Question

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

Answer 1

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

Explanation:

`=4(9-x^4y^4)`
For the moment we put the first `4` on hold.
Then we can see that
`9=3^2` and `x^4=(x^2)^2` and `y^4=(y^2)^2`

Putting this al together we get:
`3^2-(x^2y^2)^2` which is a difference of 2 squares:

`(3+x^2y^2)(3-x^2y^2)` and putting in the `4`:

`=4(3+x^2y^2)(3-x^2y^2)`

This cannot be further factorized without using radicals.

Answer 2

`-4x^4y^4+36`

`=-4(x^2y^2-3)(x^2y^2+3)`

`=-4(xy-sqrt(3))(xy+sqrt(3))(x^2y^2+3)`

`=-4(xy-sqrt(3))(xy+sqrt(3))(xy-sqrt(3)i)(xy+sqrt(3)i)`

Explanation:

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

`a^2-b^2 = (a-b)(a+b)`

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

`-4x^4y^4+36`

`=-4(x^4y^4-9)`

`=-4((x^2y^2)^2-3^2)`

`=-4(x^2y^2-3)(x^2y^2+3)`

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

`=-4((xy)^2-(sqrt(3))^2)(x^2y^2+3)`

`=-4(xy-sqrt(3))(xy+sqrt(3))(x^2y^2+3)`

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

`=-4(xy-sqrt(3))(xy+sqrt(3))((xy)^2-(sqrt(3)i)^2)`

`=-4(xy-sqrt(3))(xy+sqrt(3))(xy-sqrt(3)i)(xy+sqrt(3)i)`