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Factorization- any techniques ?

  • Thread starter Roni1985
  • Start date
  • #1
201
0

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.
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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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?
 
  • #3
33,507
5,192
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.
 
  • #4
201
0
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...
 

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