Question

How do you express $\dfrac{5x-1}{{{x}^{2}}-x-2}$ in partial fractions?

Answer 1

Hint: We first try to describe the requirement and the process of finding the partial fractions. Then we factorise the denominator of $\dfrac{5x-1}{{{x}^{2}}-x-2}$. We form the numerator of $\dfrac{5x-1}{{{x}^{2}}-x-2}$ with the factors. We simplify the equation $\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$ to find the partial fraction form.

Complete step by step solution:
We have to form the partial form of $\dfrac{5x-1}{{{x}^{2}}-x-2}$.
A brief introduction to partial fractions is required to solve the problem.
The condition for the process is to break the denominator in its factor forms and rearrange the numerator in addition or subtraction form of those exact factors in the denominator.
This process is required to break the whole expression in its simpler form.
For our given problem $\dfrac{5x-1}{{{x}^{2}}-x-2}$, we first break ${{x}^{2}}-x-2$ into factors.
Using middle-term factorisation we get
$\begin{align}
  & {{x}^{2}}-x-2 \\
 & ={{x}^{2}}-2x+x-2 \\
 & =x\left( x-2 \right)+\left( x-2 \right) \\
 & =\left( x-2 \right)\left( x+1 \right) \\
\end{align}$
Now we have to arrange the numerator $5x-1$ in the binary addition and subtraction of the same factors of ${{x}^{2}}-x-2$ which are $\left( x-2 \right),\left( x+1 \right)$.
We can do that directly or apply the form where $\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$.
We simplify the $\dfrac{A}{x-2}+\dfrac{B}{x+1}$. We get $\dfrac{A}{x-2}+\dfrac{B}{x+1}=\dfrac{A\left( x+1 \right)+B\left( x-2 \right)}{\left( x-2 \right)\left( x+1 \right)}$.
So, $\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1}=\dfrac{A\left( x+1 \right)+B\left( x-2 \right)}{\left( x-2 \right)\left( x+1 \right)}$ gives $5x-1=A\left( x+1 \right)+B\left( x-2 \right)$.
Simplifying we get $5x-1=A\left( x+1 \right)+B\left( x-2 \right)=x\left( A+B \right)+\left( A-2B \right)$.
Equating the variables and constants we get $\left( A+B \right)=5;\left( A-2B \right)=-1$.
Subtracting the equations, we get
$\begin{align}
  & \left( A+B \right)-\left( A-2B \right)=5+1 \\
 & \Rightarrow 3B=6 \\
 & \Rightarrow B=\dfrac{6}{3}=2 \\
\end{align}$
This gives $A=5-B=5-2=3$.
The partial fraction form becomes $\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{3}{x-2}+\dfrac{2}{x+1}$.

Note: The form gets trickier when there are the same factors within its power form. The indices of the numerator and the denominator decides how to form the partial fractions. If they are equal then we start with a constant term as the additional value. The forms are mentioned below