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
 A: The author's reasoning appears to be that it works in this situation so that must be what you have to do. The problem is that it only works for this particular problem. As MJD points out in the comments, Germain's identity shows $a^4+4b^4$ can be factored in the integers. In order to factor this, you need to add and subtract $4a^2b^2$. Then we get $$a^4+4b^4 = a^4+4a^2b^2+4b^4-4a^2b^2 = (a^2+2b^2)-(2ab)^2 = (a^2+2b^2-2ab)(a^2+2b^2+2ab)$$
The only situation where $4b^4\div 4 = 4b^2$ is when $b=2$ and hence $4b^4 = 64$ and $4b^2 = 16$. If you try the same technique if dividing by $4$ on $x^4+4$, which fits the form of Germain's identity with $a=x$ and $b=1$, it fails.
A: 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
$$(x^2+8)^2-(4x)^2=(x^2+8-4x)(x+8+4x)$$
A: If one starts with $(x^2+2k^2)^2=x^4+4k^2x^2+4k^4,$ the same method works to factor $x^4+4k^4$, namely add and subtract the term $4k^2x^2=(2kx)^2.$ But note the obvious relation here between the coefficient of the added/subtracted term and the constant term $4k^4$ is that one should divide that constant term by $k^2$ to get the coefficient $4k^2$ of the added/subtracted term. When $k=2$ it means we divide by $4$, but in other cases one does not use the authors rule, since only for $k=2$ does $k^2$ come out the same as the power $4$. 
