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

[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:

 

[Office_Shredder]

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

 

[epenguin]

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

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

Remember now?

 

[Nikarasu M]

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

Remember now?

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:

 

[blue_raver22]

Hello,

It seems like you are trying to factor a 3rd degree polynomial in order to use partial fractions for solving a problem in your electrical subject. This is a common technique used in many scientific fields, including engineering and physics.

To factor a 3rd degree polynomial, there are a few methods that can be used. One method is the rational root theorem, which helps identify possible rational roots of a polynomial. In this case, the rational root theorem tells us that the possible rational roots of the denominator are ±1, ±2, ±5, ±10, ±25, ±50. By trying these values and using polynomial long division, we can find that (s+1) is indeed a factor of the denominator.

Another method is to use the quadratic formula to solve for the roots of the denominator and then factor it accordingly. In this case, the quadratic formula gives us the roots of 2s^2+s+1 as -0.5±0.866i. From here, we can use the complex conjugate root theorem to factor the denominator into (s+1)(2s^2+s+1).

Both of these methods are commonly used in mathematics and can be applied to factor polynomials of any degree. I hope this helps and good luck with your studies!