# How do you factor 4x^3-12x^2-37x-15?

May 17, 2015

If $4 {x}^{3} - 12 {x}^{2} - 37 x - 15 = 0$ has an integer root then it has a corresponding factor of the form $\left(x + a\right)$ with $a$ a divisor of the constant term $- 15$. That gives possibilities of $\pm 1$, $\pm 3$, $\pm 5$ or $\pm 15$. If we try $x = 5$ we find:

$4 {x}^{3} - 12 {x}^{2} - 37 x - 15$

$= \left(4 \times 125\right) - \left(12 \times 25\right) - \left(37 \times 5\right) - 15$

$= 500 - 300 - 185 - 15 = 0$

So $x = 5$ is a root of $4 {x}^{3} - 12 {x}^{2} - 37 x - 15 = 0$ and $\left(x - 5\right)$ must be a factor of $4 {x}^{3} - 12 {x}^{2} - 37 x - 15$.

Next use synthetic division to find:

$4 {x}^{3} - 12 {x}^{2} - 37 x - 15 = \left(x - 5\right) \left(4 {x}^{2} + 8 x + 3\right)$

The remaining quadratic factor $4 {x}^{2} + 8 x + 3$ is of the form $a {x}^{2} + b x + c$ with $a = 4$, $b = 8$ and $c = 3$.

This has discriminant

$\Delta = {b}^{2} - 4 a c = {8}^{2} - \left(4 \times 4 \times 3\right) = 64 - 48 = 16 = {4}^{2}$

...a nice positive square number, so $4 {x}^{2} + 8 x + 3 = 0$ has 2 distinct rational roots, given by the formula

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 8 \pm 4}{8} = \frac{- 2 \pm 1}{2}$

Multiplying both sides by $2$ we get

$2 x = - 2 \pm 1$

Hence $\left(2 x + 1\right)$ and $\left(2 x + 3\right)$ are factors of $4 {x}^{2} + 8 x + 3$

Putting this together:

$4 {x}^{3} - 12 {x}^{2} - 37 x - 15 = \left(x - 5\right) \left(2 x + 1\right) \left(2 x + 3\right)$