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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

cgnx
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user135247
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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
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    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
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    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 Look at MJD's comment above. I think that goes to why the division by 4. It's not necessarily because it's the exponent, but because he was dividing out the form of $4b^4$ – Roger Mar 13 '14 at 17:32
  • The author's words: "The two terms $16x^2$ and $-16x^2$ appear by considering the square root of $x^4$ and by dividing the original constant by the leading exponent ($64/4=16$)". I read the original question as asking what the author means by that, and it's still not clear. – G Tony Jacobs Mar 13 '14 at 17:34
  • I agree that the phrasing is poor and the justification for that step is lacking, but I do believe that identity is the underlying reason. This step almost certainly wouldn't work if the leading exponent wasn't 4. – Roger Mar 13 '14 at 17:38
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    @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
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    @MJD I agree that your comment is valuable and interesting, and I appreciate that you posted it as a comment and not an answer. I was just mentioning that we still were in the dark about the author's phrasing. It wasn't a swipe at anyone. – G Tony Jacobs Mar 13 '14 at 17:44

3 Answers3

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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.

John Habert
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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
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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 $$(x^2+8)^2-(4x)^2=(x^2+8-4x)(x+8+4x)$$

cgnx
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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 $4$.

coffeemath
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