Kyle,
There is a discussion on factoring quartics at the end of the solution to Math Central's monthly problem, MP41 (April 2004). Below is a copy of that discussion.
Question: How does one factor a fourth degree polynomial?
The general quartic looks like
p(x) = c_{4}x^{4} + c_{3}x^{3} + c_{2}x^{2} + c_{1}x + c_{0},
or in its homogeneous form (replacing x by ^{a}/_{b} and multiplying through by b^{4}),
q(a, b) = c_{4}a^{4} + c_{3}a^{3}b + c_{2}a^{2}b^{2} + c_{1}ab^{3} + c_{0}b^{4}.
It is a fact that any quartic polynomial with real coefficients can be factored into a product of two quadratic polynomials with real coefficients. The formula has been known since the sixteenth century, and software is readily available that does the factorization for you; see, for example,
http://mathworld.wolfram.com/QuarticEquation.html
On the other hand, the factorization is practical only in special cases such as
q(a, b) = a^{4} + b^{4}.
There is no need to memorize a formula here. By symmetry the factorization must look like
a^{4} + b^{4} = (a^{2} + xab + b^{2})(a^{2} + yab + b^{2}).
To find x and y, expand the righthand side and observe that the coefficient of both a^{3}b and ab^{3} is x + y, and of a^{2}b^{2} is 2 + xy. For the equality to hold these coefficients must both be zero, so x = –y = √2. Thus,
a^{4} + b^{4} = (a^{2} + √2 ab + b^{2})(a^{2} – √2 ab + b^{2}).
To factor a^{4} + 4c^{4} (= a^{4} + (√2 c)^{4}) one can use b = √2 c in our formula; alternatively, solve a^{4} + 4c^{4} = (a^{2} + xac + 2c^{2})(a^{2} – xac + 2c^{2}) for x. Either way, one gets the formula used in our first solution. Bornsztein attached the name of Sophie Germain (1776 – 1831) to that formula, but it would be surprising had the ancient Greeks failed to know it 2000 years before Germain was born. Another nice quartic that is easily factored is p(x) = x^{4} + x^{3} + x^{2} + x + 1; as a simple exercise you might want to try factoring it yourself.
Chris
