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AoPS Wiki talk:Problem of the Day/July 16, 2011

Revision as of 20:23, 23 June 2012 by Xantos C. Guin (talk | contribs) (→‎Solution)
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Problem

AoPSWiki:Problem of the Day/July 16, 2011

Solution

Note that $81x^{4}+4x^{4}-16x^{3}+24x^{2}-16x+4$ $= 81x^4 + 4(x^4-4x^3+6x^2-4x+1)$ $= (3x)^4+4(x-1)^4$.

Then, using the Sophie Germain Identity, we have:

$(3x)^4+4(x-1)^4$ $= \left((3x)^2+2(3x)(x-1)+2(x-1)^2\right)\left((3x)^2-2(3x)(x-1)+2(x-1)^2\right)$ $= (17x^2 - 10x + 2)(5x^2 + 2x + 2)$.

The discriminant of these quadratics are $(-10)^2-4(17)(2) = -36$ and $(2)^2-4(5)(2) = -36$ which are both less than $0$.

Therefore, these quadratics cannot be factored into real linear factors, and the irreducible factorization is $(17x^2 - 10x + 2)(5x^2 + 2x + 2)$

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