Question:
In each of the following two polynomials, find the value of a, if (x - a) is a factor:
1. $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
2. $x^{5}-a^{2} x^{3}+2 x+a+1$
Solution:
(1) $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
let, $f(x)=x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
here, x - a = 0
⟹ x = a
Substitute the value of x in f(x)
$f(a)=a^{6}-a(a)^{5}+(a)^{4}-a(a)^{3}+3(a)-a+2$
$=a^{6}-a^{6}+(a)^{4}-a^{4}+3(a)-a+2$
= 2a + 2
Equate to zero
⟹ 2a + 2 = 0
⟹ 2(a + 1) = 0
⟹ a = -1
So, when (x - a) is a factor of f(x) then a = -1
(2) $x^{5}-a^{2} x^{3}+2 x+a+1$
let, $f(x)=x^{5}-a^{2} x^{3}+2 x+a+1$
here, x - a = 0
⟹ x = a
Substitute the value of x in f(x)
$f(a)=a^{5}-a^{2} a^{3}+2(a)+a+1$
= 3a + 1
Equate to zero
⟹ 3a + 1 = 0
⟹ 3a = -1
⟹ a = −13
So, when (x - a) is a factor of f(x) then a = −1/3