# 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?

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I don't know, but if you add $16x^2$ in the middle and subtract $16x^2$ at the end, it factors $(x^2+8)^2-16x^2$. and then recognize the difference of two squares. That's how I would factor it... – imranfat Mar 13 '14 at 17:15
I'd probably just note that its roots are the fourth roots of $-64$, which consist of two conjugate pairs $2\pm 2i$ and $-2\pm 2i$, and then just write down the minimal polynomials for those. – G Tony Jacobs Mar 13 '14 at 17:20
Useful here is Germain's identity: $$a^4+4b^4 = (a^2+2ab+2b^2)(a^2-2ab+2b^2).$$ In this case we have $b=2$, so $$x^4+64 = (x^2+4x+8)(x^2-4x+8).$$ – MJD Mar 13 '14 at 17:29
None of this addresses why the author said to divide $64$ by the exponent $4$, as if dividing by the exponent were some kind of general method. – G Tony Jacobs Mar 13 '14 at 17:30
@GTonyJacobs That is why I posted my remark as a comment, rather than an answer to the question, which I agree it is not. I expect the questioner to be generally interested in matters generally related to their question, and if not I expect that other readers will be interested. – MJD Mar 13 '14 at 17:41

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.

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Certainly seems like the author learned a "trick" for this problem without understanding why it works for this instance alone. – Roger Mar 13 '14 at 17:48
I'm not sure if anyone else noticed but the comment to the post in question actually factors $x^4+4$. The fact that this worked differently was obviously lost. – John Habert Mar 13 '14 at 17:55

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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Sure, but what was the author's point about dividing $64$ by $4$ to obtain that coefficient of $16$? – G Tony Jacobs Mar 13 '14 at 17:29
Thank You This makes sense and I see where the 16 is coming from now. – user135247 Mar 13 '14 at 17:42

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 $4$.

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