# How do you factor completely 45x^4 + 43x^2 y^2 - 28y^4?

Apr 6, 2016

$\left(3 x + 2 y\right) \left(3 x - 2 y\right) \left(5 {x}^{2} + 7 {y}^{2}\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 45 {x}^{4} + 43 {x}^{2} {y}^{2} - 28 {y}^{4}$

Replacing $p = {x}^{2}$ and $q = {y}^{2}$
$\textcolor{w h i t e}{\text{XXX}} = 45 {p}^{2} + 43 p q - 28 {q}^{2}$

Factoring by examining factors of $45$ and $28$
or by using the quadratic formula:
$\textcolor{w h i t e}{\text{XXX}} = \left(9 p - 4 q\right) \left(5 p + 7 q\right)$

Replacing $p$ and $q$ with ${x}^{2}$ and ${y}^{2}$ respectively
and
Noting that the first of these factors is the difference of squares
$\textcolor{w h i t e}{\text{XXX}} = \left(3 x + 2 y\right) \left(3 x + 2 y\right) \left(5 {x}^{2} + 7 {y}^{2}\right)$