How do you use the remainder theorem to find the remainder for each division $(x^{5} + 32) \div (x + 2)$ $(x^5+32)div(x+2)$ ?

Question

1615 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T11:05:13

0

the Remainder theorem states :

if a polynomial $P (x)$ $" "P(x)" "$ is divided by $(x - a)$ $" "(x-a)" "$ the remainder is $P (a)$ $" "P(a)" "$

proof:

$P (x) = (x - a) Q (x) + R$ $P(x)=(x-a)Q(x)+R$

$P(a)=cancel((a-a)Q(x))+R$

$:.P(a)=R$

$(x^5+32)-:(x+2)$

$P(x)=(x^5+32)$

to find remainder

$P(-2)=(-2)^5+32=-32+32=0$

remainder $" "=0" "$ which implies $" "(x+2)" "$ is a factor of $" " (x^5+32)$