3 ms·
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,
by eximius 13y ago
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.
- omegaham 13y agoI 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.
- Grue3 13y agoJust 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_complex_numbers http://en.wikipedia.org/wiki/Irreducible_polynomial#Real_and...