Factorizing a Complex Polynomial: Step-by-Step Guide

[snowJT]

Homework Statement

Find the LT of this...

$$\frac{s^3 - 2s^2 - 6s - 6}{s^4 + 4s^3 + 24s^2 + 40s + 100}$$
Apparently this question is solvable.. however.. the first step is nasty, I have no idea how to factor this... the ways I know won't work? Can someone help me out with factoring out the bottom?

 

[sutupidmath]

my native language is not english, so i do not really know what LT stands for?

 

[ssb]

Homework Statement

Find the LT of this...

$$\frac{s^3 - 2s^2 - 6s - 6}{s^4 + 4s^3 + 24s^2 + 40s + 100}$$
Apparently this question is solvable.. however.. the first step is nasty, I have no idea how to factor this... the ways I know won't work? Can someone help me out with factoring out the bottom?

$$\frac{s^3 - 2s^2 - 6s - 6}{(s^2+2s+10)^2}$$

 

[snowJT]

did you find a program to do it?

and LT stands for Laplas Transform

 

[ssb]

did you find a program to do it?

and LT stands for Laplas Transform

Laplace Transformation and yes I found a calculator that could factor the denominator.

 

[HallsofIvy]

my native language is not english, so i do not really know what LT stands for?

My native language is English and I didn't know what LT stood for!

You might try the "Rational Root Theorem". Any rational root of the denominator must evenly divide the constant term which is 100. Try plugging in divisors of 100 (1, 2, 5, 10, 20, 25, 50, 100) to find a factor of the denominator.

 

[Gib Z]

No real (therefore no rational) roots to the equation from the looks of ssbs factorization. Although it is quite easy to factorize:

$$s^4 + 4s^3 + 24s^2 + 40s + 100$$

Now, we can tell by simple logic the in a binomial expansion, to get an s^4 term, the first term in the factorization must be s^2. And to get a constant term of 100, the constant term in the bionomial must be 10.

So what we know is $$s^4 + 4s^3 + 24s^2 + 40s + 100 = (s^2 + ks +10)^2$$ For Some constant k.

What we do from there is to expand (s^2 + ks +10)^2.

I did it for you, because you should already know how to do it - $s^4 + 2ks^3 + (20+k^2)s^2 + 20ks + 100$. What do we do, we read off co efficients of course :) The co efficent of s^3 we already know is 4.

2k=4
k=2

$$s^4 + 4s^3 + 24s^2 + 40s + 100 =(s^2 + 2s +10)^2$$

 

[mathPimpDaddy]

Looks to me like you have partial fractions laplace transform problem. All I can say is factor and break it down into all the pieces you see on the top and then do some algebraic manipulation. This is such a annoying problem, its a patience learning problem. A very special problem for a very special person.

 

[AlephZero]

No real (therefore no rational) roots to the equation from the looks of ssbs factorization. Although it is quite easy to factorize:

$$s^4 + 4s^3 + 24s^2 + 40s + 100$$

Now, we can tell by simple logic the in a binomial expansion, to get an s^4 term, the first term in the factorization must be s^2. And to get a constant term of 100, the constant term in the bionomial must be 10.

Obviously your method happens to work for this example, But it's not "logical" to assume the both factors end with 10, just because 100 is a perfect square. If nothing else, they might have ended in -10.

Consder $$s^4 + 4s^3 + 28s^2 + 65s + 100$$ which can be factorised into $$(s^2+3s+5)(s^2+s+20)$$, but not by your logic.

 

[Gib Z]

Obviously your method happens to work for this example, But it's not "logical" to assume the both factors end with 10, just because 100 is a perfect square. If nothing else, they might have ended in -10.

Consder $$s^4 + 4s^3 + 28s^2 + 65s + 100$$ which can be factorised into $$(s^2+3s+5)(s^2+s+20)$$, but not by your logic.

Damn. I Just got Owned.