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

  • Last Updated : 11 Feb, 2021
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Question 13. Find the value of k if x – 3 is a factor of k2x3 – kx2 + 3kx – k

Solution:

Let, f(x) = k2x3 – kx2 + 3kx – k

According to factor theorem 

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

⇒ x – 3 = 0



⇒ x = 3

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

f(3) = k2(3)3 – k(3)2+ 3k(3) – k

= 27k2 – 9k + 9k – k

= 27k2 – k

= k( 27k – 1)

Equate f(3) to zero, to find k 

⇒ f(3) = 0

 ⇒ k(27k – 1) = 0

 ⇒ k = 0 and 27k – 1 = 0 

 ⇒ 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 x2 – 4 is a factor of f(x) = ax4 + 2x3 – 3x2 + bx – 4

Solution:

Given: f(x) = ax4 + 2x3 – 3x2 + bx – 4, g(x) = x2 – 4

We need to find the factors of g(x)

⇒ x2 – 4 = 0

⇒ x2 = 4

⇒ x = √4



⇒ x = ±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 

⇒ 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 

⇒ 16a + 2b = 0 

⇒ 2(8a + b) = 0

⇒ 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

⇒ 16a – 2b – 32 = 0 

⇒ 16a – 2b – 32 = 0 

⇒ 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 \alpha, \beta if (x + 1) and (x + 2) are factors of x^3+3x^2-2\alpha x+\beta

Solution:

Given: f(x)=x^3+3x^2-2\alpha x+\beta 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,

⇒ x + 1 = 0

⇒ x = -1

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

f(x)=x^3+3x^2-2\alpha x+\beta

=(-1)^3+3(-1)^2-2\alpha (-1)+\beta

=-1+3+2\alpha +\beta

= 2\alpha +\beta +2         -(1)

Let,

⇒ x + 2 = 0

⇒ x = -2

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

f(x)=x^3+3x^2-2\alpha x+\beta

=(-2)^3+3(-2)^2-2\alpha (-2)+\beta    

=-8+12+4\alpha +\beta

=4\alpha +\beta +4          -(2)

On solving eq(1) and (2) 

⇒ 2\alpha +\beta+2 -(4\alpha +\beta +4)=0

-2\alpha -2 =0

2\alpha=-2



⇒ \alpha =-1

On substituting \alpha =-1  in equation(1)

2(-1)+\beta=-2

\beta=-2+2

\beta=0

The values are \alpha =-1  and \beta=0

Question 16. Find the values of p and q so that x4 + px3 + 2x2 – 3x + q is divisible by x2 – 1

Solution:

Given: f(x) = x4 + px3 + 2x2 – 3x + q, g(x) = x2 – 1

First, we need to find the factors of x2 – 1 

⇒ x2 – 1 = 0



⇒ x2 = 1

⇒  x = ±1

⇒ (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 x4 + ax3 – 3x2 + 2x + b

Solution:

Given: f(x) = x4 + ax3 – 3x2 + 2x + b

The 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

⇒ x + 1 = 0

 ⇒ 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

 ⇒ x – 1 = 0

⇒ 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 x3 + ax2 – bx + 10 is divisible by x3 – 3x + 2, find the values of a and b 

Solution:

Given: f(x) = x3 + ax2 – bx + 10, g(x) = x3 – 3x + 2   

First we need to find the factors of g(x)

g(x) = x3 – 3x + 2

=x3 – 2x – x + 2



= x(x – 2) – 1(x – 2)

= (x – 1)(x – 2) are the factors

Let us take (x – 1)

⇒ x – 1 = 0

⇒ x = 1

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

f(1) = 13 + a(1)2 – b(1) + 10 

= 1 + a – b + 10

= a – b +11         -(1)

Let us take (x – 2)

⇒ x – 2 = 0

⇒ x = 2

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

f(2) = 23 + a(2)2 – b(2) + 10 

= 8 + 4a – 2b + 10

= 4a – 2b + 18

Equating f(2) to zero 

⇒  4a – 2b +18 = 0 

⇒ 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 

⇒ 2 – b = – 11

⇒ -b = -11 – 2

⇒ b = 13

The value is a = 2 and b = 13

Question 19. If both (x + 1) and (x – 1) are the factors of ax3 + x2 – 2x + b, Find the values of a and b 

Solution:

Given: f(x) = ax3 + x2 – 2x + b, (x + 1) and (x – 1) are the factors 

According to factor theorem, 

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

Let, x – 1 = 0

⇒ 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

⇒ 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 

⇒ a + b = 1

⇒ -a + b = -3

⇒ 2b = -2

⇒ b = -1

On substituting the b in eq (1) 

⇒ a – 1 = 1 

⇒ a = 2

The values are a = 2 and b = -1

Question 20. What must be added to x3 – 3x2 – 12x + 19 so that the result is exactly divisible by x2 + x – 6  

Solution:

Given: p(x) = x3 – 3x2 – 12x + 19, g(x) = x2 + x – 6

According 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)

⇒ f(x) = p(x) + r(x)

= f(x) = x3 – 3x2 – 12x + 19 + ax + b

We know that, g(x) = x2 + x – 6

First, we find the factors of g(x)

⇒ g(x) = x2  + x – 6

= x2 + 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

⇒ 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

⇒ 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

⇒ -3a + b = -1

⇒ 2a + b = 9

⇒ -5a = -10

⇒ a = 2

On substituting the value of a in eq(1) 

⇒ -3(2) + b = -1

⇒ -6 + b = -1

⇒ b = 5

Therefore, r(x) = ax + b

= 2x + 5

Hence, x3 – 3x2 – 12x + 19 is divisible by x2 + x – 6 when it is added by 2x + 5.

Question 21. What must be added to x3 – 6x2 – 15x + 80 so that the result is exactly divisible by x2 + x – 12

Solution:

Let p(x) = x3 – 6x2 – 15x + 80, q(x) = x2 + x – 12

According 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) = x2 + x – 12

= x2 + 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

⇒ f(3) = 33 – 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 – b 

⇒ -64 – 96 – 4a + 60 + 80 – b = 0 

⇒ 4a – b – 20 = 0

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

4a – b – 20 = 0         -(2) 

⇒ 7a – 28 = 0

⇒ a = 28/7

⇒ a = 4 

On Putting a = 4 in eq (1), we get 

⇒ -3(4) – b = -8

⇒ -b – 12 = -8

⇒ -b = -8 + 12

⇒ b = -4

On substituting a and b values in r(x)

⇒ r(x) = ax + b

⇒ 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 3x3 + x2 – 22x + 9 so that the result is exactly divisible by 3x2 + 7x – 6

Solution:

Let, p(x) = 3x3+ x2– 22x + 9 and q(x) = 3x2 + 7x – 6

According 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)

⇒ f(x) = 3x3 + x2 – 22x + 9(ax + b)

= 3x3 + x2+ x(a – 22) + b + 9

 We know that,

q(x) = 3x2 + 7x – 6

= 3x2 + 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

f(\frac23)=3(\frac23)^3+(\frac23)^2+\frac23(a-22)+b+9

3(\frac8{27})+\frac49+\frac23a-\frac{44}3+b+9

\frac{12}9+\frac23a-\frac{44}3+b+9

\frac{12+6a-132+9b+81}{9}

Equate to zero 

⇒ \frac{12+6a-132+9b+81}{9}=0

⇒ 6a + 9b – 39 = 0

⇒ 2a + 3b – 13 = 0         -(1)

Similarly, 

Let, x + 3 = 0

⇒ 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

⇒ -3a + b + 3 = 0 

Multiply the given equation by 3 

⇒ -9a + 3b + 9 = 0         -(2)

On Subtracting eq (1) from eq(2) 

⇒ -9a + 3b + 9 – 2a – 3b + 13 = 0

⇒-11a + 22 = 0

⇒-11a = -22

⇒ a = 2

On substituting value of a in eq (1) 

⇒ -3(2) + b = -3

⇒ -6 + b = -3

⇒ 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) x3 – 2ax2 + ax – 1

(ii) x5 – 3x4 – ax3 + 3ax2 + 2ax + 4

Solution:

(i) Let f(x) = x3 – 2ax2 + ax – 1

According to factor theorem

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

Let, x – 2 = 0

⇒ x = 2

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

f(2) = 23 – 2a(2)2 + a(2) – 1  

= 8 – 8a + 2a – 1

= -6a + 7

Equate f(2) to zero 

 ⇒ -6a + 7 = 0

⇒ -6a = -7

⇒ a = 7/6



So, (x – 2) is the factor of f(x)

(ii) Let f(x) = x5 – 3x4 – ax3 + 3ax2 + 2ax + 4

According to factor theorem

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

Let, x – 2 = 0

⇒ x = 2 

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

f(2) = 25 – 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

⇒ 8a – 12 = 0

⇒ a = 12/8

⇒ 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) x6 – ax5 + x4 – ax3 + 3x – a + 2   

(ii) x5 – a2x3 + 2x + a + 1

Solution:

(i) Let, f(x) = x6 – ax5 + x4 – ax3 + 3x – a + 2

Here, x – a = 0

⇒ x = a

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

f(a) = a6 – a(a)5 + (a)4 – a(a)3 + 3(a) – a + 2

= a6 – a6 + a4 – a4 + 3a – a + 2  

= 2a+2

Equate to zero

⇒ 2a + 2 = 0

⇒ 2(a + 1) = 0

⇒ a = -1

So, (x – a) is a factor of f(x)

(ii) Let, f(x) = x5 – a2x3 + 2x + a + 1

Here, x – a = 0

⇒ x = a

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

f(a) = a5 – a2(a)3 + 2(a) + a + 1

= a5 – a5 + 2a + a + 1  

= 3a + 1

Equate to zero

⇒ 3a + 1 = 0

⇒ 3a = -1

⇒ 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) x3 + ax2 – 2x + a +4 

(ii) x4 – a2x2 + 3x – a 

Solution:

(i) Let, f(x) = x3 + ax2 – 2x + a + 4

Here, x + a = 0

⇒ 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

⇒ 3a + 4 = 0

⇒ 3a = -4

⇒ a = -4/3

So, (x + a) is a factor of f(x) 

(ii) Let, f(x) = x4– a2x2 + 3x – a

Here, x + a = 0

⇒ x = -a

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

f(-a) = (-a)4 – a2(-a)2 + 3(-a) – a

= a4 – a4 – 3a – a

= -4a

Equate to zero

⇒ -4a = 0

⇒ a = 0

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

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