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

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

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Answer 1 · 2016-10-10T05:22:53

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

Explanation:

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

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

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

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

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

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

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

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

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