Factorisation

Hello, Pinsky!

Quote:

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

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

What is the procedure?

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

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

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

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

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

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

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

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