Question

What is the value of $\dfrac{{{x}^{3}}+4{{x}^{2}}+x-6}{x+2}$?

Answer 1

Hint: In this problem we need to simplify the given fraction. We can observe that the given fraction has a cubic equation as numerator and a monomial in the denominator. First, we will equate the denominator to zero and calculate the value of $x$. Now we will check whether the calculated value of $x$ is a factor of the cubic equation which is in the numerator by substituting the calculated $x$ value in the cubic equation. If we get the result as zero then we can say that the substitute value is a factor of the cubic equation. After that we will consider the cubic equation and rearrange the terms so that we can take the term $x+2$ as common and simplify the equation to get the factorial form of the cubic equation. Once we have the factorial form of the cubic equation, we will substitute this value in the given fraction to get the required result.

Complete step-by-step solution:
Given fraction $\dfrac{{{x}^{3}}+4{{x}^{2}}+x-6}{x+2}$.
In the above fraction we have $x+2$ as a denominator and ${{x}^{3}}+4{{x}^{2}}+x-6$.
Considering the denominator and equating to zero in order to find the value of $x$, then we will get
$\begin{align}
  & x+2=0 \\
 & \Rightarrow x=-2 \\
\end{align}$
Substituting $x=-2$ in the cubic equation ${{x}^{3}}+4{{x}^{2}}+x-6$, then we will get
${{x}^{3}}+4{{x}^{2}}+x-6={{\left( -2 \right)}^{3}}+4{{\left( -2 \right)}^{2}}+\left( -2 \right)-6$
Simplifying the above equation by using mathematical operations, then we will have
$\begin{align}
  & {{x}^{3}}+4{{x}^{2}}+x-6=-8+4\left( 4 \right)-2-6 \\
 & \Rightarrow {{x}^{3}}+4{{x}^{2}}+x-6=16-16 \\
 & \Rightarrow {{x}^{3}}+4{{x}^{2}}+x-6=0 \\
\end{align}$
We have the value of ${{x}^{3}}+4{{x}^{2}}+x-6$ for $x=-2$ as $0$. So, we can say that the value $x=-2$ is the root of the equation ${{x}^{3}}+4{{x}^{2}}+x-6$.
Considering the equation ${{x}^{3}}+4{{x}^{2}}+x-6$. We are going to rearrange all the terms in the above equation so that we can easily take $x+2$ as common, then we will get
$\begin{align}
  & {{x}^{3}}+4{{x}^{2}}+x-6={{x}^{3}}+2{{x}^{2}}+2{{x}^{2}}+4x-3x-6 \\
 & \Rightarrow {{x}^{3}}+4{{x}^{2}}+x-6={{x}^{2}}\left( x+2 \right)+2x\left( x+2 \right)-3\left( x+2 \right) \\
\end{align}$
Taking $x+2$ as common from the above equation, then we will get
${{x}^{3}}+4{{x}^{2}}+x-6=\left( x+2 \right)\left( {{x}^{2}}+2x-3 \right)$
Substituting the above value in the given fraction $\dfrac{{{x}^{3}}+4{{x}^{2}}+x-6}{x+2}$, then we will get
$\dfrac{{{x}^{3}}+4{{x}^{2}}+x-6}{x+2}=\dfrac{\left( x+2 \right)\left( {{x}^{2}}+2x-3 \right)}{\left( x+2 \right)}$
Cancelling the term $x+2$ which is in both numerator and denominator, then the given fraction is modified as
$ \dfrac{{{x}^{3}}+4{{x}^{2}}+x-6}{x+2}={{x}^{2}}+2x-3$

Note: We can also solve this problem in another method which is called a long polynomial division. In this method we will divide the cubic equation which is in the numerator with the monomial which is in the denominator. After getting the remainder is zero we can write the quotient as required.