Question

Factorize $x^3-23x^2+142x-120$

Answer 1

Let the given polynomial be $f\left(x\right)=x^3-23x^2+142x-120$
Now, putting $x=0$ we get,
$f\left(0\right)=0^3+23(0{)}^2+142(0)-120$
$=-120\ne 0$
So, $x=0$ can't be a root.
Similar way, put $x=1$, we get
$f\left(1\right)=1^3-23\left(1^2\right)+142\left(1\right)-120$
$=1-23+142-120$
$=143-143$
$=0$
So, $x=1$ is a root of the given polynomial.
$⟹(x-1)$ is factor of the given polynomial.

To find the other factors, apply long division method :
$x^2-22x+120x-1)\overline{x^3-23x^2+142x-120}{x}^3-x^2(-)-+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_-22x^2+142x-12022x^2+22x(-)--\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_120x-120120x-120(-)-+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_0$

Therefore, $x^2-22x+120$ is also a factor of polynomial f(x).
Now, factorize $x^2-22x+120$ as follow:
$x^2-22x+120=x^2-12x-10x+120$
$=x(x-12)-10(x-12)$
$\therefore x^2-22x+120=(x-12)(x-10)$

Hence, $(x-1)(x-12)(x-10)$ are the factors of given polynomial f(x).