Previous Exercise – Polynomials – Exercise 8.3 – Class X
Polynomials – Exercise 8.4
- 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
- Find the valule of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10)
- If (x3 + ax – bx + 10) is divisible by x2 – 3x + 2 , find the values of a and b
- If both (x – 2) and (x – 1/2)are factors of (ax2 + 5x + b), show that a = b
Polynomials – Exercise 8.4 –Solutions:
- 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
- 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
- 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
- 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