# How do you factor 15x^3+7x^2-40x+12?

May 9, 2016

$15 {x}^{3} + 7 {x}^{2} - 40 x + 12 = \left(3 x - 1\right) \left(5 x - 6\right) \left(x + 2\right)$

#### Explanation:

$f \left(x\right) = 15 {x}^{3} + 7 {x}^{2} - 40 x + 12$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $12$ and $q$ a divisor of the coefficient $15$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{15}$, $\pm \frac{2}{15}$, $\pm \frac{1}{5}$, $\pm \frac{4}{15}$, $\pm \frac{1}{3}$, $\pm \frac{2}{5}$, $\pm \frac{2}{3}$, $\pm \frac{3}{5}$, $\pm \frac{4}{5}$, $\pm 1$, $\pm \frac{6}{5}$, $\pm \frac{4}{3}$, $\pm 2$, $\pm \frac{12}{5}$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$

Trying each in turn, the first one that works is $x = \frac{1}{3}$:

$f \left(\frac{1}{3}\right) = \frac{15}{27} + \frac{7}{9} - \frac{40}{3} + 12 = \frac{5 + 7 - 120 + 108}{9} = 0$

So $\left(3 x - 1\right)$ is a factor:

$15 {x}^{3} + 7 {x}^{2} - 40 x + 12 = \left(3 x - 1\right) \left(5 {x}^{2} + 4 x - 12\right)$

To factor the remaining quadratic, use an AC method.

Find a pair of factors of $A C = 5 \cdot 12 = 60$ which differ by $B = 4$

The pair $10 , 6$ works.

Use this pair to split the middle term and factor by grouping:

$5 {x}^{2} + 4 x - 12$

$= 5 {x}^{2} + 10 x - 6 x - 12$

$= \left(5 {x}^{2} + 10 x\right) - \left(6 x + 12\right)$

$= 5 x \left(x + 2\right) - 6 \left(x + 2\right)$

$= \left(5 x - 6\right) \left(x + 2\right)$

Putting it all together:

$15 {x}^{3} + 7 {x}^{2} - 40 x + 12 = \left(3 x - 1\right) \left(5 x - 6\right) \left(x + 2\right)$