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

## Factorisation of Polynomials Exercise Ex. 6.1

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### Solution 8

Binomial has two terms in it. So binomial of degree 35 can be written as x

^{35}+ 7 .

^{100}.

**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 .

## Factorisation of Polynomials Exercise Ex. 6.2

### Solution 1

f(x) = 2x

^{3}- 13x

^{2}+ 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

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## Factorisation of Polynomials Exercise Ex. 6.3

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## Factorisation of Polynomials Exercise Ex. 6.4

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## Factorisation of Polynomials Exercise Ex. 6.5

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### Solution 11

Let p(x) = x^{3} + 13x^{2 }+ 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 x^{3} + 13x^{2} + 32x + 20 by (x + 1)

By long division

Dividend = Divisor Quotient + Remainder

x

^{3}+ 13x

^{2}+ 32x + 20 = (x + 1) (x

^{2}+ 12x + 20) + 0

= (x + 1) (x

^{2}+ 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) = x^{3} - 3x^{2} - 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 x^{3} + 3x^{2} - 9x - 5 by x + 1

By long division

x

^{3}- 3x

^{2}- 9x - 5 = (x + 1) (x

^{2}- 4 x - 5) + 0

= (x + 1) (x

^{2}- 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) = 2y^{3} + y^{2} - 2y - 1

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,

^{3}+ y

^{2}- 2y - 1

= (y - 1) (2y

^{2}+3y + 1)

= (y - 1) (2y

^{2}+2y + y +1)

= (y - 1) [2y (y + 1) + 1 (y + 1)]

= (y - 1) (y + 1) (2y + 1)

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## Factorisation of Polynomials Exercise 6.34

### Solution 12

Let p(x) = x^{2} + 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) = x^{3} + 6x^{2} + 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) = x^{3} - 3x^{2}a + 2a^{2}x + b

(x - a) is a factor of p(x).

So, p(a) = 0

a^{3} - 3a^{2}.a + 2a^{2}.a + b = 0

a^{3} - 3a^{3} + 2a^{3} + b = 0

3a^{3} - 3a^{3} + b = 0

b = 0

Hence, correct option is (a).

### Solution 15

Let p(x) = x^{140} + 2x^{151} + 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 x^{2} + mx + 14,

then at x = -2,

x^{2} + 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 x^{2} - ax - 15,

then at x = 3,

x^{2} - 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) = 2x^{2} + 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 x^{3} + 10x^{2} + mx + n,

then x = -2, x = +1 will satisfy the polynomial.

Let p(x) = x^{3} + 10x^{2} + 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).

## Factorisation of Polynomials Exercise 6.35

### Solution 27

If x + 1 is a factor of x^{n} + 1,

then, at x = -1, x^{n} + 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} = a_{7}x^{7} + a_{6}x^{6} + ......... + a_{1}x + a_{0} ....(1)

Putting x = 1 in equation (1), we have

[3(1) - 1]^{7} = a_{7} + a_{6} + ..... + a_{1} + a_{0}

So, a_{7} + a_{6} + a_{5} + ..... + a_{1} + a_{0} = 2^{7 }= 128

Hence, correct option is (c).

### Solution 30

### Solution 31

### Solution 23

### Solution 24

Let p(x) = x^{3} - 2x^{2} + ax - b, r(x) = x - 6 and q(x) = x^{2} - 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) = x^{2} - 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

x^{4} + x^{2} - 20

= x^{4} + 5x^{2} - 4x^{2} - 20

=x^{2}(x^{2} + 5) - 4(x^{2} + 5)

= (x^{2} + 5)(x^{2} - 4)

So, other factor is x^{2} - 4.

Hence, correct option is (a).

### Solution 26