# tricky algebraic manipulation

• Dec 12th 2009, 03:28 PM
jut
tricky algebraic manipulation
I need this $\frac{-18477 s-2.82843\times 10^8}{s^2+15307 s+4.\times 10^8}$ to be in this format : $\frac{s+a}{(s+a)^2+w^2}$

I can easily factor the top to get the "s+a",

$-18477\frac{s+15307 }{s^2+15307s+4*10^8}$

Now $(s+15307)^2=s^2+30614 s+234304249$

but how can I factor the bottom in such a way to get the required form? Can someone help please?
• Dec 12th 2009, 03:46 PM
Gusbob
Have you tried completing the square? Do the numbers have to be integers?

$s^2+15307s+4\times10^8 = (s^2 + 15307s + 7653.5^2) - 7653.5 ^2 + 4 \times 10^8$

$(s+ 7653.5)^2 + (4 \times 10^8 - 7653.5^2)$
• Dec 12th 2009, 04:31 PM
jut
Quote:

Originally Posted by Gusbob
Have you tried completing the square? Do the numbers have to be integers?

$s^2+15307s+4\times10^8 = (s^2 + 15307s + 7653.5^2) - 7653.5 ^2 + 4 \times 10^8$

$(s+ 7653.5)^2 + (4 \times 10^8 - 7653.5^2)$

Good idea.

So I get:

$-18477\frac{s+15307 }{(s+7653)^2+18477^2}$

which is tantalizingly close, but not in the required form.
• Dec 12th 2009, 05:32 PM
Gusbob
Sorry I didn't read the question carefully. Didn't realise the square on the bottom has to be the square of the numerator.

In this case, I don't think it will be possible to make it into the format you wish, unless 'w' can have 's' in it.

This is because we cannot make the bottom square $(s + 15307)^2$ without introducing more 's' into the denominator, which would have to be part of 'w'.
• Dec 12th 2009, 05:36 PM
jut
It's ok, thanks for the help anyway.

Yeah I was beginning to think it's not possible too. Oh well. (Speechless)