# How do you factor 13a^3 - 16a - 72?

Apr 1, 2016

$13 {a}^{3} - 16 a - 72$

$= \left(a - 2\right) \left(13 {a}^{2} + 26 a + 36\right)$

$= \left(a - 2\right) \left(\sqrt{13} a + \sqrt{13} - \sqrt{23} i\right) \left(\sqrt{13} a + \sqrt{13} + \sqrt{23} i\right)$

#### Explanation:

By the rational root theorem, any rational zeros of this quadratic will be expressible in the form $a = \frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 72$ and $q$ a divisor of the coefficient $13$ of the leading term.

So the possible rational zeros are:

$\pm \frac{1}{13} , \pm \frac{2}{13} , \pm \frac{3}{13} , \pm \frac{4}{13} , \pm \frac{6}{13} , \pm \frac{8}{13} , \pm \frac{9}{13} , \pm \frac{12}{13} , \pm 1 , \pm \frac{18}{13} , \pm \frac{24}{13} , \pm 2 , \pm \frac{36}{13} , \pm 3 , \pm 6 , \pm 8 , \pm 9 , \pm 12 , \pm 18 , \pm 24 , \pm 36 , \pm 72$

Let $f \left(a\right) = 13 {a}^{3} - 16 a - 72$

We can rule out any of the non-integral fractions, since for such fractions, $13 {a}^{3}$ will have a denominator ${13}^{2}$ in lowest terms, which will not be possible to cancel out by the other terms.

So try the integer possibilities in sequence.

We find $f \left(2\right) = 13 \cdot 8 - 16 \cdot 2 - 72 = 104 - 32 - 72 = 0$

So $a = 2$ is a zero and $\left(a - 2\right)$ a factor:

$13 {a}^{3} - 16 a - 72 = \left(a - 2\right) \left(13 {a}^{2} + 26 a + 36\right)$

The remaining quadratic factor has negative discriminant:

$\Delta = {26}^{2} - \left(4 \cdot 13 \cdot 36\right) = 676 - 1872 = - 1196$

so no linear factors with Real coefficients.

We can factor using Complex coefficients as follows:

$\left(a - 2\right) \left(13 {a}^{2} + 26 a + 36\right)$

$= \left(a - 2\right) \left(13 \left({a}^{2} + 2 a + 1\right) + 23\right)$

$= \left(a - 2\right) \left(13 {\left(a + 1\right)}^{2} + 23\right)$

$= \left(a - 2\right) \left({\left(\sqrt{13} \left(a + 1\right)\right)}^{2} - {\left(\sqrt{23} i\right)}^{2}\right)$

$= \left(a - 2\right) \left(\sqrt{13} \left(a + 1\right) - \sqrt{23} i\right) \left(\sqrt{13} \left(a + 1\right) + \sqrt{23} i\right)$

$= \left(a - 2\right) \left(\sqrt{13} a + \sqrt{13} - \sqrt{23} i\right) \left(\sqrt{13} a + \sqrt{13} + \sqrt{23} i\right)$