# How do you factor 2x^3 + 7x^2 - 3x -18?

$2 {x}^{3} + 7 {x}^{2} - 3 x - 18 = \left(x + 3\right) \left(x + 2\right) \left(2 x - 3\right)$
The $2 {x}^{3}$ term quickly becomes dominant as $| x |$ increases, so any solutions of $2 {x}^{3} + 7 {x}^{2} - 3 x - 18 = 0$ will be quite small.
Integer solutions will be factors of 18, so try $x = - 3 , - 2 , - 1 , 0 , 1 , 2 \mathmr{and} 3$. $x = - 3$ is a solution, so $\left(x + 3\right)$ is a factor. $x = - 2$ is also a solution, so $\left(x + 2\right)$ is also a factor. If the remaining factor is of the form $\left(a x + b\right)$ then looking at the coefficient of ${x}^{3}$ ($2$) and the constant term ($18$) we must have $a = 2$ and $b = - 3$.