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Thread: Factorisation

  1. #1
    Junior Member Pinsky's Avatar
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    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?
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  2. #2
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    Hello, Pinsky!

    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) $

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