# How do you use the remainder theorem to determine the remainder when the polynomial (n^4-3n^2-5n+2)/(n-2)?

Dec 30, 2016

$R = - 4$

#### Explanation:

The remainder theorem states that if a polynomial $\text{ "P(x)" }$ is divided by the linear factor $\left(x - a\right) \text{ }$then the remainder is $\text{ } P \left(a\right)$

proof;

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

then:$\text{ } P \left(x\right) = \left(x - a\right) Q \left(x\right) + R$

so:$\text{ } P \left(a\right) = \cancel{\left(a - a\right) Q \left(a\right)} + R$

$\text{ ":.P(a)=R" }$as required.

In this case we have :$\left({n}^{4} - 3 {n}^{2} - 5 n + 2\right) \div \left(n - 2\right)$

$P \left(n\right) = {n}^{4} - 3 {n}^{2} - 5 n + 2$

the divisor is $\text{ } n - 2 \implies a = 2$

$R = P \left(2\right) = {2}^{4} - 3 \times {2}^{2} - 5 \times 2 + 2$

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