Question
Factorise : $x^3-23 x^2+142 x-120$
✓
Answer
Let $p(x)=x^3-23 x^2+142 x-120$
We shall now look for all the factors of $-120 .$
Some of these are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60$.
By hit and trial, we find that $p(1)=0$. Therefore, $x-1$ is a factor of $p(x)$.
Now we see that $x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$
$=x^2(x-1)-22 x(x-1)+120(x-1)$
$=(x-1)\left(x^2-22 x+120\right)[\text { Taking }(x-1) \text { common }]$
Now $x^2-22 x+120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
$x^2-22 x+120=x^2-12 x-10 x+120$
$=x(x-12)-10(x-12)$
$=(x-12)(x-10)$
Therefore, $x^3-23 x^2-142 x-120=(x-1)(x-10)(x-12)$
We shall now look for all the factors of $-120 .$
Some of these are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60$.
By hit and trial, we find that $p(1)=0$. Therefore, $x-1$ is a factor of $p(x)$.
Now we see that $x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$
$=x^2(x-1)-22 x(x-1)+120(x-1)$
$=(x-1)\left(x^2-22 x+120\right)[\text { Taking }(x-1) \text { common }]$
Now $x^2-22 x+120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
$x^2-22 x+120=x^2-12 x-10 x+120$
$=x(x-12)-10(x-12)$
$=(x-12)(x-10)$
Therefore, $x^3-23 x^2-142 x-120=(x-1)(x-10)(x-12)$
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