Question

Using the factor theorem, factorize ${{x}^{3}}-6{{x}^{2}}+3x+10$.

Answer 1

Hint: Using hit and trial, find three number (a, b, c) such that $f\left( a \right)=0,\,f\left( b \right)=0\,and\,f\left( c \right)=0$ where $f\left( x \right)={{x}^{3}}-6{{x}^{2}}+3x+10$ and by factor theorem, write $f\left( x \right)=\left( x-a \right)\left( x-b \right)\left( x-c \right)$ to get the factorize of ${{x}^{3}}-6{{x}^{2}}+3x+10$.

Complete step-by-step answer:

We know that, according to factor theorem, if is a polynomial of degree $n\ge 1\,$and ‘a’ as any real number, then $\left( x-a \right)$ is a factor of , if$f\left( a \right)=0$.

We have to find factors of ${{x}^{3}}-6{{x}^{2}}+3x+10$ using factor theorem.

According to factor theorem-

${{x}^{3}}-6{{x}^{2}}+3x+10$= $\left( x-a \right)\left( x-b \right)\left( x-c \right)$.

If $f\left( a \right)=0$ and

$f\left( b \right)=0$

$f\left( c \right)=0$

Where $f\left( x \right)={{x}^{3}}-6{{x}^{2}}+3x+10$.

i.e. a, b, c are zeroes of the polynomial


By guessing we can see that $f\left( -1 \right)=0$

$f\left( -1 \right)={{\left( -1 \right)}^{3}}-6{{\left( -1 \right)}^{2}}+3\left( -1 \right)+10=0$

As $f\left( -1 \right)=0$, $\left( x+1 \right)$ is a factor of \[\] ${{x}^{3}}-6{{x}^{2}}+3x+10$ by

 factor theorem.

Now, let us divide ${{x}^{3}}-6{{x}^{2}}+3x+10$ by $\left( x+1 \right)$.

$\begin{align}
  & \left( x+1 \right)\overset{{{x}^{2}}-7x+10}{\overline{\left){{{x}^{3}}-6{{x}^{2}}+3x+10}\right.}} \\

 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{3}}+{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\

 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\begin{align}

  & \,\,\,\,\,\,\,\,\,\,\,\,\,-7{{x}^{2}}+3x \\

 & \\

 & \overline{\begin{align}

  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10x+10 \\

 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10x+10 \\

 & \,\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,}\,\,\,\, \\

\end{align}} \\

 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\

\end{align}} \\

\end{align}$

$\begin{align}

  & \Rightarrow {{x}^{3}}-6{{x}^{2}}+3x+10=\left( x+1 \right)\left( {{x}^{2}}-7x+10 \right).........................(i) \\

 & \\

\end{align}$

Now, let us further factorise \[~\left( {{x}^{2}}-7x+10 \right)\]

By factor theorem: \[~\left( {{x}^{2}}-7x+10 \right)\]\[=\left( x-\alpha \right)\left( x-\beta \right)\]

\[\alpha \,and\,\beta \]are the quadratic polynomial \[~\left( {{x}^{2}}-7x+10 \right)\] zeroes.

By theory of equations for quadratic equation:

\[\begin{align}

  & sum\,of\,roots=\,\dfrac{-coefficient\,of\,x}{coefficient\,of\,{{x}^{2}}}\,\,\,\,and\, \\

 & product\,of\,roots\,=\dfrac{constant\,term}{coefficient\,of\,{{x}^{2}}}\, \\

\end{align}\]

So, \[\alpha +\beta =7\,and\,\alpha \beta =10\]

By hit and trial, we can guess that \[\alpha =5\,\,and\,\,\,\beta =2\] or \[\alpha =2\,\,and\,\,\,\beta =5\] satisfies the above two equations.

So, by factor theorem-

\[~\left( {{x}^{2}}-7x+10 \right)=\left( x-2 \right)\left( x-5 \right)....................(ii)\]

From equation (i) and (ii)-

\[{{x}^{3}}-6{{x}^{2}}+8x+10=\left( x+1 \right)\left( x-2 \right)\left( x-5 \right)\]

Hence, the required factorization of \[{{x}^{3}}-6{{x}^{2}}+8x+10=\left( x+1 \right)\left( x-2 \right)\left( x-5 \right)\].


Note: Reverse of factor theorem also holds i.e. \[\left( x-a \right)\] is a factor of polynomial \[f\left( x \right)\] of degree \[n\ge 1\], then \[f\left( x \right)\] will be equal to zero.