Question

How do you factor given that $f\left( 6 \right)=0$ and $f\left( x \right)={{x}^{3}}-3{{x}^{2}}-16x-12$ . ?

Answer 1

Hint: Now we are given that $f\left( 6 \right)=0$ . Hence we have 6 is the root and $x-6$ is the factor of the given quadratic. Now we will divide the equation by $\left( x-6 \right)$ and get a quadratic. Now we will factorize the quadratic using splitting the middle term method. Hence we will split the middle terms such that the product of the two terms is multiplication of first and last term. Now on simplifying we get the factors of the equation.

Complete step by step solution:
Now we are given that $f\left( x \right)={{x}^{3}}-3{{x}^{2}}-16x-12$ and $f\left( 6 \right)=0$.
Now since $f\left( 6 \right)=0$ we have that 6 is the root of the equation.
Hence we get $x-6$ is the factor of the given equation.
Now dividing the given equation by $\left( x-6 \right)$ we get,
$\begin{align}
  & \text{ }\left( {{x}^{2}}+3x+2 \right) \\
 & \left( x-6 \right)|\overline{{{x}^{3}}-3{{x}^{2}}-16x-12} \\
 & \text{ }\!\!|\!\!\text{ }{{x}^{3}}-6{{x}^{2}} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ }3{{x}^{2}}-16x}\text{ } \\
 & \text{ }\!\!|\!\!\text{ }3{{x}^{2}}-18x \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ }2x-12} \\
 & \text{ }\!\!|\!\!\text{ }2x-12 \\
\end{align}$
Hence we get,
${{x}^{3}}-3{{x}^{2}}-16x-12=\left( x-6 \right)\left( {{x}^{2}}+3x+2 \right)$
Now we will factorize the quadratic equation ${{x}^{2}}+3x+2$ .
To do so we will split the middle term such that the product of the terms is the the multiplication of first term and last term. Hence we get.
$\Rightarrow \left( x-6 \right)\left( {{x}^{2}}+2x+x+2 \right)$
Now in the quadratic we will take x common from first two terms and 1 common from last two terms. Hence we get,
$\begin{align}
  & \Rightarrow \left( x-6 \right)\left( x\left( x+2 \right)+1\left( x+2 \right) \right) \\
 & \Rightarrow \left( x-6 \right)\left( x+1 \right)\left( x+2 \right) \\
\end{align}$
Hence we get the factorization of the equation as $\left( x-6 \right)\left( x+1 \right)\left( x+2 \right)$ .

Note: Now note that whenever we have a cubic equation we will try to get one root of the equation by substituting different values in x and hence finding one solution. Now we will use division to get the quadratic. Now we can also find the roots of quadratic equation by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . And now if $\alpha $ and $\beta $ are the roots of the equation then $x-\alpha $ and $x-\beta $ are the factors of the equation. Hence we get all the factors of the equation.