# Nasty Factor

1. Apr 21, 2007

### snowJT

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

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?

2. Apr 21, 2007

### sutupidmath

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

3. Apr 21, 2007

### ssb

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

4. Apr 21, 2007

### snowJT

did you find a program to do it?

and LT stands for Laplas Transform

5. Apr 21, 2007

### ssb

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

6. Apr 21, 2007

### HallsofIvy

Staff Emeritus
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.

7. Apr 22, 2007

### 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

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

8. Apr 22, 2007

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.

Last edited: Apr 22, 2007
9. Apr 22, 2007

### AlephZero

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.

Last edited: Apr 22, 2007
10. Apr 22, 2007

### Gib Z

Damn. I Just got Owned.