# Whats a faster way for factorizing polynomials of order 3 and above

1. Oct 6, 2007

### Ian_Brooks

1. The problem statement, all variables and given/known data

Course - Control systems engineering, chapter: design using root locus

I'm familiar with dividing a polynomial when given a factor using the remainder theorem however is there another way when only the third or fourth order equation is given and nothing else? We aren't allowed to use calculators in our exams.

Any help?

2. Relevant equations

none - rules of basic factorization, remainder theorem, if theres another than I'd be happy to learn of it.

3. The attempt at a solution

My method is quick for simple non complex numbers. However since its for a control systems exam i'm worried that they will definitely throw in a complex root set thats gonna make me wish i had my TI - 8# calculator in the exam.

my method is simply :

say we have
(s+1)(s+2)(s+3)
= (s^2 + 3s + 2 )(s+3)
= s^3 + 3s^2 + 2s + 3s^2 + 9s + 6
= s^3 + 6s^2 + 11s + 6

so now that we know what our answer should be - lets start

1) look at the final number of the polynomial without s terms - in this case its 6
we find the GCD for it - being
3 |_6_
......2

2) so ... lets substitute -3 into the above equation and see if we obtain 0.
s^3 + 6s^2 + 11s + 6 | s = -3
= -27 + 54 -33 + 6 = -60 + 60 = 0

so therefore -3 is a root so we now have
(s+3) as one of the factors

3) rewrite the original polynomial in terms of (s+3)
s^3 + 6s^2 + 11s + 6
= s^2(s+3) + 3s(s+3) + 2(s+3)
= (s+3)(s2 + 3s +2)
= (s+3)(s+1)(s+2)
<--- as its a 2nd order simple polynomial(quadratic) we can factorize by inspection

done.

But as you can see i need to first check if its a zero then split up the terms - its very traditional and I've used it since high school however I'm sure it won't work for complex roots.

can someone find a faster method? Hopefully one that incorporates complex factors?

Last edited: Oct 6, 2007
2. Oct 6, 2007

### Hurkyl

Staff Emeritus
I'm not sure what a GCD has to do with anything here. But the rational root theorem tells you that if this polynomial has a rational root, then (in lowest terms), its numerator has to divide 6: that is, it is 1, 2, 3, or 6. The choices for the denominator come from the leading coefficient; in this case, the denominator must be 1.

So, the only possibilities for a rational root are:
-6, -3, -2, -1, 1, 2, 3, 6.

Clearly, if a third degree polynomial factors over the integers, then it must have a linear factor -- and thus a rational root.

3. Oct 6, 2007

### Ian_Brooks

so that method should be fine then, how do we expand the above for complex roots - we were given a third order transfer function and the book gave solutions of the form

(s + j0.5)(s - j0.5)(2s + 9)

quite honestly i lost 'bowel control' when i saw that. Any help? Finals are in 6 weeks,

4. Oct 6, 2007

### physicsnewblol

regarding complex roots, the rule of thumb or some theorem states that if
(a - bi) is a solution, then (a + bi) is a solution as well. Break up the polynomial into its simplest, non complex roots, then take the roots with variables higher than degree 1 and break them up.

5. Oct 7, 2007

### Gib Z

The Conjugate Root theorem, which only applies when the co efficients of the polynomial are real by the way.