Question

Factorise ${{x}^{3}}-2{{x}^{2}}-5x+6$.

Answer 1

Hint: In this problem we need to factorize the given expression. We can observe that the given expression is a cubic expression. So, we will first consider the given expression as a function let us say $f\left( x \right)$. Now we will first calculate at least one root by using trial and error method in which we will calculate the values of $f\left( 0 \right)$, $f\left( \pm 1 \right)$, $f\left( \pm 2 \right)$ and so on by substituting $x=0,\pm 1,\pm 2,...$. If we will get the value of the function as zero for any $x=a$, then we can say that $a$ is the root of the given expression. Now we will rearrange the terms in the function so that we can take the term $x-a$ as common. Now the cubic equation is simplified into a quadratic equation. For the quadratic equation if possible, we can split the middle term and factorize it also, if not possible we may leave it.

Complete step by step answer:
Given expression is ${{x}^{3}}-2{{x}^{2}}-5x+6$.
Let us assume that $f\left( x \right)={{x}^{3}}-2{{x}^{2}}-5x+6$.
Calculating the value of $f\left( 0 \right)$ by substituting $x=0$ in the above function, then we will get
$\begin{align}
  & f\left( 0 \right)={{\left( 0 \right)}^{3}}-2{{\left( 0 \right)}^{2}}-5\left( 0 \right)+6 \\
 & \Rightarrow f\left( 0 \right)=6 \\
\end{align}$
The value of $f\left( 0 \right)$ is not equal to zero. So, $x=0$ is not the solution.
Now calculating the value of $f\left( 1 \right)$ by substituting $x=1$ in the function $f\left( x \right)$, then we will get
$\begin{align}
  & f\left( 1 \right)={{\left( 1 \right)}^{3}}-2{{\left( 1 \right)}^{2}}-5\left( 1 \right)+6 \\
 & \Rightarrow f\left( 1 \right)=1-2-5+6 \\
 & \Rightarrow f\left( 1 \right)=0 \\
\end{align}$
The value of $f\left( 1 \right)$ is equal to zero. So, $x=1$ is the solution of the equation and $x-1$ is the factor of the given function.
Now rearranging the terms in the given equation so that we can take the term $x-1$ as common, then we will get
$\begin{align}
  & f\left( x \right)={{x}^{3}}-{{x}^{2}}-{{x}^{2}}+x-6x+6 \\
 & \Rightarrow f\left( x \right)={{x}^{2}}\left( x-1 \right)-x\left( x-1 \right)-6\left( x-1 \right) \\
\end{align}$
Taking $x-1$ as common from the above equation, then we will have
$f\left( x \right)=\left( x-1 \right)\left[ {{x}^{2}}-x-6 \right]$
In the above equation considering the quadratic equation and comparing it with $a{{x}^{2}}+bx+c$, then we will get
$a=1$, $b=-1$, $c=-6$
The value of $ac=1\left( -6 \right)=-6$
Splitting the middle term $b$ as $2-3$ since $2\left( -3 \right)=-6$, then the quadratic equation is modified as
$\begin{align}
  & {{x}^{2}}-x-6={{x}^{2}}+2x-3x-6 \\
 & \Rightarrow {{x}^{2}}-x=6=x\left( x+2 \right)-3\left( x+2 \right) \\
 & \Rightarrow {{x}^{2}}-x-6=\left( x+2 \right)\left( x-3 \right) \\
\end{align}$
Substituting this value in the given function, then we will have
${{x}^{3}}-2{{x}^{2}}-5x+6=\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)$

Note: For this problem we have considered the quadratic equation separately and factorized it also. Sometimes we don’t have the possibility to factorize the quadratic equation, then we will leave that one and write it as a factor.