Find the roots of the cubic equation given as \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\]

Hint: Because the given equation is a cubic equation so it has three roots. We determine the first root by hit and trial method and then proceed to determine other two roots by the same method as we get roots of a quadratic equation.

Complete step-by-step answer:
We start the solution by getting one root of the equation \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\] by hit and trial method.
Checking for different values of x as \[x=1,-1,2,-2\] etc and substituting them in the given equation we will get the first root of the equation, this is called Hit and trial method
Substituting x=1 in \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\] we get that \[1-23+142-120=143-143=0\]
Hence x=1 satisfies the equation \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\]
So, we get that x=1 is one of the roots of the equation \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\]
Which implies that \[(x-1)\] is a factor of the equation \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\]
So, we now divide \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\] by \[(x-1)\] using division,

\[\Rightarrow \dfrac{{{x}^{3}}-23{{x}^{2}}+142x-120}{(x-1)}={{x}^{2}}-22x+120\] ………. (i)
Next step is to solve the equation obtained in (i) which is a quadratic equation, to get the remaining two roots.
we will solve the quadratic equation by splitting the middle term as,
  \[{{x}^{2}}-22x+120={{x}^{2}}-12x-10x+120\]

\[\Rightarrow {{x}^{2}}-22x+120=x(x-12)-10(x-12)\]

\[\Rightarrow {{x}^{2}}-22x+120=(x-12)(x-10)\]
which gives the roots as 10 and 12
Hence combining all the above obtained roots we get all the three roots of the equation \[{{x}^{3}}-23{{x}^{2}}+142x-120=0\] as \[x=1,10,12\] , which is the required answer.

Note: The possibility of the error in the question is that after obtaining the first root of the equation as x=1 we can mistakenly put \[(x+1)\] as a factor of the equation, which is wrong because when \[x=1\] becomes root then that implies \[x-1=0\] becomes a factor.