RD SHARMA Solutions for Class 9 Maths Chapter 6 - Factorisation of Polynomials
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.1
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution 8
Degree of a polynomial is the highest power of variable in the polynomial.
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .
Monomial has only one term in it. So monomial of degree 100 can be written as 7x100.
Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable. The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with .
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.2
Question 1
If f(x) = 2x3 - 13x2 + 17x + 12, find:
(i) f(2)
(ii) f(-3)
(iii) f(0)
(i) f(2)
(ii) f(-3)
(iii) f(0)
Solution 1
(i)
f(x) = 2x3 - 13x2 + 17x + 12
f(2) = 2(2)3 - 13(2)2 + 17(2) + 12
= 16 - 52 + 34 + 12
= 10
(ii)
f(-3) = 2(-3)3 - 13(-3)2 + 17(-3) + 12
= -54 - 117 - 51 + 12
= - 210
(iii)
f(0) = 2(0)3 - 13(0)2 + 17(0) + 12
= 0 - 0 + 0 + 12
=12
f(x) = 2x3 - 13x2 + 17x + 12
f(2) = 2(2)3 - 13(2)2 + 17(2) + 12
= 16 - 52 + 34 + 12
= 10
(ii)
f(-3) = 2(-3)3 - 13(-3)2 + 17(-3) + 12
= -54 - 117 - 51 + 12
= - 210
(iii)
f(0) = 2(0)3 - 13(0)2 + 17(0) + 12
= 0 - 0 + 0 + 12
=12
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Find rational roots of the polynomial f(x) = 2x3 + x2 - 7x - 6.
Solution 7
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.3
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
If the polynomials ax3 + 3x2 - 13 and 2x3 -5x + a, when divided by (x-2) leave the same remainder, find the value of a.
Solution 10
Question 11
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
Solution 11
Question 12
Solution 12
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.4
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
What must be added to 3x3 + x2 - 22x + 9 so that the result is exactly divisible by 3x2 + 7x - 6?
Solution 25
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.5
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Using factor theorem, factorize: x4 - 7x3 + 9x2 + 7x -10
Solution 4
Question 5
Using factor theorem, factorize: 3x3 - x2 - 3x + 1
Solution 5
Question 6
Using factor theorem, factorize each of the following polynomials:
x3 - 23x2 + 142x - 120
Solution 6
Question 7
Using factor theorem, factorize: y3 - 7y + 6
Solution 7
Question 8
Using factor theorem, factorize: x3 - 10x2 - 53x - 42
Solution 8
Question 9
Using factor theorem, factorize: y3 - 2y2 - 29y - 42
Solution 9
Question 10
Using factor theorem, factorize: 2y3 - 5y2 - 19y + 42
Solution 10
Question 11
x3 + 13x2 + 32x + 20
Solution 11
Let p(x) = x3 + 13x2 + 32x + 20
The factors of 20 are 
1, 2, 4, 5 ... ...
By hit and trial method
p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
= - 1 + 13 - 32 + 20
= 33 - 33 = 0
As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
By long division
We know 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)
Dividend = Divisor
Quotient + Remainderx3 + 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)
Question 12
Factorise:
x3 - 3x2 - 9x - 5
Solution 12
Let p(x) = x3 - 3x2 - 9x - 5
Factors of 5 are 
1, 5.
By hit and trial method
p(- 1) = (- 1)3 - 3(- 1)2 - 9(- 1) - 5
= - 1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial
Let us find the quotient while dividing x3 + 3x2 - 9x - 5 by x + 1
By long division
Now, Dividend = Divisor 
Quotient + Remainder
x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0
= (x + 1) (x2 - 5 x + x - 5)
= (x + 1) [(x (x - 5) +1 (x - 5)]
= (x + 1) (x - 5) (x + 1)
= (x - 5) (x + 1) (x + 1)
Quotient + Remainder x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0= (x + 1) (x2 - 5 x + x - 5)
= (x + 1) [(x (x - 5) +1 (x - 5)]
= (x + 1) (x - 5) (x + 1)
= (x - 5) (x + 1) (x + 1)
Question 13
Factorise:
2y3 + y2 - 2y - 1
Solution 13
Let p(y) = 2y3 + y2 - 2y - 1
By hit and trial method
p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
= 2 + 1 - 2 - 1= 0
So, y - 1 is a factor of this polynomial
By long division method,
p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
= 2 + 1 - 2 - 1= 0
So, y - 1 is a factor of this polynomial
By long division method,
p(y) = 2y3 + y2 - 2y - 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)
= (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)
Question 14
Using factor theorem, factorize: x3 - 2x2 - x + 2
Solution 14
Question 15
Solution 15
Question 16
Using factor theorem, factroize : x4 - 2x3 - 7x2 + 8x + 12
Solution 16
Question 17
Using factor theorem, factroize : x4 + 10x3 + 35x2 + 50x + 24
Solution 17
Question 18
Using factor theorem, factorize : 2x4 - 7x3 - 13x2 + 63x - 45
Solution 18
Chapter 6 - Factorisation of Polynomials Exercise 6.34
Question 1
If x - 2 is a factor of x2 + 3ax - 2a, then a =
(a) 2
(b) -2
(c) 1
(d) -1
Solution 1
Let p(x) = x2 + 3ax - 2a be the given polynomial.
x - 2 is a factor of p(x).
Thus, p(2) = 0
(2)2 + 3a × 2 - 2a = 0
4 + 4a = 0
a = -1
Hence, correct option is (d).
Question 2
If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =
(a) -6
(b) -7
(c) -8
(d) -10
Solution 2
Since, p(x) = x3 + 6x2 + 4x + k is exactly divisible by x + 2,
(x + 2) is a factor of p(x).
So, p(-2) = 0
i.e (-2)3 + 6(-2)2 + 4(-2) + k = 0
-8 + 24 - 8 + k = 0
24 - 16 + k = 0
8 + k = 0
k = -8
Hence, correct option is (c).
Question 3
If x - a is a factor of x3 - 3x2a + 2a2x + b, then the value of b is
(a) 0
(b) 2
(c) 1
(d) 3
Solution 3
Let p(x) = x3 - 3x2a + 2a2x + b
(x - a) is a factor of p(x).
So, p(a) = 0
a3 - 3a2.a + 2a2.a + b = 0
a3 - 3a3 + 2a3 + b = 0
3a3 - 3a3 + b = 0
b = 0
Hence, correct option is (a).
Question 4
If x140 + 2x151 + k is divisible by x + 1, then the value of k is
(a) 1
(b) -3
(c) 2
(d) -2
Solution 4
Let p(x) = x140 + 2x151 + k
Since p(x) is divisible by (x + 1),
(x + 1) is a factor of p(x).
So, p(-1) = 0
(-1)140 + 2(-1)151 + k = 0
1 + 2(-1) + k = 0
1 - 2 + k = 0
k - 1 = 0
k = 1
Hence, correct option is (a).
Question 5
If x + 2 is a factor of x2 + mx + 14, then m =
(a) 7
(b) 2
(c) 9
(d) 14
Solution 5
If x + 2 is a factor of x2 + mx + 14,
then at x = -2,
x2 + mx + 14 = 0
i.e. (-2)2 + m(-2) + 14 = 0
4 - 2m + 14 = 0
2m = 18
m = 9
Hence, correct option is (c).
Question 6
If x - 3 is a factor of x2 - ax - 15, then a =
(a) -2
(b) 5
(c) -5
(d) 3
Solution 6
x - 3 is a factor of x2 - ax - 15,
then at x = 3,
x2 - ax - 15 = 0
i.e. (3)2 - a(3) - 15 = 0
9 - 3a - 15 = 0
a = -2
Hence, correct option is (a).
Question 7
If x51 + 51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50
Solution 7
Question 8
If x + 1 is a factor of the polynomial 2x2 + kx, then k =
(a) -2
(b) -3
(c) 4
(d) 2
Solution 8
x + 1 is a factor of p(x) = 2x2 + kx
Then, p(-1) = 0
i.e. 2(-1)2 + k(-1) = 0
2 - k = 0
k = 2
Hence, correct option is (d).
Question 9
If x + α is a factor of x4 - a2x2 + 3x - 6a, then a =
(a) 0
(b) -1
(c) 1
(d) 2
Solution 9
Question 10
The value of k for which x - 1 is a factor of 4x3 + 3x2 - 4x + k, is
(a) 3
(b) 1
(c) -2
(d) -3
Solution 10
Question 11
If x + 2 and x - 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively
(a) 5 and -3
(b) 17 and -8
(c) 7 and -18
(d) 23 and -19
Solution 11
If (x + 2) and (x - 1) are factors of polynomial x3 + 10x2 + mx + n,
then x = -2, x = +1 will satisfy the polynomial.
Let p(x) = x3 + 10x2 + mx + n
Then, p(-2) = 0
(-2)3 + 10(-2)2 + m(-2) + n = 0
-8 + 40 - 2m + n = 0
32 - 2m + n = 0 ....(1)
And, p(1) = 0
(1)3 + 10(1)2 + m(1) + n = 0
1 + 10 + m + n = 0
11 + m + n = 0 ....(2)
Substracting equation (1) from equation (2), we get
-21 + 3m = 0
3m = 21
m = 7
Substituting m = 7 in equation (2),
11 + 7 + n = 0
18 + n = 0
n = -18
Hence, correct option is (c).
Chapter 6 - Factorisation of Polynomials Exercise 6.35
Question 12
Solution 12
Question 13
When x3 - 2x2 + ax - b is divided by x2 - 2x - 3, then remainder is x - 6. The values of a and b are respectively
(a) -2, -6
(b) 2 and -6
(c) -2 and 6
(d) 2 and 6
Solution 13
Let p(x) = x3 - 2x2 + ax - b, r(x) = x - 6 and q(x) = x2 - 2x - 3
Then q(x) is a factor of [p(x) - r(x)]
{because if p(x) is divided by q(x), remainder is r(x). So, [p(x) - r(x)] will be exactly divided by q(x)}
Now, q(x) = x2 - 2x - 3 = (x - 3)(x + 1)
If q(x) is a factor of [p(x) - r(x)] then (x - 3) and (x + 1) are also factors of [p(x) - r(x)]
So, at x = 3 and x = -1, p(x) - r(x) will be zero.
Now p(3) - r(3) = 0
i.e. (3)3 - 2(3)2 + a(3) - b - (3 - 6) = 0
i.e. 27 - 18 + 3a - b + 3 = 0
i.e. 3a - b + 12 = 0 ....(1)
And, p(-1) - r(-1) = 0
i.e. (-1)3 - 2(-1)2 + a(-1) - b - (-1 - 6) = 0
i.e. -1 - 2 - a - b + 7 = 0
i.e -a - b + 4 = 0 ....(2)
Subtracting equation (2) from equation (1), we get
4a + 8 = 0
a = -2
From (2), -(-2) - b + 4 = 0
b = 6
Hence, correct option is (c).
Question 14
One factor of x4 + x2 - 20 is x2 + 5. The other factor is
(a) x2 - 4
(b) x - 4
(c) x2 - 5
(d) x + 2
Solution 14
x4 + x2 - 20
= x4 + 5x2 - 4x2 - 20
=x2(x2 + 5) - 4(x2 + 5)
= (x2 + 5)(x2 - 4)
So, other factor is x2 - 4.
Hence, correct option is (a).
Question 15
If (x - 1) is a factor of polynomial f(x) but not of g(x), then it must be a factor of
(a) f(x) g(x)
(b) -f(x) + g(x)
(c) f(x) - g(x)
(d) {f(x) + g(x)} g(x)
Solution 15
Question 16
(x + 1) is a factor of xn + 1 only if
(a) n is an odd integer
(b) n is an even integer
(c) n is a negative integer
(d) n is a positive integer
Solution 16
If x + 1 is a factor of xn + 1,
then, at x = -1, xn + 1 = 0
(-1)n + 1 = 0
(-1)n = -1
(-1)n will be equal to -1 if and only if n is an odd integer.
If n is even, then (-1)n = 1
So, n should be an odd integer.
Hence, correct option is (a).
Question 17
If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + 3 + 5k, then the value of k is
(a) 0
(b) 2/5
(c) 5/2
(d) -1
Solution 17
Question 18
If (3x - 1)7 = a7x7 + a6x6 + a5x5 + ... + a1x + a0, then a7 + a6 + a5 + ... + a1 + a0 =
(a) 0
(b) 1
(c) 128
(d) 64
Solution 18
Correct option (c)
(3x - 1)7 = a7x7 + a6x6 + ......... + a1x + a0 ....(1)
Putting x = 1 in equation (1), we have
[3(1) - 1]7 = a7 + a6 + ..... + a1 + a0
So, a7 + a6 + a5 + ..... + a1 + a0 = 27 = 128
Hence, correct option is (c).
Question 19
Solution 19
Question 20
If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e, then
(a) a + c + e = b + d
(b) a + b + e = c + d
(c) a + b + c = d + e
(d) b + c + d = a + e
Solution 20
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