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) = c₄x⁴ + c₃x³ + c₂x² + c₁x + c₀,

or in its homogeneous form (replacing x by ^a/_b and multiplying through by b⁴),

q(a, b) = c₄a⁴ + c₃a³b + c₂a²b² + c₁ab³ + c₀b⁴.

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) = a⁴ + b⁴.

There is no need to memorize a formula here.  By symmetry the factorization must look like

a⁴ + b⁴ = (a² + xab + b²)(a² + yab + b²).

To find x and y, expand the right-hand side and observe that the coefficient of both a³b and ab³ is x + y, and of a²b² is 2 + xy.  For the equality to hold these coefficients must both be zero, so x = –y = √2.  Thus,

a⁴ + b⁴ = (a² + √2 ab + b²)(a² – √2 ab + b²).

To factor a⁴ + 4c⁴ (= a⁴ + (√2 c)⁴) one can use b = √2 c in our formula; alternatively, solve a⁴ + 4c⁴ = (a² + xac + 2c²)(a² – xac + 2c²) 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) = x⁴ + x³ + x² + x + 1; as a simple exercise you might want to try factoring it yourself.

Chris