Question

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divided by $x+3$.

Answer 1

Let $f\left(x\right)=x^4+4x^3-5x^2-6x+7$.

Let $R$ be the remainder obtained when $f\left(x\right)$ is divided by $x+3$.

Equate $x+3$ to $0$ and put the value of $x$ in $f\left(x\right)$.

$x+3$=0

$⟹x=-3$

$R=(-3{)}^4+4(-3{)}^3-5(-3{)}^2-6(-3)+7$

$=81-108-45+18+7$

$=-47$

Thus required remainder is $-47$