In each of the following two polynomials, find the value of a, if (x - a) is a factor:

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