Factorization- any techniques ?

[Roni1985]

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

Homework Equations

The Attempt at a Solution

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

 

[ideasrule]

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]

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 a_nxⁿ + ... + a₁x¹ + a₀ = 0, the p has to divide a₀ and q has to divide a_n.

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.

 

[Roni1985]

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 a_nxⁿ + ... + a₁x¹ + a₀ = 0, the p has to divide a₀ and q has to divide a_n.

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...