# How do you factor 2x^3 + x^2 - 4x - 3?

May 30, 2015

First notice that $x = - 1$ is a root of $2 {x}^{3} + {x}^{2} - 4 x - 3 = 0$.

That is: if you substitute $- 1$ for $x$ in the polynomial then the result is $0$

So $\left(x + 1\right)$ is a factor of $2 {x}^{3} + {x}^{2} - 4 x - 3$.

Use synthetic division by this factor...

$2 {x}^{3} + {x}^{2} - 4 x - 3 = \left(x + 1\right) \left(2 {x}^{2} - x - 3\right)$

Notice that $- 1$ is also a root of $2 {x}^{2} - x - 3 = 0$, so we have another $\left(x + 1\right)$ factor...

$\left(2 {x}^{2} - x - 3\right) = \left(x + 1\right) \left(2 x - 3\right)$

So $2 {x}^{3} + {x}^{2} - 4 x - 3 = \left(x + 1\right) \left(x + 1\right) \left(2 x - 3\right)$