Question

Factorize the following

\[{{x}^{3}}-2{{x}^{2}}-x+2\]

Answer 1

Hint: First of all, find any one root of the given equation by hit and trial method and then divide that factor with the original equation to get a quadratic factor. Now, further, factorize the quadratic factor into two linear factors to fully factorize the given equation.

Complete step-by-step answer:
In the question, we have to factorize the expression \[{{x}^{3}}-2{{x}^{2}}-x+2\]. Let us consider the given expression as f(x).

\[f\left( x \right)={{x}^{3}}-2{{x}^{2}}-x+2\]

We can see that the above equation is a 3-degree equation or we can say that it is 3 degrees. To factorize these types of equations, we need to first find its one root by hit and trial. So, by substituting the different values of x in the above equation like x = 0, 1, – 1, 2, – 2, etc. we get f(x) = 0 for x = 2. So, x = 2 is one of the roots of the given equation. So, (x – 2) is a factor of the given equation. So, we get,

\[f\left( x \right)={{x}^{3}}-2{{x}^{2}}-x+2=\left( x-2 \right)h\left( x \right)\]

Now, to get h (x), we divide f (x) by (x – 2), we get,

\[h\left( x \right)=\dfrac{f\left( x \right)}{\left( x-2 \right)}=\dfrac{{{x}^{3}}-2{{x}^{2}}-x+2}{\left( x-2 \right)}\]

\[\left( x-2 \right)\overset{{{x}^{2}}-1}{\overline{\left){\begin{align}

  & {{x}^{3}}-2{{x}^{2}}-x+2 \\

 & {{x}^{3}}-2{{x}^{2}} \\

 & \underline{-\text{ + }} \\

 & 0\text{ +}\text{ 0 }-x+2 \\

 & \text{ }-x+2 \\

 & \text{ }\underline{+\text{ }-} \\

 & \text{ 0 0} \\

\end{align}}\right.}}\]

So, we get, \[h\left( x \right)={{x}^{2}}-1\]. So, we can write

\[f\left( x \right)={{x}^{3}}-2{{x}^{2}}-x+2=\left( x-2 \right)\left( {{x}^{2}}-1 \right)\]

We can also write the above equation as

\[f\left( x \right)=\left( x-2 \right)\left[ {{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}} \right]\]

We know that \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]. By using this in the above equation, we get,

\[f\left( x \right)=\left( x-2 \right)\left( x-1 \right)\left( x+1 \right)\]

So, we have factorized \[{{x}^{3}}-2{{x}^{2}}-x+2=\left( x-2 \right)\left( x-1 \right)\left( x+1 \right)\]

Note: In this question, students can cross-check their answer by substituting x = 2, 1 and – 1 in the original equation and checking if these values of x are making that equation 0 or not. Also students can directly write f(x) = (x – 2) (x – 1) (x + 1) initially, while in the kit and trial method, they get f(x) = 0 for x = 1, – 1 and 2 because we know that a cubic equation has at most 3 factors.