[eximius]

Yes, but this is actually false, so it's probably for the best.

Anything of degree 5 or higher is not guaranteed to have solutions solved by radicals (that is, a solution that can be expressed as some rational number to some exponent). For example, x^5 - x + 1 = 0 cannot factored into linear and quadratic polynomials in this way.* The proof for the insolvability is actually quite elegant.

Even so, factoring a polynomial from degree 3 or 4 into quadratics or linear terms is hard. The most general way I can think of is using the rational root theorem and plugging a few values in.

* - You can factor using ultraradicals (yes, it's a thing), but that is far above highschoolers or undergrads, even.

[Grue3]

Just because the solutions are not guaranteed to have a closed form in radicals doesn't mean that my statement is false. In the field of real numbers any polynomial with a degree higher than two is reducible.

http://en.wikipedia.org/wiki/Irreducible_polynomial#Real_and...

[omegaham]

I remember reading (with horror) the methods for factoring cubics and quartics and thinking, "Well, I guess that's what they did before they had Newton's method and computers." Ewwwww.