Question

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-a{x}^2+6x-a,g\left(x\right)=x-a$.

Answer 1

$p\left(x\right)=x^3-a{x}^2+6x-a$

​$g\left(x\right)=x-a$

By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).

Putting x = a in p(x), we get

$p\left(a\right)=a^3-a\times a^2+6\times a-a=a^3-a^3+6a-a=5a$

∴ Remainder = 5a

Thus, the remainder when p(x) is divided by g(x) is 5a.