# How do you factor the expression 6x^4+24x^2-72x?

Apr 24, 2016

$6 {x}^{4} + 24 {x}^{2} - 72 x$

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

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

where ${x}_{1}$, ${x}_{2}$ and ${x}_{3}$ are the zeros of the cubic as follows...

#### Explanation:

Firstly, all of the terms are divisible by $6 x$, so we can separate that out as a factor:

$6 {x}^{4} + 24 {x}^{2} - 72 x = 6 x \left({x}^{3} + 4 x - 12\right)$

We can factor the remaining cubic factor using Cardano's method:

Let $x = u + v$ and solve ${x}^{3} + 4 x - 12 = 0$

${u}^{3} + {v}^{3} + \left(3 u v + 4\right) \left(u + v\right) - 12 = 0$

Add the constraint $v = - \frac{4}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} - {\left(\frac{4}{3 u}\right)}^{3} - 12 = 0$

Multiply through by $27 {u}^{3}$ to get:

$27 {\left({u}^{3}\right)}^{2} - 324 \left({u}^{3}\right) - 64 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{324 \pm \sqrt{{324}^{2} + 4 \cdot 27 \cdot 64}}{54}$

$= \frac{324 \pm \sqrt{111888}}{54}$

$= \frac{324 \pm 12 \sqrt{777}}{54}$

$= \frac{162 \pm 6 \sqrt{777}}{27}$

Since the square root is Real, this gives us a Real value for ${u}^{3}$.so since the derivation was symmetric in $u$ and $v$ we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find the Real root:

${x}_{1} = \frac{1}{3} \left(\sqrt[3]{162 + 6 \sqrt{777}} + \sqrt[3]{162 - 6 \sqrt{777}}\right)$

and Complex roots:

${x}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{162 + 6 \sqrt{777}} + {\omega}^{2} \sqrt[3]{162 - 6 \sqrt{777}}\right)$

${x}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{162 + 6 \sqrt{777}} + \omega \sqrt[3]{162 - 6 \sqrt{777}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$

Then:

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