How do you factor -4x^4y^4+36 −4x4y4+36?
2 Answers
We first rewrite into
Explanation:
For the moment we put the first
Then we can see that
Putting this al together we get:
This cannot be further factorized without using radicals.
-4x^4y^4+36−4x4y4+36
=-4(x^2y^2-3)(x^2y^2+3)=−4(x2y2−3)(x2y2+3)
=-4(xy-sqrt(3))(xy+sqrt(3))(x^2y^2+3)=−4(xy−√3)(xy+√3)(x2y2+3)
=-4(xy-sqrt(3))(xy+sqrt(3))(xy-sqrt(3)i)(xy+sqrt(3)i)=−4(xy−√3)(xy+√3)(xy−√3i)(xy+√3i)
Explanation:
I will use the difference of squares identity a few times, so here it is:
a^2-b^2 = (a-b)(a+b)a2−b2=(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
-4x^4y^4+36−4x4y4+36
=-4(x^4y^4-9)=−4(x4y4−9)
=-4((x^2y^2)^2-3^2)=−4((x2y2)2−32)
=-4(x^2y^2-3)(x^2y^2+3)=−4(x2y2−3)(x2y2+3)
If we allow irrational coefficients we can go a little further:
=-4((xy)^2-(sqrt(3))^2)(x^2y^2+3)=−4((xy)2−(√3)2)(x2y2+3)
=-4(xy-sqrt(3))(xy+sqrt(3))(x^2y^2+3)=−4(xy−√3)(xy+√3)(x2y2+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−√3)(xy+√3)((xy)2−(√3i)2)
=-4(xy-sqrt(3))(xy+sqrt(3))(xy-sqrt(3)i)(xy+sqrt(3)i)=−4(xy−√3)(xy+√3)(xy−√3i)(xy+√3i)
