# Factoring a 3rd degree poly to get a start on partial fractions

1. Jun 12, 2013

### Nikarasu M

Hello all,

I'm working through old exams for an electrical subject (no solutions given) and I think I've gone wrong somewhere and been left with something I'd like to learn how to work with anyway:

$\frac{50}{(s+\frac{1}{s}+1)^2-s^2}$

$\frac{50}{2s+3+\frac{2}{s}+\frac{1}{s^2}}\times \frac{s^2}{s^2}$

$\frac{50s^2}{2s^3+3s^2+2s+1}$

$\frac{50s^2}{(s+1)(2s^2+s+1)}$​

From the last step I can do partial fractions and inverse Laplace etc...

But I got that factorisation via matlab, I can see it obviously works, and given a guess at a factor of the denominator of the 2nd to last line I can see that '+1' would be involved so a next guess might be '(s+1)' and then I'd see if I get zero remainder with polynomial long division, and I'd win in this case, but it was just a lucky guess.

Ok, time for the question: what is the direct method for doing this ?

I left the earlier working up the case we could hijack the math earlier in the process.

:tongue2:

2. Jun 12, 2013

### Office_Shredder

Staff Emeritus
There is a formula for solutions to the cubic polynomial but nobody uses it in practice. Instead they use the fact that if there is a rational root to $ax^3 + bx^2 + cx + d$ where a,b,c,d are all integers, then the root has to be of the form +/-m/n where m and n are integers, coprime, and m divides d, and n divides a (this is called the rational root theorem)

For your cubic, the only possible rational roots are 1,-1, 1/2 and -1/2 from this. So you can just try them all and see that -1 is a root and factor out an s+1

3. Jun 13, 2013

### epenguin

The thing to have seen was that this is just the standard "difference of two squares"
(a2 - b2) = (a - b)(a + b)

Remember now?

4. Jun 14, 2013

### Nikarasu M

wha?

I've missed a lot of basic stuff at university (sort of jumped ahead via some credit I perhaps shouldn't have got) and school was a long time ago so a lot of these rules of thumbs and tricks go way over my head...

Although here I am doing a masters in engineering - you might say 'lol' - if you were that way inclined :tongue2: