Question

Find the remainder when $x^3$ – a$x^2+6x–a$ is divided by $x–a$.

Answer 1

Remainder theorem:

If $P\left(x\right)$ is polynomial and it is divided by $(x-a)$ then the remainder is $P\left(a\right)$.

Let $P\left(x\right)=x^3$ – a$x^2+6x–a$

It is given that, $x^3$ – a$x^2+6x–a$ is divided by $x–a$.

$\Rightarrow P\left(a\right)$ is remainder.

Now, lets substitute $x=a$ in $x^3$ – a$x^2+6x–a$

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

$\Rightarrow P\left(a\right)=a^3-a^3+5a$

$\Rightarrow P\left(a\right)=5a$

Therefore, the remainder when $x^3$ – a$x^2+6x–a$ is divided by $x–a$ is $5a$.