How do you use the remainder theorem to find the remainder for the division $(n^4-n^3-10n^2+4n+24)div(n+2)$ ?

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Answer 1 · 2016-10-10T05:22:53

$f(-2) = 0""$
There is no remainder and $(n+2)$ is a factor of $n^4 -n^3-10n^2 +4n+24$

Let $f(n) =n^4 -n^3-10n^2 +4n+24$

Setting $n+2 = 0" "$ gives $n=-2$

Substitute $n=-2" into " f (n)""$ to find the remainder.

$f(-2) = (-2)^4 -(-2)^3-10(-2)^2 +4(-2)+24$

$f(-2) = 16+8-40-8+24 =0$

As this is 0, it means there is no remainder.

This tells us that $n+2$ is a factor of $n^4 -n^3-10n^2 +4n+24$ and therefore divides into it exactly without leaving a remainder.

How do you use the remainder theorem to find the remainder for the division (n^4-n^3-10n^2+4n+24)div(n+2)(n^4-n^3-10n^2+4n+24)div(n+2)?