Question
Using factor theorem, factorize the following polynomials:
$x^3 - 23x^2 + 142x - 120$

Answer

Let $p(x) = x^3 - 23x^2 + 142x - 120$
We shall now look for all the factors of -120. Some of these are $1,\pm2,\pm3,\pm,\pm4,\pm5,\pm6,\pm8\\\pm12,\pm15,\pm20,\pm24,\pm30,\pm,60$
By trial, we find that $p(1) = 0.$
So $x - 1 $is a factor of $p(x).$
Now, we see that $x^3 - 23x^2 + 142x - 120 = x^3 - x^2- 22x^2 + 22x +120x - 120$
$= x^2(x - 1) - 22x(x - 1) + 120(x - 1)$
$= (x - 1)(x^2 - 22x + 120)$ [Taking (x - 1) common]
We could have also got this by dividing p(x) by $x - 1.$
Now $x^2 - 22x + 120$ can be factorised either by splitting the middle term or by using the factor theorem.
By splitting the middle term, we have:
$x^2 - 22x + 120 = x^2 - 12x - 10x+ 120$
$= x(x - 12) - 10(x - 12)$
$= (x - 12)(x - 10)$
So,
$x^3 - 23x^2 + 142x - 120 = (x - 1)(x - 10)(x - 12)$

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