Question

How do you factor \[{{x}^{3}}-4{{x}^{2}}+x+6\]?

Answer 1

Hint: First take all the terms to the L.H.S. and assume the function as \[f\left( x \right)\]. Now, find one root of the function with the help of hit and trial method. With the help of this root form a quadratic polynomial and factorize it by using the middle term split method. Before using the middle term split method check if the obtained quadratic polynomial can be factored by finding its discriminant. If the value of discriminant is less than 0 then it cannot be factored.

Complete step by step answer:
Here, we have been provided with the polynomial \[{{x}^{3}}-4{{x}^{2}}+x+6\] and we are asked to factorize it. So, let us use this polynomial as,
\[\Rightarrow f\left( x \right)={{x}^{3}}-4{{x}^{2}}+x+6\]
Clearly, we can see that \[f\left( x \right)\] is a cubic polynomial and therefore to factorize it we need to find its first factor by hit – and – trial method. We need to find such a value of x so that the value of \[f\left( x \right)\] becomes 0.
Now, substituting x = -1 randomly in \[f\left( x \right)\], we get,
\[\begin{align}
  & \Rightarrow f\left( -1 \right)={{\left( -1 \right)}^{3}}-4{{\left( -1 \right)}^{2}}+\left( -1 \right)+6 \\
 & \Rightarrow f\left( -1 \right)=-6+6 \\
 & \Rightarrow f\left( -1 \right)=0 \\
\end{align}\]
So, x = -1 is a root of \[f\left( x \right)\]. Therefore, \[\left( x+1 \right)\] is a factor of \[f\left( x \right)\].
So, the function \[f\left( x \right)\] can be written as: -
\[\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{2}}\left( x+1 \right)-5x\left( x+1 \right)+6\left( x+1 \right) \\
 & \Rightarrow f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-5x+6 \right) \\
\end{align}\]
Now, we have formed a quadratic polynomial \[{{x}^{2}}-5x+6\] and we have to factorize it. But first let us check if it can be factored or not. For this we need to find the discriminant value.
\[\Rightarrow \] Discriminant = \[{{b}^{2}}-4ac\]
Here, b = -5 = coefficient of x
\[\Rightarrow \] a = 1 = coefficient of \[{{x}^{2}}\]
\[\Rightarrow \] c = 6 = constant term
\[\Rightarrow \] Discriminant = \[{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 6 \right)=1>0\]
Therefore, the quadratic polynomial can be factored. So, using the middle term split method, we have,
\[\begin{align}
  & \Rightarrow f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-3x-2x+6 \right) \\
 & \Rightarrow f\left( x \right)=\left( x+1 \right)\left[ x\left( x-3 \right)-2\left( x-3 \right) \right] \\
\end{align}\]
Taking \[\left( x-3 \right)\] common, we get,
\[\begin{align}
  & \Rightarrow f\left( x \right)=\left( x+1 \right)\left( x-3 \right)\left( x-2 \right) \\
 & \Rightarrow {{x}^{3}}-4{{x}^{2}}+x+6=\left( x+1 \right)\left( x-3 \right)\left( x-2 \right) \\
\end{align}\]
Hence, the above expression represents the factored form of the given polynomial.

Note:
One may note that we cannot directly use the middle term split method or any method or any method to factorize the given cubic polynomial. First, we need to find one root and factor using the hit – and – trial method, so that it can be further converted into the product of a linear polynomial and a quadratic polynomial. If we have to find the roots of the polynomial \[f\left( x \right)\] then we will substitute the obtained factored form with 0 and find the three values of x.