# How do you factor 9x^5+6x^4-12x-8?

Jun 13, 2016

$9 {x}^{5} + 6 {x}^{4} - 12 x - 8 = \left(3 {x}^{4} - 4\right) \left(3 x + 2\right)$

= $\left(\sqrt[2]{3} {x}^{2} + 2\right) \left(\sqrt[4]{3} x + \sqrt{2}\right) \left(\sqrt[4]{3} x - \sqrt{2}\right) \left(3 x + 2\right)$

#### Explanation:

$9 {x}^{5} + 6 {x}^{4} - 12 x - 8$

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

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

Now no further factors are there if we the domain is rational numbers.

However, if the domain is extended to real i.e. include irrational numbers too, above can be factorized further as below - as $\left({x}^{4} - {a}^{4}\right) = \left({x}^{2} + {a}^{2}\right) \left(x + a\right) \left(x - a\right)$.

$\left(3 {x}^{4} - 4\right) \left(3 x + 2\right)$

= $\left({\left(\sqrt[4]{3} x\right)}^{4} - {\left(\sqrt{2}\right)}^{4}\right) \left(3 x + 2\right)$

= $\left({\left(\sqrt[4]{3} x\right)}^{2} + {\left(\sqrt{2}\right)}^{2}\right) \left(\left(\sqrt[4]{3} x\right) + \left(\sqrt{2}\right)\right) \left(\left(\sqrt[4]{3} x\right) - \left(\sqrt{2}\right)\right) \left(3 x + 2\right)$

= $\left(\sqrt[2]{3} {x}^{2} + 2\right) \left(\sqrt[4]{3} x + \sqrt{2}\right) \left(\sqrt[4]{3} x - \sqrt{2}\right) \left(3 x + 2\right)$