# RD SHARMA Solutions for Class 9 Maths Chapter 6 - Factorisation of Polynomials

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## Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.1

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 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 .
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

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
Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 ## Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.3

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 ## Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.4

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 Solution 15 Solution 16 Solution 17 Solution 18 Solution 19 Solution 20 Solution 21 Solution 22 Solution 23 Solution 24 Solution 25 ## Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.5

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 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)
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)

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(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)

Solution 14 Solution 15 Solution 16 Solution 17 Solution 18 ## Chapter 6 - Factorisation of Polynomials Exercise 6.34

Solution 12

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).

Solution 13

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).

Solution 14

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).

Solution 15

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).

Solution 16

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).

Solution 17

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).

Solution 18 Solution 19

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).

Solution 20 Solution 21 Solution 22

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

Solution 27

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).

Solution 28 Solution 29

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 = 2= 128

Hence, correct option is (c).

Solution 30 Solution 31 Solution 23 Solution 24

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).

Solution 25

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).

Solution 26 ### STUDY RESOURCES

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