Check my answer please (factoring)

  • Thread starter mlowery
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In summary: Well, (x-u) is a constant times f(u), or f(x-u) = f(u). In summary, Mitch is suggesting that you should try rewriting the problem in a more symmetric form.
  • #1
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My teacher assigned us this problem (it is for extra credit, but I'll take what I can get):
[tex]P = a(b^4-c^4)+b(a^4-c^4)+c(a^4-b^4)[/tex]

My work:
1. [tex]P = a( (b^2)^2 - (c^2)^2 )+b( (a^2)^2 - (c^2)^2 ) + c( (a^2)^2 - (b^2)^2 )[/tex]

2. [tex]P = a[ (b^2+c^2) (b+c) (b-c) ] + b[ (a^2+c^2) (a+c) (a-c) ] + c[ (a^2+b^2) (a+b) (a-b) ][/tex]

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Final Answer
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[tex]P = a[ (b^2+c^2) (b+c) (b-c) ] + b[ (a^2+c^2) (a+c) (a-c) ] + c[ (a^2+b^2) (a+b) (a-b) ][/tex]

Can someone please double check my work. As far as I can see, that is the simplest form to which it can be factored.

Thanks,
Mitch
 
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  • #2
If I wanted the answer to be "factored", I would think that it should be in the form of a product of sums. Factoring, that is, is the process of turning a sum of products into a product of sums.

Just looking at the problem, perhaps the teacher meant to write:

[tex]P = a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)[/tex]

This is a bit more symmetric, and reminds one of a classic SU(3) symmetry, which should give hints on how to factor it. For example, note that if a=b, then P=0, so (a-b) must be a factor.

If, on the other hand, the teacher meant what he wrote, then you should consider the effect of making the substitution b => -b'. In other words, (a+b) is a factor for the problem you've written.

Carl
 
  • #3
I'm almost certain the cyclically symmetric form what what was intended.

Mitch, do you understand what factoring means ? It means that the given expression be written entirely as a (non-trivial) product of terms - each of these terms called a factor. What you've arrived at is a sum of terms and hence is not what is asked for.

Look at CarlB's suggestion - do you understand it ? If you have some polynomial (in one variable, say) function, f(x), and if f(u) = 0, then what can one say about (x-u) ?
 

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