + 40

(ii) By using the identity (x + a)(x + b) = x2 + (a + b)x + ab
(x + 8)(x − 10) = x2 + (8 − 10)x + (8)(−10)
= x2 − 2x − 80

(iii) By using the identity (3x + 4)(3x − 5) = 9 (x + 43 ) (x − 3 )
5

(x + a)(x + b) = x2 + (a + b)x + ab
9 (x + 43 ) (x − 53 ) = 9 [x2 + ( 43 − 53 ) x + ( 43 ) (− 53 )]
9 [x2 − 13 x − 20
9
]
9x2 − 3x − 20

(iv) By using the identity (x + y)(x − y) = x2 − y 2
2
(y 2 + 32 ) (y 2 − 32 ) = (y 2 ) − ( 32 )
2

y 4 − 94

(v) By using the identity (x + y)(x − y) = x2 − y 2
(3 − 2x)(3 + 2x) = (3)2 − (2x)2
9 − 4x2

Page : 48 , Block Name : Exercise 2.5

Q2 Evaluate the following products without multiplying directly:

(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96

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Answer. (i) 103 × 107 = (100 + 3)(100 + 7)
= (100)2 + (3 + 7)100 + (3)(7)
By using the identity (x + a)(x + b) = x2 + (a + b)x + ab where
x = 100, a = 3, and b = 7]
= 10000 + 1000 + 21
= 11021

(ii) 95 × 96 = (100 − 5)(100 − 4)
= (100)2 + (−5 − 4)100 + (−5)(−4)
By using the identity (x + a)(x + b) = x2 + (a + b)x + ab where
x = 100, a = −5, and b = −4]
= 10000 − 900 + 20
= 9120

(iii) 104 × 96 = (100 + 4)(100 − 4)
(100)2 − (4)2 [(x + y)(x − y) = x2 − y 2 ]
10000 − 16
9984

Page : 48 , Block Name : Exercise 2.5

Q3 Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y 2
(ii) 4y 2 − 4y + 1
y2
(iii) x2 − 100

Answer. (i) 9x2 + 6xy + y 2 = (3x)2 + 2(3x)(y) + (y)2
= (3x + y)(3x + y)

(ii) 4y 2 − 4y + 1 = (2y)2 − 2(2y)(1) + (1)2
= (2y − 1)(2y − 1) [x2 − 2xy + y 2 = (x − y)2 ]

2
(iii) x2 − 100 = x2 − ( 10 )
y2 y

= (x + 10 ) (x − 10 )
y y
[x2 − y 2 = (x + y)(x − y)]

Page : 48 , Block Name : Exercise 2.5

Q4 Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)2
(ii) (2x − y + z)2
(iii) (−2x + 3y + 2z)2
(iv) (3a − 7b − c)2

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(v) (−2x + 5y − 3z)2
2
(vi) [ 14 a − 12 b + 1]

Answer. (i) It is known that ,
(x + y + z)2 = x2 + y 2 + z 2 + 2xy + 2yz + 2zx
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x2 + 4y 2 + 16z 2 + 4xy + 16yz + 8xz

(ii) (2x − y + z)2 = (2x)2 + (−y)2 + (z)2 + 2(2x)(−y) + 2(−y)(z) + 2(z)(2x)
= 4x2 + y 2 + z 2 − 4xy − 2yz + 4xz

(iii) (−2x + 3y + 2z)2
= (−2x)2 + (3y)2 + (2z)2 + 2(−2x)(3y) + 2(3y)(2z) + 2(2z)(−2x)
= 4x2 + 9y 2 + 4z 2 − 12xy + 12yz − 8xz

(iv) (3a − 7b − c)2
= (3a)2 + (−7b)2 + (−c)2 + 2(3a)(−7b) + 2(−7b)(−c) + 2(−c)(3a)
= 9a2 + 49b2 + c2 − 42ab + 14bc − 6ac

(v) (−2x + 5y − 3z)2
= (−2x)2 + (5y)2 + (−3z)2 + 2(−2x)(5y) + 2(5y)(−3z) + 2(−3z)(−2x)
= 4x2 + 25y 2 + 9z 2 − 20xy − 30yz + 12xz

2
(vi) [ 14 a − 12 b + 1]
2 2
( 14 a) + (− 12 b) + (1)2 + 2 ( 14 a) (− 12 b) + 2 (− 12 b) (1) + 2 ( 14 a) (1)
1 2
= 16 a + 14 b2 + 1 − 14 ab − b + 12 a

Page : 49 , Block Name : Exercise 2.5

Q5 Factorise:
(i) 4x2 + 9y 2 + 16z 2 + 12xy − 24yz − 16xz
(ii) 2x2 + y 2 + 8z 2 − 2√2xy + 4√2yz − 8xz

Answer. It is known that
(x + y + z)2 = x2 + y 2 + z 2 + 2xy + 2yz + 2zx
(i) 4x2 + 9y 2 + 16z 2 + 12xy − 24yz − 16xz
= (2x)2 + (3y)2 + (−4z)2 + 2(2x)(3y) + 2(3y)(−4z) + 2(2x)(−4z)
= (2x + 3y − 4z)2
= (2x + 3y − 4z)(2x + 3y − 4z)

(ii) 2x2 + y 2 + 8z 2 − 2√2xy + 4√2yz − 8xz
(−√2x)2 + (y)2 + (2√2z)2 + 2(−√2x)(y) + 2(y)(2√2z) + 2(−√2x)(2√2z)

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(−√2x + y + 2√2z)2
(−√2x + y + 2√2z)(−√2x + y + 2√2z)

Page : 49 , Block Name : Exercise 2.5

Q6 Write the following cubes in expanded form
(i) (2x + 1)3
(ii) (2a − 3b)3
3
(iii) [ 2 x + 1]
3

3
(iv) [x − 23 y]

Answer. It is known that
(a + b)3 = a3 + b3 + 3ab(a + b)
And (a − b)3 = a3 − b3 − 3ab(a − b)

(i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x2 + 6x
= 8x3 + 12x2 + 6x + 1
(ii)(2a − 3b)3 = (2a)3 − (3b)3 − 3(2a)(3b)(2a − 3b)
= 8a3 − 27b3 − 18ab(2a − 3b)
= 8a3 − 27b3 − 36a2 b + 54ab2

(iii)
3 3
[ 32 x + 1] = [ 32 x] + (1)3 + 3 ( 32 x) (1) ( 32 x + 1)

= 27
8
x3 + 1 + 92 x ( 32 x + 1)
= 27
8
x3 + 1 + 27
4
x2 + 92 x
= 27
8
x3 + 27
4
x2 + 92 x + 1

(iv)
3 3
[x − 23 y] = x3 − ( 23 y) − 3(x) ( 23 y) (x − 23 y)

y − 2xy (x − 23 y)
8 3
= x3 − 27
8 3
= x3 − 27 y − 2x2 y + 43 xy 2

Page : 49 , Block Name : Exercise 2.5

Q7 Evaluate the following using suitable identities:
(i)(99)3 (ii)(102)3 (iii) (998)3

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer.
It is known that,
(a + b)3 = a3 + b3 + 3ab(a + b)
and (a − b)3 = a3 − b3 − 3ab(a − b)
(i)(99)3 = (100 − 1)3
= (100)3 − (1)3 − 3(100)(1)(100 − 1)
= 1000000 − 1 − 300(99)
= 1000000 − 1 − 29700
= 970299
(ii)(102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208
( iii) (998)3 = (1000 − 2)3
= (1000)3 − (2)3 − 3(1000)(2)(1000 − 2)
= 1000000000 − 8 − 6000(998)
= 1000000000 − 8 − 5988000
= 1000000000 − 5988008
= 994011992

Page : 49 , Block Name : Exercise 2.5

Q8 Factorise each of the following:
(i) 8a3 + b3 + 12a2 b + 6ab2 ((ii) 8a3 − b3 − 12a2 b + 6ab2
(iii) 27 − 125a3 − 135a + 225a2 (iv) 64a3 − 27b3 − 144a2 b + 108ab2
1
(v) 27p3 − 216 − 92 p2 + 14 p

Answer.
It is known that,
(a + b)3 = a3 + b3 + 3a2 b + 3ab2
and (a − b)3 = a3 − b3 − 3a2 b + 3ab2
(i) 8a3 + b3 + 12a2 b + 6ab2
= (2a)3 + (b)3 + 3(2a)2 b + 3(2a)(b)2
= (2a + b)3
= (2a + b)(2a + b)(2a + b)
( ii) 8a3 − b3 − 12a2 b + 6ab2
= (2a)3 − (b)3 − 3(2a)2 b + 3(2a)(b)2
= (2a − b)3
= (2a − b)(2a − b)(2a − b)

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(iii) 27 − 125a3 − 135a + 225a2
= (3)3 − (5a)3 − 3(3)2 (5a) + 3(3)(5a)2
= (3 − 5a)3
= (3 − 5a)(3 − 5a)(3 − 5a)
( iv) 64a3 − 27b3 − 144a2 b + 108ab2
= (4a)3 − (3b)3 − 3(4a)2 (3b) + 3(4a)(3b)2
= (4a − 3b)3
= (4a − 3b)(4a − 3b)(4a − 3b)
3 1 9 2 1
(v)27p − 216 − 2 p + 4 p
3 2
= (3p)3 − ( 16 ) − 3(3p)2 ( 16 ) + 3(3p)( 16 )
3
= (3p − 16 )

= (3p − 16 ) (3p − 16 ) (3p − 16 )

Page : 49 , Block Name : Exercise 2.5

Q9 Verify:
(i) x3 + y 3 = (x + y) (x2 − xy + y 2 )
(ii) x3 − y 3 = (x − y) (x2 + xy + y 2 )

Answer.
(i) It is known that,
(x + y)3 = x3 + y 3 + 3xy(x + y)
x3 + y 3 = (x + y)3 + 3xy(x + y)
= (x + y) [(x + y)2 − 3xy]
= (x + y) (x2 + y 2 + 2xy − 3xy)
= (x + y) (x2 + y 2 + 2xy − 3xy)
= (x + y) (x2 − xy + y 2 )
(ii) It is known that,
(x − y)3 = x3 − y 3 − 3xy(x − y)
x3 − y 3 = (x − y)3 + 3xy(x − y)
= (x − y) [(x − y)2 + 3xy]
= (x − y) (x2 + y 2 + 3xy)
= (x − y) (x2 + y 2 + xy)
= (x − y) (x2 + y 2 + xy)
= (x − y) (x2 + xy + y 2 )

Page : 49 , Block Name : Exercise 2.5

Q10 Factorise each of the following:

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

(i) 27y 3 + 125z 3
(ii) 64m3 − 343n3
[Hint: See question] 9.

Answer.
(i) 27y 3 + 125z 3
= (3y)3 + (5z)3
= (3y + 5z) [(3y)2 + (5z)2 − (3y)(5z)] [∵ a3 + b3 = (a + b) (a2 + b2 − ab)]
= (3y + 5z) [9y 2 + 25z 2 − 15yz]
(ii) 64m3 − 343n3
= (4m)3 − (7n)3
= (4m − 7n) [(4m)2 + (7n)2 + (4m)(7n)] [∵ a3 − b3 = (a − b) (a2 + b2 + ab)]
= (4m − 7n) [16m2 + 49n2 + 28mn]

Page : 49 , Block Name : Exercise 2.5

Q11 Factorise: 27x3 + y 3 + z 3 − 9xyz

Answer. It is known that,
x3 + y 3 + z 3 − 3xyz = (x + y + z) (x2 + y 2 + z 2 − xy − yz − zx)
∴ 27x3 + y 3 + z 3 − 9xyz
= (3x)3 + (y)3 + (z)3 − 3(3x)(y)(z)(z) − z(3x)
= (3x + y + z) [9x2 + y 2 + z 2 − 3xy − yz − 3xz]

Page : 49 , Block Name : Exercise 2.5

Q12 Verify that x3 + y 3 + z 3 − 3xyz = 12 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2 ]

Answer. It is known that
x3 + y 3 + z 3 − 3xyz = (x + y + z) (x2 + y 2 + z 2 − xy − yz − zx)
= 12 (x + y + z) [2x2 + 2y 2 + 2z 2 − 2xy − 2yz − 2zx]
= 12 (x + y + z) [(x − y)2 + 2z 2 + z 2 − 2yz) + (x2 + z 2 − 2zx)
= 12 (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2 ]

Page : 49 , Block Name : Exercise 2.5

Q13 If x + y + z = 0, show that x3 + y 3 + z 3 − 3xyz

Answer. It is known that,

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

x3 + y 3 + z 3 − 3xyz = (x + y + z) (x2 + y 2 + z 2 − xy − yz − zx)
Put x + y + z = 0
x3 + y 3 + z 3 − 3xyz = (0) (x2 + y 2 + z 2 − xy − yz − zx)
x3 + y 3 + z 3 − 3xyz = 0
x3 + y 3 + z 3 = 3xyz

Page : 49 , Block Name : Exercise 2.5

Q14 Without actually calculating the cubes, �nd the value of each of the following:
(i)(−12)3 + (7)3 + (5)3
(ii) (28)3 + (−15)3 + (−13)3

Answer. ( i )
(−12)3 + (7)3 + (5)3
Let x = −12, y = 7, and z = 5
It can be observed that,
x + y + z = −12 + 7 + 5 = 0
It is known that if x + y + z = 0 , then
x3 + y 3 + z 3 = 3xyz
∴ (−12)3 + z 3 = 3xyz
∴ (−12)3 + z 3 = 3xyz
= −1260

(ii)
(28)3 + (−15)3 + (−13)3
Let x = 28, y = −15, and z = −13
It can be observed that,
x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0
It is known that if x + y + z = 0, then
x3 + y 3 + z 3 = 3xyz
∴ (28)3 + (−15)3 + (−13)3 = 3(28)(−15)(−13)
= 16380

Page : 49 , Block Name : Exercise 2.5

Q15 Give possible expressions for the length and breadth of each of the following rectangles, in
which their areas are given:
Area: 25a2 −35a+12 Area: 35y 2 +13y−12
I II

Answer.

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Book : Mathematics Ncert Solutions | Chapter-2 Maths

Area = Length × Breadth
The expression given for the area of the rectangle has to be factorised. One of its
factors will be its length and the other will be its breadth.
(i) 25a2 − 35a + 12 = 25a2 − 15a − 20a + 12
= 5a(5a − 3) − 4(5a − 3)
= (5a − 3)(5a − 4)
Therefore, possible length = 5a − 3
And, possible breadth = 5a − 4
(ii) 35y 2 + 13y − 12 = 35y 2 + 28y − 15y − 12
= 7y(5y + 4) − 3(5y + 4)
= (5y + 4)(7y − 3)
Therefore, possible length = 5y + 4
And, possible breadth = 7y − 3

Page : 49 , Block Name : Exercise 2.5

Q16 What are the possible expressions for the dimensions of the cuboids whose volumes are given
below?
Volume: 3x2 −12x Volume: 12ky 2 +8ky−20k
I II

Answer. Volume of cuboid = Length × Breadth × Height The expression given for the volume of the
cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will
be its height.
(i) 3x2 − 12x = 3x(x − 4)
One of the possible solutions is as follows.
Length = 3, Breadth = x, Height = x − 4
(ii) 12ky 2 + 8ky − 20k = 4k (3y 2 + 2y − 5)
= 4k [3y 2 + 5y − 3y − 5]
= 4k[y(3y + 5) − 1(3y + 5)]
= 4k(3y + 5)(y − 1)
One of the possible solutions is as follows.
Length = 4kk , Breadth = 3y + 5, Height = y − 1

Page : 50 , Block Name : Exercise 2.5

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