# How do you long divide (2n^3 + 0n^2 - 14n + 12) /(n + 3)?

Aug 7, 2015

$2 \left(n - 2\right) \left(n - 1\right)$

#### Explanation:

Assume $n + 3$ is a factor for the numerator and infer the other factor:
$2 {n}^{3} - 14 n + 12 = \left(n + 3\right) \left(a {n}^{2} + b n + c\right) =$
$a {n}^{3} + \left(b + 3 a\right) {n}^{2} + \left(c + 3 b\right) n + 3 c$

This gives the result:
$a = 2$
$b + 3 a = b + 6 = 0 \implies b = - 6$
$c + 3 b = c - 18 = - 14 \implies c = 4$
$3 c = 12$

Therefore $n + 3$ is a factor and we have:
$\frac{2 {n}^{3} - 14 n + 12}{n + 3} = \frac{\cancel{\left(n + 3\right)} \left(2 {n}^{2} - 6 n + 4\right)}{\cancel{n + 3}} =$
$2 \left({n}^{2} - 3 n + 2\right) = 2 \left(n - 2\right) \left(n - 1\right)$