How do you use the remainder theorem to determine the remainder when the polynomial $\frac{n^{4} - 3 n^{2} - 5 n + 2}{n - 2}$ $(n^4-3n^2-5n+2)/(n-2)$ ?

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How do you use the remainder theorem to determine the remainder when the polynomial $\frac{n^{4} - 3 n^{2} - 5 n + 2}{n - 2}$ $(n^4-3n^2-5n+2)/(n-2)$ ?

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Answer 1 · 2016-12-30T14:36:15

$R = - 4$ $R=-4$

Explanation:

The remainder theorem states that if a polynomial $P (x)$ $" "P(x)" "$ is divided by the linear factor $(x - a)$ $(x-a)" "$ then the remainder is $P (a)$ $" "P(a)$

proof;

let $P (x)$ $" "P(x) " "$ be divided by $(x - a)$ $" "(x-a)" "$ to give quotient $Q (x)$ $" "Q(x)" "$ and remainder $R$ $" "R$

then: $P (x) = (x - a) Q (x) + R$ $" "P(x)=(x-a)Q(x)+R$

so: $" "P(a)=cancel((a-a)Q(a))+R$

$" ":.P(a)=R" "$ as required.

In this case we have : $(n^4-3n^2-5n+2)-:(n-2)$

$P(n)=n^4-3n^2-5n+2$

the divisor is $" "n-2=>a=2$

$R=P(2)=2^4-3xx2^2-5xx2+2$

$R=P(2)=16-12-10+2=-4$

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How do you use the remainder theorem to determine the remainder when the polynomial $\frac{n^{4} - 3 n^{2} - 5 n + 2}{n - 2}$ $(n^4-3n^2-5n+2)/(n-2)$ ?

1 Answer

Explanation:

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Explanation:

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How do you use the remainder theorem to determine the remainder when the polynomial $\frac{n^{4} - 3 n^{2} - 5 n + 2}{n - 2}$ $(n^4-3n^2-5n+2)/(n-2)$ ?