# Factor Completely.

Again, this question if from my final practice exam.

Factor Completely. $$81x^4-256y^4$$

I'm able to get this far, How do I know which of the two factors should be factored further. $$(9x^2+16y^2)(9x^2-16y^2)$$

Answer Key: $$(3x-2y)(3x+2y)(9x^2+4y^2)$$

• The one that is a difference of squares is always factorable. – abiessu May 19 '16 at 18:18
• (a+b)*(a-b) = a^2 - b^2 – piepi May 19 '16 at 18:20
• So, the Answer is $$(3x-2y)(3x+2y)(9x^2+16y^2)?$$ – Gavriel May 19 '16 at 18:24
• The answer key is not correct. – wgrenard May 19 '16 at 18:31
• The key is wrong, it should be $(3x-4y)(3x+4y)(9x^2+16x^2)$. – abiessu May 19 '16 at 18:32

A difference of squares is always factorable into two binomial factors. This immediately tells you that $9x^2 - 16y^2$ can be factored further.
A sum of squares cannot be factored into two binomials, and in your case $9x^2 + 16y^2$ has no common factor to pull out, nor can it be factored in any other way.
• Yes it is. You should use the difference of squares formula for the $9x^2 - 16y^2$ factor, and you must leave the other factor as it is. – wgrenard May 19 '16 at 18:31
• The answer key is wrong. However, it’s not quite true that the sum of two squares is unfactorable. Consider $x^4+4$ which certainly is a sum of two squares, yet is equal to $(x^2-2x+2)(x^2+2x+2)$ – Lubin May 19 '16 at 18:32