# How do you factor -4x^4y^4+36 ?

Nov 28, 2015

We first rewrite into $36 - 4 {x}^{4} {y}^{4}$ and we can immediately take out the factor $4$

#### Explanation:

$= 4 \left(9 - {x}^{4} {y}^{4}\right)$
For the moment we put the first $4$ on hold.
Then we can see that
$9 = {3}^{2}$ and ${x}^{4} = {\left({x}^{2}\right)}^{2}$ and ${y}^{4} = {\left({y}^{2}\right)}^{2}$

Putting this al together we get:
${3}^{2} - {\left({x}^{2} {y}^{2}\right)}^{2}$ which is a difference of 2 squares:

$\left(3 + {x}^{2} {y}^{2}\right) \left(3 - {x}^{2} {y}^{2}\right)$ and putting in the $4$:

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

This cannot be further factorized without using radicals.

Nov 29, 2015

$- 4 {x}^{4} {y}^{4} + 36$

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

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

$= - 4 \left(x y - \sqrt{3}\right) \left(x y + \sqrt{3}\right) \left(x y - \sqrt{3} i\right) \left(x y + \sqrt{3} i\right)$

#### Explanation:

I will use the difference of squares identity a few times, so here it is:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Since I would like to keep the higher degree terms on the left, I choose to separate out a factor of $- 4$ first:

$- 4 {x}^{4} {y}^{4} + 36$

$= - 4 \left({x}^{4} {y}^{4} - 9\right)$

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

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

If we allow irrational coefficients we can go a little further:

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

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

If we allow Complex coefficients we can get a little further still:

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

$= - 4 \left(x y - \sqrt{3}\right) \left(x y + \sqrt{3}\right) \left(x y - \sqrt{3} i\right) \left(x y + \sqrt{3} i\right)$