Factorize the given expression using the factor theorem. The expression is:
\[{{x}^{3}}-23{{x}^{2}}+142x-120\]

Hint: First, find all the factors of the constant term i.e. -120. Then for factor a, if p(a) =0 , then (x-a) is the factor of the polynomial p(x).

Complete step-by-step answer:
In the question, we have to factorize the polynomial \[p\left( x \right)={{x}^{3}}-23{{x}^{2}}+142x-120\].
So for that we have to find the factors of this polynomial \[p\left( x \right)\] using the factor theorem.
Now, \[x-a\] is the factor of the polynomial \[p\left( x \right)\] if \[p\left( a \right)=0\], and this is called the factor theorem.
Now, we will find the factors of -120. So the factor of -120 are:
\[\pm 1,\pm \;2,\;\pm 3,\pm \;4,\;\pm 5,\;\pm 6,\;\pm 8,\;\pm 10,\;\pm 12,\pm \;15,\;\pm 20,\;\pm 24,\pm \;30,\pm \;40,\pm \;60,\;\pm 120\]
Next, we will first take 1, and check if \[p\left( 1 \right)\] is zero or not, if we get \[p\left( 1 \right)=0\], then \[x-1\]will be the factor of this polynomial \[p\left( x \right)={{x}^{3}}-23{{x}^{2}}+142x-120\]
Now, the value of \[p\left( 1 \right)\] is found as follows:
\[\begin{align}
  & \Rightarrow p\left( 1 \right)={{\left( 1 \right)}^{3}}-23{{\left( 1 \right)}^{2}}+142\left( 1 \right)-120 \\
 & \Rightarrow p\left( 1 \right)=0 \\
\end{align}\]
So we get \[p\left( 1 \right)=0\] which means that \[x-1\] is a factor of the polynomial \[{{x}^{3}}-23{{x}^{2}}+142x-120\].
Next, we can write the polynomial \[p\left( x \right)={{x}^{3}}-23{{x}^{2}}+142x-120\], as \[p(x)=\left( x-1 \right)\dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{x-1}\]
Next we will find the expression of \[\dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{x-1}\], and this is done as follows:
\[\begin{align}
  & \Rightarrow \dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{x-1} \\
 & \Rightarrow \dfrac{{{x}^{3}}-{{x}^{2}}-22{{x}^{2}}+22x+120x-120}{x-1} \\
 & \Rightarrow \dfrac{{{x}^{2}}(x-1)-22x(x-1)+120(x-1)}{x-1} \\
 & \Rightarrow {{x}^{2}}-22x+120 \\
\end{align}\]
So we can write the polynomial as
\[\begin{align}
  & \Rightarrow p(x)=\left( x-1 \right)\dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{x-1} \\
 & \Rightarrow p(x)=\left( x-1 \right)({{x}^{2}}-22x+120) \\
\end{align}\]
Next, we will split the middle term of the quadratic expression\[({{x}^{2}}-22x+120)\], so as to get the factors. This is done as follows:
\[\begin{align}
  & \Rightarrow ({{x}^{2}}-22x+120) \\
 & \Rightarrow \left( {{x}^{2}}-10x \right)+\left( -12x+120 \right) \\
 & \Rightarrow x\left( x-10 \right)-12\left( x-10 \right) \\
 & \Rightarrow \left( x-10 \right)\left( x-12 \right) \\
\end{align}\]
So finally the polynomial is written in the factor form as follows:
\[\begin{align}
  & \Rightarrow p\left( x \right)={{x}^{3}}-23{{x}^{2}}+142x-120 \\
 & \Rightarrow p(x)=\left( x-1 \right)({{x}^{2}}-22x+120) \\
 & \Rightarrow p(x)=\left( x-1 \right)\left( x-10 \right)\left( x-12 \right) \\
\end{align}\]
So the required factor form is \[p(x)=\left( x-1 \right)\left( x-10 \right)\left( x-12 \right)\].

Note: Make sure there is no calculation error while finding the value of \[p\left( 1 \right)\], here we have to replace \[x=1\] in the polynomial \[p\left( x \right)\].