# Factorization- any techniques ?

## Homework Statement

I'm trying to factorize a characteristic equation from my ODEs class, and I am having a problem when it comes to 4th, 5th or higher order differential equations.
Say this one:
r5-3r4+3r3-3r2+2r=0

## The Attempt at a Solution

Tried a lot of things but couldn't reach the result.

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ideasrule
Homework Helper
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
Mentor
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?
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.

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.
Thank you guys for the replies. it was helpful...