NCERT Solutions for Class 9 Maths Chapter-2 Polynomials
In the second chapter of class 9 Maths, one enters the fascinating domain of algebraic expressions and their properties. Our NCERT solutions are carefully prepared to make sure that students go through this chapter step by step so that they understand completely the meaning of polynomials and their applications.
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Students can access the NCERT Solutions for Class 9 Maths Chapter-2 Polynomials. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 2.1
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i)
(ii) 
(iii) 
(iv) 
(v) 
(i)
We can observe that in the polynomial , we have x as the only variable and the powers of x in each term are a whole number.
Therefore, we conclude that is a polynomial in one variable.
(ii)
We can observe that in the polynomial , we have y as the only variable and the powers of y in each term are a whole number.
Therefore, we conclude that is a polynomial in one variable.
(iii)
We can observe that in the polynomial, we have t as the only variable and the powers of t in each term are not a whole number.
Therefore, we conclude thatis not a polynomial in one variable.
(iv)
We can observe that in the polynomial, we have y as the only variable and the powers of y in each term are not a whole number.
Therefore, we conclude thatis not a polynomial in one variable.
(v)
We can observe that in the polynomial, we have x, y and t as the variables and the powers of x, y and t in each term is a whole number.
Therefore, we conclude that is a polynomial but not a polynomial in one variable.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
The binomial of degree 35 can be 
The binomial of degree 100 can be 
Classify the following as linear, quadratic and cubic polynomials.
(i) x2+ x
(ii) x – x3
(iii) y + y2+4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.
(iii) The degree of y + y2 + 4 is 2. So, it is a quadratic polynomial.
(iv) The degree of 1 + x is 1. So, it is a linear polynomial.
(v) The degree of 3t is 1. So, it is a linear polynomial.
(vi) The degree of r2 is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x3 is 3. So, it is a cubic polynomial.
Write the coefficients of 
in each of the following :
(i)
(ii)
(iii)
(iv)
(i)
The coefficient ofin the polynomialis 1.
(ii)
The coefficient ofin the polynomialis -1.
(iii)
The coefficient ofin the polynomialis
.
(iv)
The coefficient ofin the polynomial is 0.
Write the degree of each of the following polynomials :
(i)
(ii)
(iii)
(iv)
3
(i)
We know that the degree of a polynomial is the highest power of the variable in the polynomial.
We can observe that in the polynomial, the highest power of the variable x is 3.
Therefore, we conclude that the degree of the polynomialis 3.
(ii)
We know that the degree of a polynomial is the highest power of the variable in the polynomial.
We can observe that in the polynomial, the highest power of the variable y is 2.
Therefore, we conclude that the degree of the polynomialis 2.
(iii)
We know that the degree of a polynomial is the highest power of the variable in the polynomial.
We observe that in the polynomial, the highest power of the variable t is 1.
Therefore, we conclude that the degree of the polynomialis 1.
(iv) 3
We know that the degree of a polynomial is the highest power of the variable in the polynomial.
We can observe that in the polynomial 3, the highest power of the assumed variable x is 0.
Therefore, we conclude that the degree of the polynomial 3 is 0.
Exercise 2.2
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Let p(x) = 5x – 4x2 + 3
(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
(ii) p(-1) = 5(-1) – 4(-1)2 + 3
= – 5x – 4x2 + 3 = -9 + 3 = -6
Thus, the value of 5x – 4x2 + 3 at x = -1 is -6.
(iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3
= 10 – 16 + 3 = -3
Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
(i) Given that p(y) = y2 – y + 1.
∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴p(0) = 2 + 0 + 2(0)2 – (0)3
= 2 + 0 + 0 – 0=2
P(1) = 2 + 1 + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p( 2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) Given that 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
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
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.
(i) We have, p(x) = x + 5. Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
Thus, zero of x + 5 is -5.
(ii) We have, p(x) = x – 5.
Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5
Thus, zero of x – 5 is 5.
(iii) We have, p(x) = 2x + 5. Since, p(x) = 0
⇒ 2x + 5 =0
⇒ 2x = -5
⇒ x = -5/2
Thus, zero of 2x + 5 is -5/2.
(iv) We have, p(x) = 3x – 2. Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = 2/3
Thus, zero of 3x – 2 is 2/3
(v) We have, p(x) = 3x. Since, p(x) = 0
⇒ 3x = 0 ⇒ x = 0
Thus, zero of 3x is 0.
(vi) We have, p(x) = ax, a ≠ 0.
Since, p(x) = 0 => ax = 0 => x-0
Thus, zero of ax is 0.
(vii) We have, p(x) = cx + d. Since, p(x) = 0
⇒ cx + d = 0 ⇒ cx = -d ⇒ x= -d/c
Thus, zero of cx + d is -d/c.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –1/3
(ii) p (x) = 5x – π, x = 4/5
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – 
(vii) P (x) = 3x2 – 1, x = – 
,
(viii) p (x) = 2x + 1, x = 1/2
(i) We have , p(x) = 3x + 1
(ii) We have, p(x) = 5x – π
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.
(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).
(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.
(vi) We have, p(x) = lx + m
(vii) We have, p(x) = 3x2 – 1
(viii) We have, p(x) = 2x + 1
Since, 
≠ 0, so, x = 
is not a zero of 2x + 1.
Exercise 2.3
Find the remainder when x3+3x2+3x+1 is divided by
(i) x+1
(ii) x-1/2
(iii) x
(iv) 
(v) 5+2x
(i) x+1
We need to find the zero of the polynomial
x+1=0 => x = -1
While applying the remainder theorem, we need to put the zero of the polynomial x+1 in the polynomial x3+3x2+3x+1, to get
p(x) = x3+3x2+3x+1
p(0)
= -1+3-3+1
= 0
Therefore, we conclude that on dividing the polynomial x3+3x2+3x+1 by x+1, we will get the remainder as 0.
(ii)x=-1/2
We need to find the zero of the polynomial x-1/2.
x-1/2 = 0 => x = 1/2
While applying the remainder theorem, we need to put the zero of the polynomial x-1/2 in the polynomial x3+3x2+3x+1, to get
p(x) = x3+3x2+3x+1
= p(1/2) = (1/2)3 + 3(1/2)2 + 3(1/2) + 1
= 1/8 + 3(1/4) + 3/2 +1
= 1+6+12+8/8
= 27/8
Therefore, we conclude that on dividing the polynomial x3+3x2+3x+1 by x-1/2 , we will get the remainder as 27/8.
(iii)
We need to find the zero of the polynomial x.
x=0
While applying the remainder theorem, we need to put the zero of the polynomial x in the polynomial x3+3x2+3x+1, to get
p(x) = x3+3x2+3x+1
= p(0) = (0)3 + 3(0)2 + 3(0) +1
=0+0+0+1
=1
Therefore, we conclude that on dividing the polynomial x3+3x2+3x+1 by x, we will get the remainder as 1.
(iv)
We need to find the zero of the polynomial.
While applying the remainder theorem, we need to put the zero of the polynomial in the polynomial x3+3x2+3x+1, to get
p(x) = x3+3x2+3x+1
= 
Therefore, we conclude that on dividing the polynomial x3+3x2+3x+1 by, we will get the remainder as .
(v) 5+2x
We need to find the zero of the polynomial 5+2x.
5+2x = 0 => x = - 5/2
While applying the remainder theorem, we need to put the zero of the polynomial 5+2x in the polynomial x3+3x2+3x+1, to get
p(x) = x3+3x2+3x+1
p(-5/2) = (-5/2)3 + 3(-5/2)2 + 3(-5/2) +1
= -125/8 + 3(25/4) - 15/2 + 1
= -125/8 + 75/4 - 15/2 + 1
= -125 + 150 - 60 + 8 / 8
= -27/4
Therefore, we conclude that on dividing the polynomial x3+3x2+3x+1 by 5+2x, we will get the remainder as -27/4.
Find the remainder when x3 - ax2 + 6x - a is divided by x-a.
We need to find the zero of the polynomial x-a.
x-a = 0 => x = a
While applying the remainder theorem, we need to put the zero of the polynomial x-a in the polynomial x3 - ax2 + 6x - a, to get
p(x) = x3 - ax2 + 6x - a
p(a) = (a)3 - a(a)2 + 6(a) - a
= a3 - a3 + 6a - a
= 5a
Therefore, we conclude that on dividing the polynomial x3 - ax2 + 6x - a byx-a , we will get the remainder as 5a.
Check whether 7+3x is a factor of 3x3+7x.
We know that if the polynomial 7 + 3x is a factor of 3x3+7x, then on dividing the polynomial 3x3+7x by 7 + 3x, we must get the remainder as 0.
We need to find the zero of the polynomial 7 + 3x.
7 + 3x = 0 => x = - 7/3
While applying the remainder theorem, we need to put the zero of the polynomial 7 + 3x inthe polynomial 3x3+7x, to get
p(x) = 3x3+7x
= 3( - 7/3)3 + 7(- 7/3)
= 3( -343/27) - 49/3
= - 343/9 - 49/3
= - 343 - 147/9
= - 490/9
We conclude that on dividing the polynomial 3x3+7x by 7 + 3x, we will get the remainder as -490/9, which is not 0.
Therefore, we conclude that 7 + 3x is not a factor of 3x3+7x.
Exercise 2.4
Determine which of the following polynomials has (x +1) a factor.
(i) x3+x2+x +1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 – x2 – (2 +√2 )x + √2
The zero of x + 1 is -1.
(i) Let p (x) = x3 + x2 + x + 1
∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x3 + x2 + x + 1.
(ii) Let p (x) = x4 + x3 + x2 + x + 1
∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1.
(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 .
∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1.
(iv) Let p (x) = x3 – x2 – (2 + √2) x + √2
∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.
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
(i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).
(ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).
(iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3
∴ p(3) = (3)3 – 4(3)2 + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6 = 0
⇒ p(3) = 0, so g(x) is a factor of p(x).
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 – √2 x + 1
(iv) p (x) = kx2 – 3x + k
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.
(i) Here, p(x) = x2 + x + k
Since, p(1) = (1)2 +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.
(ii) Here, p (x) = 2x2 + kx + √2
Since, p(1) = 2(1)2 + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)
(iii) Here, p (x) = kx2 – √2 x + 1
Since, p(1) = k(1)2 – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1
(iv) Here, p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k = 3/4
Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
(i) We have,
12x2 – 7x + 1 = 12x2 – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x2 -7x + 3 = (2x – 1) (x + 3)
(ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)
(iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2)
(iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)
Factorise
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
(i) We have, x3 – 2x2 – x + 2
Rearranging the terms, we have x3 – x – 2x2 + 2
= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)
= [(x)2 – (1)2](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a2 – b2) = (a + b)(a-b)]
Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)
(ii) We have, x3 – 3x2 – 9x – 5
= x3 + x2 – 4x2 – 4x – 5x – 5 ,
= x2 (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)
(iii) We have, x3 + 13x2 + 32x + 20
= x3 + x2 + 12x2 + 12x + 20x + 20
= x2(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x2 + 12x + 20)
= (x + 1)(x2 + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x3 + 13x2 + 32x + 20
= (x + 1)(x + 2)(x + 10)
(iv) We have, 2y3 + y2 – 2y – 1
= 2y3 – 2y2 + 3y2 – 3y + y – 1
= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y2 + 3y + 1)
= (y – 1)(2y2 + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y3 + y2 – 2y – 1
= (y – 1)(y + 1)(2y +1)
Exercise 2.5
Factorize :
(i)
(ii)
(i)
The expression 
We can observe that, we can apply the identity
Therefore, we conclude that after factorizing the expression
(ii)
We need to factorize the expression
The expression 
We can observe that, we can apply the identity
Therefore, we conclude that after factorizing the expression
Use suitable identities to find the following products :
(i)
(ii)
(iii)
(iv)
(v)
(i)
We know that
We need to apply the above identity to find the product
Therefore, we conclude that the product
(ii)
We know that
We need to apply the above identity to find the product
Therefore, we conclude that the product
(iii)
We know that
We need to apply the above identity to find the product
Therefore, we conclude that the product
(iv)
We know that
We need to apply the above identity to find the product
Therefore, we conclude that the product
(v)
We know that
We need to apply the above identity to find the product
Therefore, we conclude that the product
Evaluate the following products without multiplying directly :
(i)
(ii)
(iii)
(i)
We can observe that, we can apply the identity
=10000+1000+21
=11021
Therefore, we conclude that the value of the product
(ii)
We can observe that, we can apply the identity
=10000-900+20
=9120
Therefore, we conclude that the value of the product
(iii)
We can observe that, we can apply the identity
=10000-16
=9984
Therefore, we conclude that the value of the product
Factorize the following using appropriate identities :
(i)
(ii)
(iii)
(i)
We can observe that, we can apply the identity
(ii)
We can observe that, we can apply the identity
(iii)
We can observe that, we can apply the identity
Expand each of the following, using suitable identities :
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
We know that
We need to apply the above identity to expand the expression
(ii)
We know that
We need to apply the above identity to expand the expression
(iii)
We know that
We need to apply the above identity to expand the expression
(iv)
We know that
We need to apply the above identity to expand the expression
(v)
We know that
We need to apply the above identity to expand the expression
(vi)
We know that
Write the following cubes in expanded form :
(i)
(ii)
(iii)
(iv)
(i)
We know that
Therefore, the expansion of the expression
(ii)
We know that
Therefore, the expansion of the expression
(iii)
We know that
Therefore, the expansion of the expression
(iv)
We know that
Therefore, the expansion of the expression
Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3
(i) We have, 99 = (100 -1)
∴ 993 = (100 – 1)3
= (100)3 – 13 – 3(100)(1)(100 -1)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299
(ii) We have, 102 =100 + 2
∴ 1023 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208
(iii) We have, 998 = 1000 – 2
∴ (998)3 = (1000-2)3
= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992
Factorise each of the following
(i) 8a3 +b3 + 12a2b+6ab2
(ii) 8a3 -b3-12a2b+6ab2
(iii) 27-125a3 -135a+225a2
(iv) 64a3 -27b3 -144a2b + 108ab2
(i) 8a3 +b3 +12a2b+6ab2
= (2a)3 + (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)
(ii) 8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a – b) (2a – b) (2a – b)
(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a3 -27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a3 – b3 – 3 ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)
Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
Hence, verified.
(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
Hence, verified.
Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint See question 9]
(i) We know that
x3 + y3 = (x + y)(x2 – xy + y2)
We have, 27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)
(ii) We know that
x3 – y3 = (x – y)(x2 + xy + y2)
We have, 64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)
Factorise 27x3 +y3 +z3 -9xyz.
We have,
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using the identity,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)
Verify that
x3 +y3 +z3 – 3xyz = 
R.H.S
= 
= 
= 
= 
= 2 x 
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S.
Hence, verified.
If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Since, x + y + z = 0
⇒ x + y = -z (x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y3 + z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z3 = 3xyz
Without actually calculating the cubes, find the value of each of the following
(i) (- 12)3 + (7)3 + (5)3
(ii) (28)3 + (- 15)3 + (- 13)3
(i) We have, (-12)3 + (7)3 + (5)3
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)]
= 3[-420] = -1260
(ii) We have, (28)3 + (-15)3 + (-13)3
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13)
= 3(5460) = 16380
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a2 – 35a + 12
(ii) Area 35y2 + 13y – 12
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).
(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x2 – 12x = 3(x2 – 4x)
= 3 x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).
(ii) We have, 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).
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