Polynomials – Exercise 8.4 – Class X

Previous Exercise – Polynomials – Exercise 8.3 – Class X


Polynomials – Exercise 8.4

  1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x3 – 3x2 + 6x – 20 ; g(x) = x – 2

(ii) p(x) = 2x4 + x3 + 4x2 – x – 7 ; g(x) = x + 2

(iii) p(x) = 3x4 + 3x2 – 2x2 – 9x – 12 ; g(x) = x – 1/2

(iv) p(x) = 3x3 + x2 – 20x + 12 ; g(x) = 3x – 2

(v) p(x) = 2x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x2 – 3

  1. Find the valule of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10)
  2. If (x3 + ax – bx + 10) is divisible by x2 – 3x + 2 , find the values of a and b
  3. If both (x – 2) and (x – 1/2)are factors of (ax2 + 5x + b), show that a = b

Polynomials – Exercise 8.4 –Solutions:

  1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x3 – 3x2 + 6x – 20 ; g(x) = x – 2

Solution:

Let p(x) = x3 – 3x2 + 6x – 20

By factor theorem, (x – 2) is a factor of p(x) if p(2) = 0

p(2) = 23 – 3(2)2 + 6(2) – 20

= 8 – 12 + 12 – 20

= – 12

⸫ (x – 2) is not a factor of p(x) = x3 – 3x2 + 6x – 20


(ii) p(x) = 2x4 + x3 + 4x2 – x – 7 ; g(x) = x + 2

Solution:

Let p(x) = 2x4 + x3 + 4x2 – x – 7

By factor theorem, (x + 2) is a factor of p(x) if p(-2) = 0

p(-2) = 2(-2)4 + (-2)3 + 4(-2)2 – (-2) – 7

= 32 – 8 + 16 + 2 – 7

= 35

⸫ (x + 2) is not a factor of p(x) = 2x4 + x3 + 4x2 – x – 7


 (iii) p(x) = 3x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x – 1/2

Solution:

Let p(x) = 3x4 + 3x3 – 2x2 – 9x – 12

By factor theorem, (x – 1/2) is a factor of p(x) if p(1/2) = 0

p(1/2) = 3(1/2)4 + 3(1/2)3 – 2(1/2)2 – 9(1/2) – 12

= 3(1/16) + 3(1/8) – 2(1/4) – 9(1/2) – 12

= – 263/16

⸫ (x – 1/2) is not a factor of p(x) = 3x4 + 3x2 – 2x2 – 9x – 12


 (iv) p(x) = 3x3 + x2 – 20x + 12 ; g(x) = 3x – 2

Solution:

Let p(x) = 3x3 + x2 – 20x + 12

By factor theorem, (3x – 2) is a factor of p(x) if p(2/3) = 0

p(2/3) = 3(2/3)3 + (2/3)2 – 20(2/3) + 12

= 0

⸫ (3x – 2) is a factor of p(x) = 3x3 + x2 – 20x + 12


 (v) p(x) = 2x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x2 – 3

Solution:

Let p(x) = 2x4 + 3x3 – 2x2 – 9x – 12

By factor theorem, (x2 – 3) is a factor of p(x) if p(√3) = 0

p (√3) = 2 (√3)4 + 3 (√3)3 – 2 (√3)2 – 9 (√3) – 12

= 2x3x3 + 3×3√3 – 2×3 – 9√3 – 12

= 18 + 9√3 – 6 – 9√3 – 12

= 0

⸫ (x2 – 3) is a factor of p(x) = 2x4 + 3x3 – 2x2 – 9x – 12


  1. Find the value of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10)

Solution:

Let p(x) = x3 – 3x2 + ax – 10

By factor theorem, (x – 5) is a factor of p(x) if p(5) = 0

p(x) = x3 – 3x2 + ax – 10 = 0

53 – 3(52) + a(5) – 10 = 0

125 – 75 + 5a – 10 = 0

40 + 5a = 0

5a = – 40

a = –40/5 = – 8


  1. If (x3 + ax2 – bx + 10) is divisible by x2 – 3x + 2 , find the values of a and b

Solution:

p(x) = x3 + ax2 – bx + 10

g(x) = x2 – 3x + 2

g(x) = x2 –x – 2x + 2

g(x) = x(x – 1)-2(x – 1)

g(x) = (x – 1)(x – 2)

p(x) = (x3 + ax2 – bx + 10) is divisible by g(x) = (x – 1)(x – 2) , then we have p(2) = 0 and p(1) = 0

⇒ p(2) = p(1)

We have to find the values of a and b.

p(x) = x3 + ax2 – bx + 10

p(2) = (2)3 + a(2)2 – b(2) + 10 = 8 + 4a – 2b + 10 = 4a – 2b + 18 = 2a – b + 9 = 0

p(1) = (1)3 + a(1)2 – b(1) + 10 = 1 + a – b + 10 = a – b + 11 = 0

Since p(2) = p(1)

2a – b + 9 = a – b + 11

2a – a = 11 – 9

a = 2

Substitute the value of a in p(2) = 0

p(2) = a – b + 11 = 0

⇒ 2 – b + 11 = 0

-b = -11 – 2 = – 13

b = 13

Therefore, a = 2 and b = 13


  1. If both (x – 2) and (x – 1/2)are factors of (ax2 + 5x + b), show that a = b

Solution:

p(x) = ax2 + 5x + b

By factor theorem, (x – 2) and (x – 1/2) is a factor of p(x) then p(2) = p(1/2) = 0

p(2) = a(2)2 + 5(2) + b = 4a + b + 10 = 0

p(1/2) = a(1/2)2 + 5(1/2) + b = a/4 + 5/2 + b = 0

Since p(2) = p(1/2)

4a + b + 10 = a/4 + 5/2 + b

4a + 10 = a/4 + 5/2

4a – a/4 = 5/2 – 10

16a – a/4 = 5-20/2

15a/4 = 15/2

a/2 = 1

a = 2

Substitute the value of p(2) = 4a + b + 10 = 0

4a + b + 10 = 0

8 + b + 10 = 0

b = -18


Next exercise – Polynomials – Exercise 8.5 – Class X