Quandaries and Queries


Name: Kyle

Level: Secondary (Alg. 2)

Who?: Teacher

How do I factor y4 + y2 +1?? I think the answer is (y2 + y + 1)(y2 - y + 1), but I'm not sure how to get that...




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,


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.