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Using factor theorem, factorize each of the following polynomials:$x^3 - 23x^2 + 142x - 120$
Given:
Given expression is $x^3 - 23x^2 + 142x - 120$.
To do:
We have to find the given polynomial using factor theorem.
Solution:
Let $f(x)=x^{3}-23 x^{2}+142 x-120$
The factors of the constant term $-120$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12,\pm 15, \pm 20, \pm 24, \pm 60, \pm 120$
Let $x=1$, this implies,
$f(1)=(1)^{3}-23(1)^{2}+142(1)-120$
$=1-23+142-120$
$=143-143$
$=0$
Therefore $x-1$ is a factor of $f(x)$
Dividing $f(x)$ by $x-1$, we get,
$x - 1$) $x ^ { 3 } - 2 3 x ^ { 2 } + 1 4 2 x - 1 2 0$( $x ^ { 2 } - 2 2 x + 1 2 0$
$x^{3}-x^{2}$
--------------------------------
$-22 x^{2}+142 x$
$-22 x^{2}+22 x$
------------------------------
$120 x-120$
$120 x-120$
----------------
0
Therefore,
$f(x)=(x-1)(x^{2}-22 x+120)$
$=(x-1)(x^{2}-10 x-12 x+120)$
$=(x-1)[x(x-10)-12(x-10)]$
$=(x-1)(x-10)(x-12)$
Hence, $f(x)=(x-1)(x-10)(x-12)$.
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