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

Question

1580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-28T23:33:27

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

#=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 · 2015-11-29T00:21:18

#-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)#

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)#