# How do you factor completely: 12x^5 + 6x^3 + 8x^2?

Jul 14, 2015

$12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 2 {x}^{2} \left(6 {x}^{3} + 3 x + 4\right)$
$= 12 {x}^{2} \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where ${x}_{1} , {x}_{2} , {x}_{3}$ are defined below.

#### Explanation:

$12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 2 {x}^{2} \left(6 {x}^{3} + 3 x + 4\right)$

Let $f \left(x\right) = 6 {x}^{3} + 3 x + 4$.

This is way too messy to solve, but for the record...

Use Cardano's method to solve $f \left(x\right) = 0$.

Let $x = u + v$

$f \left(x\right) = 6 {\left(u + v\right)}^{3} + 3 \left(u + v\right) + 4$

$= 6 {u}^{3} + 6 {v}^{3} + \left(18 u v + 3\right) \left(u + v\right) + 4$

Let $v = - \frac{1}{6} u$.

Then $18 u v + 3 = 0$ and

$f \left(x\right) = 6 {u}^{3} - \frac{1}{36 {u}^{3}} + 4$

If $f \left(x\right) = 0$ then:

$6 {u}^{3} - \frac{1}{36 {u}^{3}} + 4 = 0$

Multiply through by $36 {u}^{3}$ to get:

$216 {\left({u}^{3}\right)}^{2} + 144 \left({u}^{3}\right) - 1 = 0$

${u}^{3} = \frac{- 144 \pm \sqrt{{144}^{2} + 4 \cdot 216}}{2 \cdot 216}$

$= \frac{- {2}^{4} {3}^{2} \pm \sqrt{{2}^{8} {3}^{4} + {2}^{5} {3}^{3}}}{{2}^{4} {3}^{3}}$

$= - \frac{1}{3} \pm \frac{{2}^{2} 3 \sqrt{{2}^{4} {3}^{2} + 2 \cdot 3}}{{2}^{4} {3}^{3}}$

$= - \frac{1}{3} \pm \frac{\sqrt{150}}{36}$

$= - \frac{1}{3} \pm \frac{5 \sqrt{6}}{36}$

Since this derivation has been symmetric in $u$ and $v$:

Let $u = \sqrt{- \frac{1}{3} + \frac{5 \sqrt{6}}{36}}$

and $v = \sqrt{- \frac{1}{3} - \frac{5 \sqrt{6}}{36}}$

The roots of $f \left(x\right) = 0$ are:

${x}_{1} = u + v$
${x}_{2} = \omega u + {\omega}^{2} v$
${x}_{3} = {\omega}^{2} u + \omega v$

where $\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$

${x}_{1}$ is the real root of $f \left(x\right) = 0$
${x}_{2}$ and ${x}_{3}$ are a pair of complex conjugate roots.

Hence $f \left(x\right) = 6 \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

and $12 {x}^{5} + 6 {x}^{3} + 8 {x}^{2} = 12 {x}^{2} \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$