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

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

As Lubin points out below, there are sums of squares that are factorable, however, so a sum of squares should not be immediately discounted as not factorable.

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  • $\begingroup$ Does that mean that the answer key is wrong? $\endgroup$ – Gavriel May 19 '16 at 18:28
  • $\begingroup$ 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. $\endgroup$ – wgrenard May 19 '16 at 18:31
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    $\begingroup$ 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 $(x^2-2x+2)(x^2+2x+2)$ $\endgroup$ – Lubin May 19 '16 at 18:32
  • $\begingroup$ @Lubin Good point. Thanks for mentioning. $\endgroup$ – wgrenard May 19 '16 at 18:53

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