Question

Factorize the following algebraic expression, ${{x}^{3}}-23{{x}^{2}}+142x-120$

Answer 1

Hint: We need to use the hit and trial method to factorize a cubic polynomial. When we find one of its factors we can divide the polynomial with this factor to get a quadratic and then we can factorize the obtained quadratic.

Complete step-by-step answer:

First of all we will use a hit and trial method. Let us first try with 1. When we substitute x=1 in the given cubic polynomial we have,
${{1}^{3}}-23({{1}^{2}})-142(1)-120=0$

Therefore, 1 is the solution of this cubic polynomial i.e. $x-1$ is a factor of this polynomial.

Now we will divide ${{x}^{3}}-23{{x}^{2}}+142x-120$ with $x-1$ .
$\dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{x-1}={{x}^{2}}-22x+120$

Therefore, we can write ${{x}^{3}}-23{{x}^{2}}+142x-120=(x-1)({{x}^{2}}-22x+120)$
We can factorize the obtained quadratic by splitting its middle term.
 ${{x}^{2}}-22x+120={{x}^{2}}-12x-10x+120$

Taking x common from first two terms and and -10 common from last two terms we have,
$\begin{align}
  & {{x}^{2}}-12x-10x+120=x(x-12)-10(x-12) \\
 & =(x-10)(x-12) \\
\end{align}$

Therefore, we can write
$\begin{align}
  & {{x}^{3}}-23{{x}^{2}}+142x-120=(x-1)({{x}^{2}}-22x+120) \\
 & =(x-1)(x-12)(x-10) \\
 & \Rightarrow {{x}^{3}}-23{{x}^{2}}+142x-120=(x-1)(x-12)(x-10) \\
\end{align}$

Hence, we have factorized the given cubic polynomial.

Note: If x=1 had not satisfied the given polynomial then we would have tried x=-1 then x=2 and so on till it does not get very complicated. When doing hit and trial normally we don’t go beyond +3 and -2 but we may need to. And after finding one root we don’t need to do a hit and trial again. We can divide the obtained factorial to get a quadratic and we are sure to find the roots of a quadratic if not by factorization then by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .