# How do you use the remainder theorem to find the remainder for the division (x^4+5x^3-14x^2)div(x-2)?

Sep 8, 2016

$f \left(2\right) = 0$ There is no remainder.
$x - 2$ is a factor.

#### Explanation:

If we have $f \left(x\right) = {x}^{4} + 5 {x}^{3} - 14 {x}^{2}$

Let $x - 2 = 0 \rightarrow x = 2$

The remainder of ${x}^{4} + 5 {x}^{3} - 14 {x}^{2} \div \left(x - 2\right)$
is found from $f \left(2\right)$

$f \left(2\right) = {2}^{4} + 5 {\left(2\right)}^{3} - 14 {\left(2\right)}^{2}$

=$16 + 40 - 56 = 0$

The fact that the remainder is 0, (ie there is no remainder), means that $\left(x - 2\right)$ is a factor of ${x}^{4} + 5 {x}^{3} - 14 {x}^{2}$