Using factor theorem, factorize of the polynomials:

Solution:

Let, $f(x)=x^{3}-23 x^{2}+142 x-120$

The constant term in f(x) is -120

The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120

Let, x - 1 = 0

=> x = 1

$f(1)=(1)^{3}-23(1)^{2}+142(1)-120$

= 1 - 23 + 142 - 120

= 0

So, (x – 1) is the factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By long division,

$x^{2}-22 x+120$

$x-1 x^{3}-23 x^{2}+142 x-120$

$x^{3}-x^{2}$

$(-)(+)$

$-22 x^{2}+142 x$

$-22 x^{2}+22 x$

(+) (-)

$120 x-120$

$120 x-120$

(-) (+)

0

$\Rightarrow x^{3}-23 x^{2}+142 x-120=(x-1)\left(x^{2}-22 x+120\right)$

Now,

$x^{2}-22 x+120=x^{2}-10 x-12 x+120$

= x(x – 10) – 12(x – 10)

= (x – 10) (x – 12)

Hence, $x^{3}-23 x^{2}+142 x-120=(x-1)(x-10)(x-12)$