# How do you simplify (x ^ 4 - 16y ^ 4)/( ( x² + 4y² ) ( x-2y ))?

Jul 27, 2016

x + 2y

#### Explanation:

Begin by factorising the numerator, which is a $\textcolor{b l u e}{\text{difference of squares}}$ and in general factorises as

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

${x}^{4} = {\left({x}^{2}\right)}^{2} \text{ and } 16 {y}^{4} = {\left(4 {y}^{2}\right)}^{2}$

$\Rightarrow a = {x}^{2} \text{ and } b = 4 {y}^{2}$

Substitute a and b into (A)

$\Rightarrow {x}^{4} - 16 {y}^{4} = \left({x}^{2} - 4 {y}^{2}\right) \left({x}^{2} + 4 {y}^{2}\right)$

rArr((x^2-4y^2)cancel((x^2+4y^2)))/((cancel((x^2+4y^2))(x-2y)

Now ${x}^{2} - 4 {y}^{2} \text{ is also a " color(blue)"difference of squares}$

${x}^{2} = {\left(x\right)}^{2} \text{ and " 4y^2=(2y)^2rArra=x" and } b = 2 y$

substitute a and b into (A)

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

$\Rightarrow \frac{\cancel{\left(x - 2 y\right)} \left(x + 2 y\right)}{\cancel{\left(x - 2 y\right)}} = x + 2 y$