2. Polynomials Mathematics class 9 exercise Exercise 2.5
2. Polynomials Mathematics class 9 exercise Exercise 2.5 ncert book solution in english-medium
NCERT Books Subjects for class 9th Hindi Medium
Exercise 2.1
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) y2 + √2
(iii)3√t + t√2
(iv) y +2/y
(v) x10 + y3 + t50
Solution:
(i) 4x2 – 3x + 7
It is polynomial in one variable. Because the power of variable is natural number.
(ii) y2 + √2
It is polynomial in one variable. Because the power of variable is natural number.
(iii)3√t + t√2
It is not a polynomial in one variable. Because the power of variable is not a natural number, it is a fractional number.
(iv) y +2/y
It is not a polynomial in one variable.
(v) x10 + y3 + t50
It is not a polynomial in one variable. while It is a polynomial in three variable.
Q2. Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x
(ii) 2 – x2 + x3 (
iii) π/2 x2 + x
(iv) √2x −1
Solution:
(i) 2 + x2 + x
Coefficient of x2 = 1
(ii) 2 – x2 + x3
Coefficient of x2 = –1
(iii) π/2 x2 + x
Coefficient of x2 = π/2
(iv) √2x −1
Coefficients of x2 = 0 [Because There is no x2 So coefficient will be 0]
Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
A binomial of degree 35
⇒ 2x35 + 5y
Note: Binomial means the expression of two terms. Like x + 5, 3a - 2b, 3t + 7 etc.
A monomial of degree 100
⇒ 3y100
Note: Monomial means the expression of single term, like 3x, 5t, y, 3xy etc.
Q4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
(ii) 4 – y2
(iii) 5t – √7
(iv) 3
Solution:
(i) 5x3 + 4x2 + 7x
Ans: The degree of polynomial = 3 [Highest power]
(ii) 4 – y2
Ans: The degree of polynomial = 2
(iii) 5t – √7
Ans: The degree of polynomial = 1
(iv) 3
Ans: The degree of polynomial = 0
[Note: There is no variable therefore degree = 0]
Q5. 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) 7x2
Solution:
(i) x2 + x
Ans: Quadratic polynomial
(ii) x – x3
Ans: Cubic polynomial
(iii) y + y2 + 4
Ans: Quadratic polynomial
(iv) 1 + x
Ans : Linear polynomial
(v) 3t
Ans : Linear polynomial
(vi) r2
Ans: Quadratic polynomial
(vii) 7x2
Ans: Cubic polynomial
Exercise 2.2
2. Polynomials
Exercise 2.2
Q1. Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Solution:
(i) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 0 is given by
P(0) = 5(0) - 4(0)2 + 3
= 0 - 0 + 3
= 3
(ii) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 1 is given by
P(1) = 5(1) - 4(1)2 + 3
= 5 - 4 + 3
= 4
(iii) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 2 is given by
P(2) = 5(2) - 4(2)2 + 3
= 10 -16 + 3
= - 9
Q2.Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t 2 – t 3
(iii) p(x) = x3
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) p(y) = y2 - y + 1
P(0) = (0)2- 0 + 1
=1
P(1) = (1)2- 1 + 1
= 1 - 1 + 1
= 1
P(2) = (2)2- 2 + 1 = 4 - 2 + 1 = 3
(ii) p(t) = 2 + t + 2t2- t3
P(0) = 2 + 0 + 2(0)2- (0)3
=2
P(1) = 2 + 1 + 2(1)2 - (1)3
= 4
P(2) = 2 + 2 + 2(2)2- (2)3
= 4 + 8 - 8
= 4
(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(1) =0
P(2)= (2-1) (2+1)=1(3) =3
Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
Solution:
Q4. Find the zero of the polynomial in each of the following cases:
(i) P(x) = x + 5
(ii) P(x) = x – 5
(iii) Px) = 2x + 5
(iv) P(x) = 3x – 2
(v) P(x) = 3x
(vi) P(x) = ax, a ≠ 0
Solution (i) :
(i) P(x) = x + 5
⇒ x + 5 = 0
⇒ x = - 5
The zero of the polynomial is - 5.
Solution (ii) :
(ii) P(x) = x – 5
⇒ x – 5 = 0
⇒ x = 5
The zero of the polynomial is 5.
The zero of the polynomial is - 5/2.
(iv) P(x) = 3x - 2
3x - 2 = 0
The zero of the polynomial is 2/3.
Exercise 2.3
Chapter 2. Polynomials
Exercise 2.3
Solution:
(i)By remainder theorem, the required remainder is equal to p(x) = (-1)
P (-1) = x3+ 3x2+ 3x+1
= (-1)3 + 3 (-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1 = 0
Required remainder is p (-1) = 0
Required remainder is p (1/2) = 0
(iv) By remainder theorem, the required remainder is equal to p(x) = - π
P(x) = x3+ 3x2+ 3x+1
P(π) = (- π)3 + 3(-π)2 + 3(-π) + 1
= (-π)3 + 3π2 + 3π + 1
Required remainder is p (π) = 0
Required remainder is p (- 5/2) = 0
Q.2. find the remainder when x3- ax2 + 6x - a is divided by x- a.
Solution: Let P(x) = x3- ax2+ 6x- a . P(x) = a.
By remainder theorem
P (a) = (a)3- a(a)2+ 6(a) - a
=a3-a3 + 6a - a
Remainder = 5a
Q3. Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
g(x) = 7 + 3x = 0
⇒ 3x = - 7
P(x) ≠ 0
Therefore, 7 + 3x is not a factor of P(x).
Exercise 2.4
Chapter 2. Polynomials
Exercise 2.4
Q.1. 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 
Solution:
(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 a factor of p(x).
P(x) = x3 + x2 + x + 1
P(−1)= (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 − 1
= 0
∵ P(x) = 0
Hence, x + 1 is 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 a factor 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
As P(x) ≠ 0, (− 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(x) = x4+ 3x3+ 3x2+ x + 1
P(−1) = (−1)4+ 3(−1)3+ 3(−1)2+ (−1) + 1
= 1 − 3 + 3 − 1 + 1
= 1
As P(x) ≠ 0, (−1)
Therefore (x+1) is not a factor of this polynomial .
Q.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 − 4 x2 + x + 6, g(x) = x − 3
Solution:
(i) 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
∵ P(x) = 0
Hence, g(x) = x + 1 is a factor of the given 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(x) ≠ 0,
Hence, g(x) = x + 2 is not a factor of the given polynomial.
(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then P(3) must be 0.
P(x) = x3− 4 x2+ x + 6
P(3) = (3)3 − 4(3)2 + 3 + 6
= 27 −36 + 9
= 0
Hence, g(x) = x − 3 is a factor of the given polynomial.
Solution:
: If x − 1 is a factor of polynomial p(x), then P(1) must be 0.
(i) P(x) = x2+ x + k
P(1) = (1)2+ 1 + k
= 1+1+
= 2+k
k =−2
(iv) P(x) = kx2-3x + k
P(1) = k(1)2-3(1) + k
= k-3+k
2k = 3
K = 3/2
Question 4: Factorise:
(i) 12x2− 7x + 1 (ii) 2x2+ 7x + 3
(iii) 6x2+ 5x – 6 (iv) 3x2− x − 4
Solution:
(i) 12x2− 7x + 1 we can find 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 find 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 find 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 find 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)
Question 5. Factorize:
(i) x3− 2x2− x + 2 (ii) x3+ 3x2−9x − 5
(iii) x3+ 13x2+ 32x + 20 (iv) 2y3+ y2− 2y – 1
Solution:
(i) Let P(x) = x3− 2x2− x + 2 all the factor are there. These are ± 1, ± 2.
By trial method, P (1) = (1)3− 2(1)2− 1 + 2
= 1 − 2 − 1+ 2
= 0 Therefore, (x − 1) is factor of polynomial p(x)
Let us find the quotient on dividing x3− 2x2− x + 2 by x − 1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3− 2x2− x + 2 = (x – 1) ( X2– x – 2) + 0
= (x – 1) (x2–2x+x–2)
= (x – 1) [x (x–2) + 1(x–2)]
= (x – 1) (x + 1) (x – 2)
(ii) Let p(x) = x3 – 3x2−9x – 5 all the factor are there. These are ± 1, ± 2.
By trial method, p (–1) = (–1)3– 3(1)2− 9(1) – 5
= –1– 3–9–5 =0
Therefore (x+1) is the factor of polynomial p(x).
. Let us find the quotient on dividing x3– 3x2−9x – 5 by x+1.
By long division method
Now,
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 + 1) (x – 5)
(iii) Let p(x) = x3+ 13x2+ 32x + 20 all the factor are there.
These are ± 1, ± 2, ± 3, ± 4.
By trial method, p (–1) = (–1)3+13(–1)2+ 32(–1) +20
= –1+13–32+20 = 0
Therefore (x+1) is the factor of polynomial p(x).
Let us find the quotient on dividing x3+ 13x2+ 32x + 20 by x+1
By long division method
Now,
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 + 2) (x + 10)
(iv) Let p(y) = 2y3+ y2− 2y – 1 all the factor are there. These are ± 1, ± 2.
By trial method, p (1) =2(1)3 + (1)2 – 2(1) – 1
=2 + 1 – 2 – 1 =0
Therefore (y–1) is the factor of polynomial p(y).
Let us find the quotient on dividing 2y3+ y2− 2y – 1 by y–1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
2y3+ y2− 2y −1 =(y − 1) (2y2+3y + 1)
= (y − 1) (2y2+2y
= (y − 1) [2y (y+1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)
Exercise 2.5
Algebraic Identities:
∵ ∴
(1) (x + y)2 = x2 + 2xy + y2
(2) (x - y)2 = x2 - 2xy + y2
(3) x2 - y2 = (x + y) (x - y)
(4) (x + a) (x + b) = x2 + (a + b)x + ab
(5) (x + y)3 = x3 + 3x2y + 3xy2 + y3
(6) (x - y)3 = x3 - 3x2y + 3xy2 - y3
(7) x3 + y3 = (x + y) (x2 - xy + y2)
(8) x3 - y3 = (x - y) (x2 + xy + y2)
(9) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(10) x3 + y3 + z3 - 3xyz = ( x + y + z) (x2 + y2 + z2 - xy - yz - zx)
Exercise 2.5
Q1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x + 4) (x + 10)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(x + 4) (x + 10) = x2 + (4 + 10)x + (4)(10)
= x2 + 14x + 40
(ii) (x + 8) (x – 10)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(x + 8) (x – 10) = x2 + [8 + (-10)]x + (8)(-10)
= x2 - 2x - 80
(iii) (3x + 4) (3x – 5)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(3x + 4) (3x – 5) = (3x)2 + [4 + (-5)]3x + (4)(-5)
= 9x2 - 3x - 20
Using identity; (x + y) (x - y) = x2 - y2
(v) (3 – 2x) (3 + 2x)
Using identity; (x + y) (x - y) = x2 - y2
(3 – 2x) (3 + 2x) = (3)2 - (2x)2
= 9 - 4x2
Q2. Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) 103 × 107 = (100 + 3) (100 + 7)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(100 + 3) (100 + 7) = (100)2 + (3 + 7)100 + 3×7
=10000 + 1000 + 21
= 11021
(ii) 95 × 96 = (90 + 5) (90 + 6)
Using identity; (x + a) (x + b) = x2 + (a + b)x + ab
(90 + 5) (90 + 6) = (90)2 + (5 + 6)90 + 5×6
=8100 + 990 + 30
= 9120
(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity; (x + y) (x - y) = x2 - y2
(100)2 - (4)2
=10000 - 16
= 9984
3. Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
Solution:
(i) 9x2 + 6xy + y2
= (3x)2 + 2.3x.y + (y)2 [ ∵ x2 + 2xy + y2 = (x + y)2]
∴ = (3x + y)2
= (3x + y) (3x + y)
(ii) 4y2 - 4y + 1
= (2y)2 - 2.2y.1 + (1)2 [ ∵ x2 - 2xy + y2 = (x - y)2]
∴ = (2y - 1)2
= (2y - 1) (2y - 1)
[ ∵ x2 - y2 = (x + y) (x - y) ]
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
(v) (–2x + 5y – 3z)2
Solution:
(i) (x + 2y + 4z)2
Here let as a = x, b = 2y, c = 4z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx
(ii) (2x – y + z)2
Here let as a = 2x, b = - y, c = z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (2x – y + z)2 = (2x)2 + (- y)2 + (z)2 + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)
= 4x2 + y2 + z2 - 4xy - 2yz + 4zx
(iii) (–2x + 3y + 2z)2
Here let as a = - 2x, b = 3y, c = 2z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (–2x + 3y + 2z)2
= (– 2x)2 + (3y)2 + (2z)2 + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx
(iv) (3a – 7b – c)2
Here let as x = 3a, y = – 7b, z = – c and putting the values of x, y and z in the
Identity (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
∴ (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
Here let as a = - 2x, b = 5y, c = –3z and putting the values of a, b and c in the
Identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (–2x + 5y – 3z)2
= (– 2x)2 + (5y)2 + (– 3z)2 + 2(–2x)(5y) + 2(5y)(– 3z) + 2(– 3z)(–2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx
Q5. Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Solution:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)
[∵ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 ]
= (2x + 3y + 4z)2
= (2x + 3y + 4z) (2x + 3y + 4z)
Q6. Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
Solution:
(i) (2x + 1)3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
(2x + 1)3 = (2x)3 + 3(2x)2(1) + 3(2x)(1)2 + (1)3
= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3
[Using identity (x - y)3 = x3 - 3x2y + 3xy2 - y3]
(2a - 3b)3 = (2a)3 - 3(2a)2(3b) + 3(2a)(3b)2 - (3b)3
= 8a3 - 36a2b + 54ab2 - 27b3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
Q7. Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3
= (100 - 1)3
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
(100 - 1)3 = (100)3 - 3(100)2(1) + 3(100)(1)2 - (1)3
= 1000000 - 30000 + 300 - 1
= 1000300 - 30001
= 970299
(ii) (102)3
= (100 + 2)3
[using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3]
(100 + 2)3 = (100)3 + 3(100)2(2)+ 3(100)(2)2 + (2)3
= 1000000 + 60000 + 1200 + 8
= 1061208
(iii) (998)3
= (1000 - 2)3
[using identity (a - b)3 = a3 - 3a2b + 3ab2 - b3]
(1000 - 2)3 = (1000)3 - 3(1000)2(2)+ 3(1000)(2)2 - (2)3
= 1000000000 - 6000000 + 12000 - 8
= 1000012000 - 6000008
= 994011992
Q8. Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a2 – b2 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
Solution:
(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2
[Using identity x3 + y3 + 3x2y + 3xy2 = (x + y)3 ]
= (2a)3 +(b)3 + 3(2a)2(b) + 3(2a)(b)2 = (2a + b)3
= (2a + b)(2a + b)(2a + b)
(ii) 8a2 – b2 – 12a2b + 6ab2
= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (2a)3 - (b)3 - 3(2a)2(b) + 3(2a)(b)2 = (2a - b)3
= (2a - b)(2a - b)(2a - b)
(iii) 27 – 125a3 – 135a + 225a2
= (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (3)3 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2= (3 - 5a)3
= (3 - 5a)(3 - 5a)(3 - 5a)
(iv) 64a3– 27b3 – 144a2b + 108ab2
= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2 = (4a - 3b)3
= (4a - 3b)(4a - 3b)(4a - 3b)
[Using identity x3 - y3 - 3x2y + 3xy2 = (x - y)3 ]
Q9. Verify:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
RHS = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y (x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
∵ LHS = RHS Verified
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
RHS = (x - y) (x2 + xy + y2)
x(x2 + xy + y2) - y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
∵ LHS = RHS Verified
Q10. Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27y3 + 125z3
= (3y)3 + (5z)3
[Using identity x3 + y3 = (x + y) (x2 – xy + y2) ]
(3y)3 + (5z)3 = (3y + 5y) [(3y)2 - (3y)(5z) + (5z)2]
= (3y + 5y) (9y2 - 15yz + 25z2)
(ii) 64m3 – 343n3
Solution:
(ii) 64m3 – 343n3
= (4m)3 – (7n)3
[Using identity x3 – y3 = (x – y) (x2 + xy + y2) ]
(4m)3 – (7n)3 = (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n) (16m2 + 28mn + 49n2)
Q11. Factorise : 27x3 + y3 + z3 – 9xyz
Solution:
= (3x)3 + (y)3 + (z)3 - 9xyz
∵ x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)
Using identity:
= (3x + y + z) ((3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x))
= (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3zx)
Q12. Verify that:
x3 + y3 + z3 - 3xyz =½ (x + y + z) [(x -y)2 + (y - z)2 + (z - x)2]
LHS = ½(x + y + z) [x2 - 2xy + y2 + y2 - 2yz + z2 + z2 - 2xz + x2]
= ½(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz)
= ½ × 2(x + y + z)(x2 + y2 + z2 - xy - yz - xz)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
= x3 + y3 + z3 - 3xyz [Using Identity]
LHS = RHS
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