# How do you factor y= x^4 + 5x^2 - 6 ?

##### 2 Answers
Dec 5, 2015

$y = \left({x}^{2} + 6\right) \left(x + 1\right) \left(x - 1\right)$

#### Explanation:

Given $y = {x}^{4} + 5 {x}^{2} - 6$

We notice that all terms containing the variable $x$ have ${x}^{2}$ as a factor,
so to simplify things initially we will replace ${x}^{2}$ with $z$

$\textcolor{w h i t e}{\text{XXX}} y = {z}^{2} + 5 z - 6$
which can easily be factored as:
$\textcolor{w h i t e}{\text{XXX}} y = \left(z + 6\right) \left(z - 1\right)$

Restoring ${x}^{2}$ back in place of $z$
$\textcolor{w h i t e}{\text{XXX}} y = \left({x}^{2} + 6\right) \left({x}^{2} - 1\right)$

The first factor, $\left({x}^{2} + 6\right)$, has no obvious sub-factors
but the second, $\left({x}^{2} - 1\right)$, is the difference of squares with sub-factors $\left(x + 1\right) \left(x - 1\right)$

giving
$\textcolor{w h i t e}{\text{XXX}} y = \left({x}^{2} + 6\right) \left(x + 1\right) \left(x - 1\right)$

Dec 5, 2015

$y = \left({x}^{2} + 6\right) \left(x + 1\right) \left(x - 1\right)$

#### Explanation:

Perhaps it will help if I rewrite this expression for $y$ in a slightly different way:

$y = {\left({x}^{2}\right)}^{2} + 5 \left({x}^{2}\right) - 6$

Let $u = {x}^{2}$. Then we have

$y = {u}^{2} + 5 u - 6$

This is something we can factor pretty easily.

$y = \left(u + 6\right) \left(u - 1\right)$

Just substitute back to get $y$ in terms of $x$:

$y = \left({x}^{2} + 6\right) \left({x}^{2} - 1\right)$

And now, the last step is to factor the second bit which is a difference of two squares:

$y = \left({x}^{2} + 6\right) \left(x - 1\right) \left(x + 1\right)$