# Factorize 7 y^3 (9 y-4) + 9 y - 4 ?

Nov 9, 2016

$\left(9 y - 4\right) \left(y + {7}^{- \frac{1}{3}}\right) \left({7}^{\frac{1}{3}} - {7}^{\frac{2}{3}} y + 7 {y}^{2}\right)$

#### Explanation:

$7 {y}^{3} \left(9 y - 4\right) + 9 y - 4 = \left(9 y - 4\right) \left(1 + 7 {y}^{3}\right)$

but any odd order polynomial has at least one real root. In this case is

$y = - {7}^{- \frac{1}{3}}$

So

$1 + 7 {y}^{3} = \left(y + {7}^{- \frac{1}{3}}\right) \left(a {y}^{2} + b y + c\right)$

equating for all $y$ we get the conditions

$\left\{\begin{matrix}1 - \frac{c}{7} ^ \left(\frac{1}{3}\right) = 0 \\ \frac{b}{7} ^ \left(\frac{1}{3}\right) + c = 0 \\ \frac{a}{7} ^ \left(\frac{1}{3}\right) + b = 0 \\ 7 - a = 0\end{matrix}\right.$ and finally

$7 {y}^{3} \left(9 y - 4\right) + 9 y - 4 = \left(9 y - 4\right) \left(y + {7}^{- \frac{1}{3}}\right) \left({7}^{\frac{1}{3}} - {7}^{\frac{2}{3}} y + 7 {y}^{2}\right)$

$\left(9 y - 4\right) \left(7 {y}^{3} + 1\right)$

#### Explanation:

We have:

$7 {y}^{3} \left(9 y - 4\right) + 9 y - 4$

I'm going to rewrite this to group together the final two terms:

$7 {y}^{3} \left(9 y - 4\right) + \left(9 y - 4\right)$

and now factor out the $\left(9 y - 4\right)$:

$\left(9 y - 4\right) \left(7 {y}^{3} + 1\right)$

We can't do anything with either term that will make this neater or easier to understand, so I'll leave it here.