The original poster is attempting to factorize a characteristic equation from their ordinary differential equations (ODEs) class, specifically a fifth-order polynomial equation: r5 - 3r4 + 3r3 - 3r2 + 2r = 0. They express difficulty in handling higher-order differential equations.
The discussion includes various perspectives on factorization techniques, with some participants providing insights into the Rational Root Theorem and its application. There is no explicit consensus on a single method, but multiple approaches are being explored.
Participants are discussing the factorization of a polynomial characteristic equation in the context of ODEs, with a focus on rational roots and methods for checking them. The original poster has expressed challenges specifically with higher-order equations.
Beg to differ with you ideasrule, at least as far as rational roots are concerned. The Rational Root Theorem says that if p/q is a solution of the equation anxn + ... + a1x1 + a0 = 0, the p has to divide a0 and q has to divide an.ideasrule said:There's no technique; you just guess and check, or plug the equation into your graphing calculator to find the roots. Have you tried r=1?
Mark44 said:Beg to differ with you ideasrule, at least as far as rational roots are concerned. The Rational Root Theorem says that if p/q is a solution of the equation anxn + ... + a1x1 + a0 = 0, the p has to divide a0 and q has to divide an.
For the equation r5 - 3r4 + 3r3 -3r2 + 2r = 0,
the left side can be factored, giving r(r4 - 3r3 + 3r2 -3r + 2) = 0.
If the fourth degree factor has any rational roots, the possible candidates are +/-1 or +/-2. Each of these can be checked using either long division or synthetic division, which is essentially the same thing but a lot more efficient. As it turns out, using this theorem, the original fifth-degree polynomial can be factored into five linear factors, giving five solutions to the characteristic equation.