# Factorisation

• May 14th 2008, 06:32 AM
Pinsky
Factorisation
I have $\displaystyle s^3+12s^2+21s+10$ and a have to show it as factors.
The resut should be $\displaystyle (s+1)^2(s+10)$.

What is the procedure?
• May 14th 2008, 07:07 AM
Soroban
Hello, Pinsky!

Quote:

Factor: .$\displaystyle f(s) \:=\:s^3+12s^2+21s+10$

Answer: .$\displaystyle (s+1)^2(s+10)$.

What is the procedure?

We are expected to be familiar with two theorems . . .

[1] The rational roots of a polynomial are of the form $\displaystyle \frac{n}{d}$
. . .where $\displaystyle n$ is a factor of the constant term and $\displaystyle d$ is a factor of the leading coefficient.

Our polynomial has a constant term 10 with factors: .$\displaystyle \pm1,\:\pm2,\:\pm5,\:\pm10$
. . and leading coefficient 1 with factors: . $\displaystyle \pm1$
Hence, the only possible rational roots are: .$\displaystyle \pm1,\:\pm2,\:\pm5,\:\pm10$

[2] If $\displaystyle f(a) = 0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle f(x).$

We find that: .$\displaystyle f(\text{-}1) \:=\:(\text{-}1)^3 + 12(\text{-}1)^2 + 21(\text{-}1) + 10 \:=\:0$
. . Hence, $\displaystyle (s+1)$ is a factor.

Using long or synthetic division: .$\displaystyle f(s) \:=\:(s+1)(s^2+11s + 10)$

. . And that quadratic factors: .$\displaystyle s^2+11s + 10 \:=\:\overbrace{(s+1)(s+10)}$

Therefore: .$\displaystyle f(s) \;=\;(s+1)^2(s+10)$