Question

How do you factor $f\left( x \right)={{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x+208$?

Answer 1

Hint: Now to factorize the equation we will first try to find the root of the equation by substituting different values of x. Hence we will find a root of the equation. Now if $\alpha $ is the root of the equation then we have $x-a$ is the factor of the equation. Now we will divide the given equation with the obtained factor. Hence we will get a cubic equation. Now we will again find the root of the cubic equation by substituting different values of x. Now again we will divide the cubic equation with the obtained factor. Hence we get a quadratic equation. Now we will find the roots of the quadratic equation with the help of formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we will find all the factors of the given equation.

Complete step by step solution:
Now to factorize the given expression we must first guess two roots of the equation.
Now we will substitute different values of x and check for which value we have ${{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x+208=0$
Now we can see that if we substitute x = - 4 we get, $\Rightarrow {{\left( -4 \right)}^{4}}+4\times {{\left( -4 \right)}^{3}}-3\times {{\left( -4 \right)}^{2}}+40\times \left( -4 \right)+208=0$
We know that if x = -4 is the root of the equation then $x-\left( -4 \right)=x+4$ is the factor of the equation.
Now let us divide the whole equation by $\left( x+4 \right)$
$\begin{align}
  & \text{ }\left( {{x}^{3}}-3x+52 \right) \\
 & \left( x+4 \right)|\overline{{{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x-208} \\
 & \text{ }\!\!|\!\!\text{ }{{x}^{4}}+4{{x}^{3}} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ }-3{{x}^{2}}+40x}\text{ } \\
 & \text{ }\!\!|\!\!\text{ }-3{{x}^{2}}-12x \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ }52x-12} \\
 & \text{ }\!\!|\!\!\text{ 52}x-12 \\
\end{align}$
$\Rightarrow \left( {{x}^{3}}-3x+52 \right)\left( x+4 \right)={{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x+208$
Now we have formed a cubic equation. Hence we will now try to guess the root of the cubic equation.
Now we can see that x = -4 is also the root of the cubic equation as $-{{4}^{3}}-3\left( -4 \right)+52=0$
Hence we have $\left( x+4 \right)$ is also the factor of the obtained cubic equation.
Now we will divide the obtained cubic equation with $\left( x+4 \right)$.
\[\begin{align}
  & \text{ }\left( {{x}^{2}}-4x+13 \right) \\
 & \left( x+4 \right)|\overline{{{x}^{3}}-3x+52} \\
 & \text{ }\!\!|\!\!\text{ }{{x}^{3}}+4{{x}^{2}} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ }-4{{x}^{2}}-3x+52}\text{ } \\
 & \text{ }\!\!|\!\!\text{ }-4{{x}^{2}}-16x \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ 13}x+52} \\
 & \text{ }\!\!|\!\!\text{ 13}x+52 \\
\end{align}\]
Hence we get,
$\left( {{x}^{2}}-4x+13 \right){{\left( x+4 \right)}^{2}}={{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x+208$
Now to factorize the quadratic equation we will find the roots of the equation.
Now to find the roots of the equation we will use the formula find the roots of quadratic equation which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now comparing the quadratic equation ${{x}^{2}}-4x+13$ with $a{{x}^{2}}+bx+c$ a = 1, b = -4 and c = 13.
Hence using the formula we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 13 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{16-52}}{2} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{-36}}{2} \\
 & \Rightarrow x=\dfrac{4\pm 6i}{2} \\
\end{align}$
Hence $x=2+3i$ and $x=2-3i$ are three roots of the equation.
Hence we have $\left( x-\left( 2+3i \right) \right)$ and $\left( x-\left( 2-3i \right) \right)$ are the factors of the equation ${{x}^{2}}-4x+13$
Hence the factors of the given equation are $\left( x-\left( 2+3i \right) \right)\left( x-\left( 2-3i \right) \right){{\left( x+4 \right)}^{2}}={{x}^{4}}+4{{x}^{3}}-3{{x}^{2}}+40x+208$
Hence we have the factors of the equation.

Note: Now note that the roots of the quadratic equation can be real roots or complex roots. To check the nature of roots we have defined discriminant $D={{b}^{2}}-4ac$ . Now if $D>0$ then the roots are real and distinct. Now if $D=0$ then the roots are real and repeating and if the value of $D<0$ then the rots are complex roots.