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Alright, so apparently I've factored this out wrong...

$x^6 + 64 =$ $x^6 + 2^6$

Then I continued, using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to get ... $$(x^2)^3 + 64 = (x^2)^3 + 4^3 = (x^2 + 4)(x^4 - 4x^2 + 16)$$

How is this incorrect?

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It is correct ;) –  Zeta.Investigator Sep 9 '12 at 20:24
    
Do you have an alternative answer - ie the one you are supposed to have got? –  Mark Bennet Sep 9 '12 at 20:25
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The first part is on correct - you can't say $x^6=2^6$ –  Thomas Andrews Sep 9 '12 at 20:26
    
@ThomasAndrews: I edited that bit out because I thought it did not make sense and the previous edit obscured the question. But that might have been the error, and if it was, for the record we had $x^6+2^6=x^6=2^6$ at one stage, and your comment was spot on. –  Mark Bennet Sep 9 '12 at 20:30
    
You can always check that you've factored correctly by multiplying the factored expression and checking that it agrees with your original expression. –  Austin Mohr Sep 9 '12 at 20:35
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3 Answers 3

up vote 7 down vote accepted

This is correct but you are probably asked to continue with the identity $$ x^4-4x^2+16=x^4+8x^2+16-12x^2=(x^2+4)^2-12x^2=a^2-b^2, $$ for some $a$ and $b$ I will let you discover. The final factorisation of $x^6+64$ over the field $\mathbb R$ is the product of three polynomials $x^2+px+q$ with $p^2\lt4q$.

Recall that, together with the polynomials of degree $1$, these are the only irreducible polynomials over $\mathbb R$, hence every real polynomial is a multiple of the product of some degree $2$ polynomials $x^2+px+q$ with $p^2\lt4q$ and some degree $1$ polynomials $x-c$.

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You are fine. You can go further with the $x^4-4x^2+16$ term, but it isn't clean or easy to find.

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Don't think you can factor this over the reals: $x^4-4x^2 + 16 = (x^2-2)^2 + 12 = (x^2-2 \pm 2i\sqrt{3})$. –  gt6989b Sep 9 '12 at 20:36
    
A simple way to continue would be to treat it as a quadratic equation in the variable $x^2$. –  rschwieb Sep 9 '12 at 20:37
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@gt6989b: You can factor (over the reals) any polynomial of degree $\gt 2$ with real coefficients. –  André Nicolas Sep 9 '12 at 20:38
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@gt6989b: but you can factor both of those (still over $\mathbb C$)and find a way to recombine them such that all coefficients are real. –  Ross Millikan Sep 9 '12 at 20:39
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A different approach is to first factorize it over complex numbers: $$x^6+2^6=\prod_{k=0}^5(x-2\cdot e^{\pi i(1/6+k/3)}) $$ And then pair conjugate roots, yielding: $$\prod_{k=0}^2(x-2\cdot e^{\pi i(1/6+k/3)})(x-2\cdot e^{-\pi i(1/6+k/3)})=\prod_{k=0}^2(x^2-4\cos(\pi(1/6+k/3))+4)$$ Which equals $$(x^2-2\sqrt 3x+4)(x^2+4)(x^2+2\sqrt 3x+4)$$ This method will work for any polynomial over $\bf R$ you can factor over $\bf C$ into linear factors (such a factorization always exists, but sometimes it might be impossible to express it in the way you would like).

The reason it will work is that for any real polynomial, it's non-real complex roots occur always occur in conjugate pairs, and for a nonreal complex $z$ we have that $(x-z)(x-\overline z)=x-2\Re(z)+\lvert z\rvert^2$ and the latter is indecomposable over $\bf R$.

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