# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.4 | Set 2

### Question 13. Find the value of k if x – 3 is a factor of k^{2}x^{3} – kx^{2 }+ 3kx – k

**Solution:**

Let, f(x) = k

^{2}x^{3}– kx^{2 }+ 3kx – kAccording to factor theorem

If x – 3 is the factor of f(x) then f(3) = 0

â‡’ x – 3 = 0

â‡’ x = 3

On substituting the value of x in f(x), we get

f(3) = k

^{2}(3)^{3}– k(3)^{2}+ 3k(3) – k= 27k

^{2}– 9k + 9k – k= 27k

^{2}– k= k( 27k – 1)

Equate f(3) to zero, to find k

â‡’ f(3) = 0

â‡’ k(27k – 1) = 0

â‡’ k = 0 and 27k – 1 = 0

â‡’ k = 0 and k = 1/27

When k = 0 and 1/27, (x – 3) will be the factor of f(x)

### Question 14. Find the value of a and b, if x^{2 }– 4 is a factor of f(x) = ax^{4} + 2x^{3 }– 3x^{2 }+ bx – 4

**Solution:**

Given:f(x) = ax^{4}+ 2x^{3 }– 3x^{2 }+ bx – 4, g(x) = x^{2 }– 4We need to find the factors of g(x)

â‡’ x

^{2 }– 4 = 0â‡’ x

^{2 }= 4â‡’ x = âˆš4

â‡’ x = Â±2

(x – 2) and (x + 2) are the factors

According to factor theorem

If (x – 2) and (x + 2) are the factors of f(x)

the result of f(2) and f(-2) should be zero

Let, x – 2 = 0

â‡’ x = 2

On substituting the value of x in f(x), we get

f(2) = a(2)

^{4 }+ 2(2)^{3 }– 3(2)^{2 }+ b(2) – 4= 16a + 2(8) – 3(4) + 2b – 4

= 16a + 2b + 16 – 12 – 4

= 16a + 2b

Equate the value of f(2) to zero

â‡’ 16a + 2b = 0

â‡’ 2(8a + b) = 0

â‡’ 8a + b = 0 -(1)

Let, x + 2 = 0

x = -2

On substituting the value of x in f(x), we get

f(-2) = a(-2)

^{4}+ 2(-2)^{3}– 3(-2)^{2 }+ b(-2) – 4= 16a + 2(-8) – 3(4) – 2b – 4

= 16a – 16 – 12 – 2b – 4

= 16a – 2b – 32

Equate the value of f(-2) to zero

â‡’ 16a – 2b – 32 = 0

â‡’ 16a – 2b – 32 = 0

â‡’ 8a – b = 16 -(2)

On solving equation (1) and (2)

8a + b = 0

8a – b = 16

16a = 16

a = 1

On substituting the value of a in eq (1), we get

8(1) + b = 0

b = -8

The values are a = 1 and b = -8

### Question 15. Find if (x + 1) and (x + 2) are factors of

**Solution:**

Given:and factors are (x + 1) and (x + 2)According to factor theorem,

If they are the factors of f(x) then results of f(-2) and f(-1) should be zero.

Let,

â‡’ x + 1 = 0

â‡’ x = -1

On substituting the value of x in f(x), we get

=

=

= 2\alpha +\beta +2 -(1)

Let,

â‡’ x + 2 = 0

â‡’ x = -2

On substituting the value of x in f(x), we get

=

-(2)

On solving eq(1) and (2)

â‡’

â‡’

â‡’

â‡’

On substituting in equation(1)

â‡’

â‡’

â‡’

The values are and

### Question 16. Find the values of p and q so that x^{4 }+ px^{3 }+ 2x^{2 }– 3x + q is divisible by x^{2 }– 1

**Solution:**

Given:f(x) = x^{4 }+ px^{3 }+ 2x^{2 }– 3x + q, g(x) = x^{2 }– 1First, we need to find the factors of x

^{2 }– 1â‡’ x

^{2 }– 1 = 0â‡’ x

^{2 }= 1â‡’ x = Â±1

â‡’ (x + 1) and (x – 1)

According to factor theorem

If x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let, us take x + 1 = 0

x = -1

On substituting the value of x in f(x), we get

f(-1) = (-1)

^{4 }+ p(-1)^{3 }+ 2(-1)^{2 }– 3(-1) + q= 1 – p + 2 + 3 + q

= -p + q + 6 -(1)

Let us take, x – 1 = 0

x = 1

On substituting the value of x in f(x), we get

f(1) = (1)

^{4 }+ p(1)^{3 }+ 2(1)^{2 }– 3(1) + q= 1 + p + 2 – 3 + q

= p + q -(2)

On solving eq(1) and (2), we get

-p + q = -6

p + q = 0

2q = -6

q = -3

On substituting q value in eq(2), we get

p + q = 0

p – 3 = 0

p = 3

The value of p = 3 and q = -3

### Question 17. Find the values of a and b so that (x + 1) and (x – 1) are the factors of x^{4 }+ ax^{3 }– 3x^{2 }+ 2x + b

**Solution:**

Given:f(x) = x^{4 }+ ax^{3 }– 3x^{2 }+ 2x + bThe factors are (x + 1) and (x – 1)

According to factor theorem

If x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let, us take x + 1

â‡’ x + 1 = 0

â‡’ x = -1

On substituting the value of x in f(x), we get

f(-1) = (-1)

^{4 }+ a(-1)^{3 }– 3(-1)^{2 }+ 2(-1) + b= 1 – a – 3 – 2 + b

= -a + b – 4 -(1)

Let, us take x – 1

â‡’ x – 1 = 0

â‡’ x = 1

On substituting the value of x in f(x), we get

f(1) = (1)

^{4 }+ a(1)^{3 }– 3(1)^{2 }+ 2(1) + b= 1 + a – 3 + 2 + b

= a + b -(2)

On solving equation (1) and (2)

-a + b = 4

a + b = 0

2b = 4

b = 2

On substituting value of b in eq (2), we get

a + 2 = 0

a = -2

The values are a = -2 and b = 2

### Question 18. If x^{3 }+ ax^{2 }– bx + 10 is divisible by x^{3 }– 3x + 2, find the values of a and b

**Solution:**

Given:f(x) = x^{3 }+ ax^{2 }– bx + 10, g(x) = x^{3}– 3x + 2First we need to find the factors of g(x)

g(x) = x

^{3 }– 3x + 2=x

^{3 }– 2x – x + 2= x(x – 2) – 1(x – 2)

= (x – 1)(x – 2) are the factors

Let us take (x – 1)

â‡’ x – 1 = 0

â‡’ x = 1

On substituting the value of x in f(x), we get

f(1) = 1

^{3 }+ a(1)^{2 }– b(1) + 10= 1 + a – b + 10

= a – b +11 -(1)

Let us take (x – 2)

â‡’ x – 2 = 0

â‡’ x = 2

On substituting the value of x in f(x), we get

f(2) = 2

^{3 }+ a(2)^{2 }– b(2) + 10= 8 + 4a – 2b + 10

= 4a – 2b + 18

Equating f(2) to zero

â‡’ 4a – 2b +18 = 0

â‡’ 2a – b + 9 = 0 -(2)

On solving eq(1) and (2), we get

a – b = -11

2a – b = -9

a = 2

On substituting the value of a in equation (1), we get

â‡’ 2 – b = – 11

â‡’ -b = -11 – 2

â‡’ b = 13

The value is a = 2 and b = 13

### Question 19. If both (x + 1) and (x – 1) are the factors of ax^{3 }+ x^{2 }– 2x + b, Find the values of a and b

**Solution:**

Given:f(x) = ax^{3 }+ x^{2 }– 2x + b, (x + 1) and (x – 1) are the factorsAccording to factor theorem,

If x = -1 & 1 are factors of f(x) then f(1) = 0 and f(-1) = 0

Let, x – 1 = 0

â‡’ x = 1

On substituting the value of x in f(x), we get

f(1) = a(1)

^{3 }+ (1)^{2 }– 2(1) + b= a +1 – 2 + b

= a + b – 1 -(1)

Let, x + 1 = 0

â‡’ x = -1

On substituting the value of x in f(x), we get

f(-1) = a(-1)

^{3 }+ (-1)^{2 }– 2(-1) + b= -a + 1 + 2 + b

= -a + b + 3 -(2)

On solving equation (1) and (2), we get

â‡’ a + b = 1

â‡’ -a + b = -3

â‡’ 2b = -2

â‡’ b = -1

On substituting the b in eq (1)

â‡’ a – 1 = 1

â‡’ a = 2

The values are a = 2 and b = -1

### Question 20. What must be added to x^{3 }– 3x^{2 }– 12x + 19 so that the result is exactly divisible by x^{2 }+ x – 6

**Solution:**

Given:p(x) = x^{3 }– 3x^{2 }– 12x + 19, g(x) = x^{2 }+ x – 6According to division algorithm when p(x) is divided by g(x),

the remainder will be the linear expression in x

Let, r(x) = ax + b is added to p(x)

â‡’ f(x) = p(x) + r(x)

= f(x) = x

^{3 }– 3x^{2 }– 12x + 19 + ax + bWe know that, g(x) = x

^{2 }+ x – 6First, we find the factors of g(x)

â‡’ g(x) = x

^{2 }+ x – 6= x

^{2 }+ 3x – 2x – 6= x(x + 3) – 2(x + 3)

= (x – 2)(x + 3)

According to factor theorem

If (x – 2) & (x + 3) are factors of f(x) then f(-3) = 0 and f(2) = 0

Let, x + 3 = 0

â‡’ x = -3

On substituting the value of x in f(x), we get

f(-3) = (-3)

^{3 }– 3(-3)^{2 }– 12(-3) + 19 + a(-3) + b= -27 – 27 – 3a + 24 + 19 + b

= -3a + b +1 -(1)

Let, x – 2 = 0

â‡’ x = 2

On substituting the value of x in f(x), we get

f(2) = (2)

^{3 }– 3(2)^{2 }– 12(2) + 19 + a(2) + b= 8 – 12 + 2a – 24 + b

= 2a + b – 9 -(2)

On solving eq(1) and eq (2), we get

â‡’ -3a + b = -1

â‡’ 2a + b = 9

â‡’ -5a = -10

â‡’ a = 2

On substituting the value of a in eq(1)

â‡’ -3(2) + b = -1

â‡’ -6 + b = -1

â‡’ b = 5

Therefore, r(x) = ax + b

= 2x + 5

Hence, x

^{3 }– 3x^{2 }– 12x + 19 is divisible by x^{2 }+ x – 6 when it is added by 2x + 5.

### Question 21. What must be added to x^{3 }– 6x^{2 }– 15x + 80 so that the result is exactly divisible by x^{2 }+ x – 12

**Solution:**

Let p(x) = x

^{3 }– 6x^{2 }– 15x + 80, q(x) = x^{2 }+ x – 12According to algorithm, when p(x) is divided by q(x) the remainder is a linear expression in x.

So, let r(x) = ax + b is subtracted from p(x), so that p(x) – q(x) is divisible by q(x)

Let, f(x) = p(x) – q(x)

q(x) = x

^{2 }+ x – 12= x

^{2 }+ 4x – 3x – 12= x(x + 4) – 3(x + 4)

=(x – 3)(x + 4)

Clearly, (x – 3) and (x + 4) are factors of q(x)

So, f(x) is divisible by q(x) if (x – 3) and (x + 4) are factors of q(x)

According to factor theorem

f(-4) = 0 and f(3) = 0

â‡’ f(3) = 3

^{3 }– 6(3)^{2 }– 3(a + 15) + 80 – b = 0= 27 – 54 – 3a – 45 + 80 – b

= -3a – b + 8 -(1)

Similarly,

f(-4) = 0

f(-4) = (-4)

^{3 }– 6(-4)^{2 }– 4(a + 15) + 80 – bâ‡’ -64 – 96 – 4a + 60 + 80 – b = 0

â‡’ 4a – b – 20 = 0

On subtracting eq (1) and eq (2), we get

4a – b – 20 = 0 -(2)

â‡’ 7a – 28 = 0

â‡’ a = 28/7

â‡’ a = 4

On Putting a = 4 in eq (1), we get

â‡’ -3(4) – b = -8

â‡’ -b – 12 = -8

â‡’ -b = -8 + 12

â‡’ b = -4

On substituting a and b values in r(x)

â‡’ r(x) = ax + b

â‡’ 4x – 4

Hence, p(x) is divisible by q(x), if r(x) = 4x – 4 is subtracted from it.

### Question 22. What must be added to 3x^{3 }+ x^{2 }– 22x + 9 so that the result is exactly divisible by 3x^{2 }+ 7x – 6

**Solution:**

Let, p(x) = 3x

^{3}+ x^{2}– 22x + 9 and q(x) = 3x^{2 }+ 7x – 6According to divisible theorem, when p(x) is divided by q(x), the reminder is linear equation in x.

Let, r(x) = ax + b is added to p(x), so that p(x) + r(x) is divisible by q(x)

f(x) = p(x) + r(x)

â‡’ f(x) = 3x

^{3 }+ x^{2 }– 22x + 9(ax + b)= 3x

^{3 }+ x^{2}+ x(a – 22) + b + 9We know that,

q(x) = 3x

^{2 }+ 7x – 6= 3x

^{2 }+ 9x – 2x – 6= 3x(x + 3) – 2(x + 3)

= (3x – 2)(x + 3)

So, f(x) is divisible by q(x) if (3x – 2) and (x + 3) are the factors of f(x)

From factor theorem

f(2/3) = 0 and f(-3) = 0

Let, 3x – 2 = 0

3x = 2

x = 2/3

=

=

=

Equate to zero

â‡’

â‡’ 6a + 9b – 39 = 0

â‡’ 2a + 3b – 13 = 0 -(1)

Similarly,

Let, x + 3 = 0

â‡’ x = -3

f(-3) = 3(-3)

^{3 }+ (-3)^{2 }– 3(a – 22) + b + 9= -81 + 9 – 3a + 66 + b + 9

= -3a + b + 3

Equate to zero

â‡’ -3a + b + 3 = 0

Multiply the given equation by 3

â‡’ -9a + 3b + 9 = 0 -(2)

On Subtracting eq (1) from eq(2)

â‡’ -9a + 3b + 9 – 2a – 3b + 13 = 0

â‡’-11a + 22 = 0

â‡’-11a = -22

â‡’ a = 2

On substituting value of a in eq (1)

â‡’ -3(2) + b = -3

â‡’ -6 + b = -3

â‡’ b = 3

Put the values in r(x)

r(x) = ax + b

= 2x + 3

Hence, p(x) is divisible by q(x), if r(x) = 2x + 3 is added to it.

### Question 23. If x – 2 is a factor of each of the following two polynomials, find the value of a in each case:

### (i) x^{3 }– 2ax^{2 }+ ax – 1

### (ii) x^{5 }– 3x^{4 }– ax^{3 }+ 3ax^{2 }+ 2ax + 4

**Solution:**

(i)Let f(x) = x^{3 }– 2ax^{2 }+ ax – 1According to factor theorem

If (x-2) is a factor of f(x) then f(2) = 0

Let, x – 2 = 0

â‡’ x = 2

On substituting the value of x in f(x), we get

f(2) = 2

^{3 }– 2a(2)^{2 }+ a(2) – 1= 8 – 8a + 2a – 1

= -6a + 7

Equate f(2) to zero

â‡’ -6a + 7 = 0

â‡’ -6a = -7

â‡’ a = 7/6

So, (x – 2) is the factor of f(x)

(ii)Let f(x) = x^{5 }– 3x^{4 }– ax^{3 }+ 3ax^{2 }+ 2ax + 4According to factor theorem

If (x – 2) is a factor of f(x) then f(2) = 0

Let, x – 2 = 0

â‡’ x = 2

On substituting the value of x in f(x), we get

f(2) = 2

^{5 }– 3(2)^{4 }– a(2)^{3 }+ 3a(2)^{2 }+ 2a(2) + 4= 32 – 48 – 8a + 12 + 4a + 4

= 8a -12

Equate f(2) to zero

â‡’ 8a – 12 = 0

â‡’ a = 12/8

â‡’ a = 3/2

So, (x – 2) is a factor of f(x)

### Question 24. In each of the following two polynomials, find the value of a, if (x – a) is a factor:

### (i) x^{6 }– ax^{5 }+ x^{4 }– ax^{3 }+ 3x – a + 2

### (ii) x^{5 }– a^{2}x^{3 }+ 2x + a + 1

**Solution:**

(i)Let, f(x) = x^{6 }– ax^{5 }+ x^{4 }– ax^{3 }+ 3x – a + 2Here, x – a = 0

â‡’ x = a

On substituting the value of x in f(x), we get

f(a) = a

^{6 }– a(a)^{5 }+ (a)^{4 }– a(a)^{3 }+ 3(a) – a + 2= a

^{6 }– a^{6 }+ a^{4 }– a^{4 }+ 3a – a + 2= 2a+2

Equate to zero

â‡’ 2a + 2 = 0

â‡’ 2(a + 1) = 0

â‡’ a = -1

So, (x – a) is a factor of f(x)

(ii)Let, f(x) = x^{5 }– a^{2}x^{3 }+ 2x + a + 1Here, x – a = 0

â‡’ x = a

On substituting the value of x in f(x), we get

f(a) = a

^{5 }– a^{2}(a)^{3 }+ 2(a) + a + 1= a

^{5 }– a^{5 }+ 2a + a + 1= 3a + 1

Equate to zero

â‡’ 3a + 1 = 0

â‡’ 3a = -1

â‡’ a = -1/3

So, (x – a) is a factor of f(x)

### Question 25. In each of the following two polynomials, find the value of a, if (x + a) is a factor:

### (i) x^{3} + ax^{2} – 2x + a +4

### (ii) x^{4} – a^{2}x^{2} + 3x – a

**Solution:**

(i)Let, f(x) = x^{3 }+ ax^{2 }– 2x + a + 4Here, x + a = 0

â‡’ x = – a

On substituting the value of x in f(x), we get

f(-a) = (-a)

^{3}+ a(-a)^{2}– 2(-a) + a + 4= 3a + 4

Equate to zero

â‡’ 3a + 4 = 0

â‡’ 3a = -4

â‡’ a = -4/3

So, (x + a) is a factor of f(x)

(ii)Let, f(x) = x^{4}– a^{2}x^{2 }+ 3x – aHere, x + a = 0

â‡’ x = -a

On substituting the value of x in f(x), we get

f(-a) = (-a)

^{4 }– a^{2}(-a)^{2 }+ 3(-a) – a= a

^{4 }– a^{4 }– 3a – a= -4a

Equate to zero

â‡’ -4a = 0

â‡’ a = 0

So, (x + a) is a factor of f(x)

## Please

Loginto comment...