Question

Can someone help me factor s^3+5s^2+12s+8?

Answer 1

Okay an easier question how do you get it to the form (s+1)(s^2+4s+8)

Answer 2

It can be factored as (s - a)(s - b)(s - c), so s=a or s=b or s=c will make it zero.
If you know one zero of s^3+5s^2+12s+8, you can find the others by dividing.
To find a simple zero, just try some numbers. YOu'll see that s=-2 is a zero, so:\[s^3+5s^2+12s+8=(s+2)(2nd-degree-polynomial)\]is a partial factorisation.

To find what 2nd degree polynomial it is, you can use synthetic division.

Answer 3

I got two complex roots

Answer 4

(s+1)(s+2+j2)(s+2-j2)

Answer 5

Is there a method to getting it to (s+1)(s^2+4s+8)? without guessing to find a simple zero

Answer 6

Sorry, typo! s = -1 is a zero, so you can factor out s+1.
The 2nd degree polynomial turns out to be s²+4s+8.
If you try to factor this further, you'll see the discriminant is negative, so there are no real zeros anymore...

Answer 7

sorry how'd you get s=-1?

Answer 8

See @allsmiles's answer. (Although I would use i instead of j)

Answer 9

because it goes to 0 if you use -1?

Answer 10

ahh I think I'm just slow but I still don't see how you ound out to use -1

Answer 11

do you make the equation equal to 0 and then solve for s?

Answer 12

yes, try it!

Answer 13

I did I can't solve it you just get s^3+5s^2+12s+8=0

Answer 14

I moved the 8 to the other side and factored the s out but I don't think thats the way to go about it

Answer 15

What you have to realize is, in general you cannot easily factor a 3rd degree polynomial.
If you are asked to do it, you can bet there are very "simple" numbers involved, like 1, -1, 2, -2 etc.
In the case of s^3+5s^2+12s+8, trying s=-1 gives 0, so that is why you get (s+1)(2nd degree polynomial)

Answer 16

alright I guess that will have to do for now thanks for your help

Answer 17

Do you know synthetic division,? It sounds much more difficult than it is, actually!
With it, you can get (s+1)(s²+4s+8) from s³+5s²+12s+8