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

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

1. $x^{3}-2 a x^{2}+a x-1$

2. $x^{5}-3 x^{4}-a x^{3}+3 a x^{2}+2 a x+4$

Solution:

(1) Let $f(x)=x^{3}-2 a x^{2}+a x-1$

from factor theorem

if (x – 2) is the factor of f(x) the f(2) = 0

let, x – 2 = 0

⟹ x = 2

Substitute x value in f(x)

$f(2)=2^{3}-2 a(2)^{2}+a(2)-1$

= 8 – 8a + 2a – 1

= – 6a + 7

Equate f(2) to zero

⟹ – 6a + 7 = 0

⟹ – 6a = -7

⟹ a= 76

When, (x – 2) is the factor of f(x) then a = 76

(2) Let, $f(x)=x^{5}-3 x^{4}-a x^{3}+3 a x^{2}+2 a x+4$

from factor theorem

if (x – 2) is the factor of f(x) the f(2) = 0

let, x – 2 = 0

⟹ x = 2

Substitute x value in f(x)

$f(2)=2^{5}-3(2)^{4}-a(2)^{3}+3 a(2)^{2}+2 a(2)+4$

= 32 – 48 – 8a + 12 + 4a + 4

= 8a – 12

Equate f(2) to zero

⟹ 8a – 12 = 0

⟹ 8a = 12

⟹ a = 12/8

= 3/2

So, when (x – 2) is a factor of f(x) then a = 32