Question

Using the factor theorem factorise ${{x}^{3}}-6{{x}^{2}}+11x-6$.

Answer 1

Hint: The factor theorem states that if we put $x=k$ in the given polynomial $p\left( x \right)$ and we get $f\left( k \right)=0$, then $\left( x-k \right)$ is a factor of the given polynomial. Since we are not given any zero of the given polynomial, we need to guess it. Therefore, we will substitute $x=1$ to get zero and using the factor theorem $\left( x-1 \right)$ will be obtained as the first factor of the given polynomial. For obtaining the other factors, we will divide the given polynomial by $\left( x-1 \right)$ to get a quadratic quotient. Finally using the middle term splitting technique on the quadratic quotient, the polynomial will be completely factored.

Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6$
Let us substitute $x=1$ in the above polynomial to get
\[\begin{align}
  & \Rightarrow p\left( 1 \right)={{\left( 1 \right)}^{3}}-6{{\left( 1 \right)}^{2}}+11\left( 1 \right)-6 \\
 & \Rightarrow p\left( 1 \right)=1-6+11-6 \\
 & \Rightarrow p\left( 1 \right)=0 \\
\end{align}\]
Since we have got $p\left( 1 \right)=0$, using the factor theorem we can say that $\left( x-1 \right)$ is the factor of the given polynomial $p\left( x \right)$. In order to factorise it, we divide the given polynomial by $\left( x-1 \right)$ as shown.
$x-1\overset{{{x}^{2}}-5x+6}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+11x-6 \\
 & \underline{{{x}^{3}}-{{x}^{2}}} \\
 & -5{{x}^{2}}+11x-6 \\
 & \underline{-5{{x}^{2}}+5x} \\
 & 6x-6 \\
 & \underline{6x-6} \\
 & \underline{0} \\
\end{align}}\right.}}$
From the above division, we got the quotient ${{x}^{2}}-5x+6$. Therefore, we can write the given polynomial as
\[\Rightarrow p\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)\]
Using the middle term splitting technique, we split the middle term of the quadratic quotient as
\[\Rightarrow p\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-2x-3x+6 \right)\]
Now, taking \[x\] and $-3$ common from the first and the last two terms of the quadratic factor, we get
\[\begin{align}
  & \Rightarrow p\left( x \right)=\left( x-1 \right)\left[ x\left( x-2 \right)-3\left( x-2 \right) \right] \\
 & \Rightarrow p\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
\end{align}\]
Hence, we have factorized the given polynomial as \[\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\].

Note: We can use the hit and trial method to guess all the three factors of the given polynomial to directly write the factored form. Also, we can apply the hit and trial method for factoring the quadratic quotient obtained above, instead of using the middle term splitting method.