<s>[INST] Factor $x^4 + 64$ I found a page where this problem was solved but his english is broken so its difficult to understand his explanation. His first step was to divide the constant, 64, by the exponent 4. What is his reasoning behind this step?
http://www.wyzant.com/resources/answers/163/factor_x_4_64
 [/INST] 1)He added and then subtracted $16x^2$, so the value wouldn't change
$$x^4 + 64 = x^4 + 16x^2 - 16x^2 + 64$$
2)He used the fact that $(A+B)^2=A^2+2AB+B^2$, in your case $A=x^2$ and $B=8$ and the formula was used "backwards" to get $(A+B)^2$ back
$$x^4 + 64 = x^4 + 16x^2 + 64- 16x^2=(x+8)^2-(4x)^2$$
3)$A^2-B^2=(A-B)(A+B)$, where in your case $A=x^2+8$ and $B=4x$, so we have
$$(x^2+8)^2-(4x)^2=(x^2+8-4x)(x+8+4x)$$
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