Question

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

Answer 1

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

By remainder theorem, when p(x) is divided by ( x-a), then the remainder = $p\left(a\right)$.

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

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

∴ Remainder = $5a$

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