Introduction to Partial Fraction Expansion/Decomposition (PFE) (Part 2)
(This page is Part 2. Click here for Part 1.)
A Simple Example: PFE with Distinct Linear Factors
This simple example shows the basic process of Partial Fraction Expansion. The way to handle situations other than distinct linear factors is discussed in subsequent sections.
Find the Partial Fraction Expansion of $\,\frac{5x-2}{2x^2 - x - 1}$
Step 1: Check That the Degree of the Numerator is Strictly Less Than the Degree of the Denominator
The numerator, $\,5x-2\,,$ has degree $\,1\,.$
The denominator, $\,2x^2 -x - 1\,,$ has degree $\,2\,.$
$\,1\,$ is strictly less than $\,2\,.$ Check!
Step 2: Completely Factor the Denominator into Linear Factors and Irreducible Quadratics
$$\cssId{s12}{2x^2 - x - 1 = (2x+1)(x-1)}$$(This denominator has distinct linear factors.)
Foolproof Factorization Method: If you don't quickly see the factorization, use the quadratic formula to find that $-\frac 12\,$ and $\,1\,$ are the zeros of $\,2x^2 - x - 1\,.$ So, $\,x+\frac 12\,$ and $\,x-1\,$ are factors. Supply the constant factor (the leading coefficient) yourself.
Step 3: Different Types of Factors in the Denominator Give Rise to Different Term(s) in the Partial Fraction Expansion
Each distinct linear factor $\,ax+b\,$ gives rise to a term of the form $\,\displaystyle\frac{A}{ax+b}\,$ in the PFE:
- The factor $\,2x+1\,$ gives rise to a term of the form $\,\frac{A}{2x+1}\,.$
- The factor $\,x-1\,$ gives rise to a term of the form $\,\frac{B}{x-1}\,.$
Together:
$$ \begin{align} \cssId{s26}{\frac{5x-2}{2x^2 - x - 1}} &\cssId{s27}{\ =\ \color{blue}{\frac{5x-2}{(2x+1)(x-1)}}}\cr\cr &\cssId{s28}{\ =\ \color{green}{\frac{A}{2x+1} + \frac{B}{x-1}}} \end{align} $$To complete the process, we must find the unknown constants $\,A\,$ and $\,B\,.$
Step 4: Solve For the Unknown Constants
(Repeated from above, for your convenience:)
$$ \color{blue}{\frac{5x-2}{(2x+1)(x-1)}} = \color{green}{\frac{A}{2x+1} + \frac{B}{x-1}} $$(Now, clear fractions:)
$$ \begin{align} &\cssId{s32}{\color{blue}{\frac{5x-2}{(2x+1)(x-1)}}} \cssId{s33}{\cdot \color{red}{(2x+1)(x-1)}}\cr\cr &\quad\cssId{s34}{= \left[\color{green}{\frac{A}{2x+1} + \frac{B}{x-1}}\right]} \cssId{s35}{\color{red}{(2x+1)(x-1)}} \end{align} $$$$ \cssId{s36}{5x - 2 = A(x-1) + B(2x+1)}\ \ \cssId{s37}{(*)} $$
Clear fractions:
- Multiply both sides of the equation by the original denominator (in factored form)
- Cancel
When there are only distinct linear factors, you don't need to solve a system of equations. Hooray! (Keep reading!)
Choose $\,x = 1\,.$ Substitute this value into equation (*). This makes $\,x-1\,$ equal zero, and hence $\,A\,$ disappears:
$$ \begin{gather} \cssId{s46}{5(1) - 2 = A(1-1) + B(2\cdot 1+1)}\cr\cr \cssId{s47}{3 = 3B}\cr\cr \cssId{s48}{B = 1} \end{gather} $$Choose $\,x = -\frac 12\,.$ Substitute this value into equation (*). This makes $\,2x+1\,$ equal zero, and hence $\,B\,$ disappears:
$$ \begin{gather} \cssId{s51}{5(-\textstyle{\frac 12}) - 2 = A(-{\textstyle\frac12}-1) + B(2\cdot -{\textstyle\frac 12}+1)}\cr\cr \cssId{s52}{-\frac 92 = -\frac 32A}\cr\cr \cssId{s53}{A = 3} \end{gather} $$Here's the rationale for what we just did:
- Equation (*) must be true for all real numbers $\,x\,.$
- Therefore, equation (*) must be true for any particular value of $\,x\,.$
- Choose values that make one of the unknowns disappear!
Step 5: Report the Answer. Spot-Check to Gain Confidence in your Result.
Substitute in the values of $\,A\,$ and $\,B\,$ found in the prior step:
$$ \cssId{s61}{\frac{5x-2}{2x^2 - x - 1} = \frac 3{2x+1} + \frac 1{x-1}} $$A quick check for an easy value of $\,x\,$ doesn't take much time, and catches many mistakes. When $\,x = 0\,$ is available, it's often easiest to use this value for your spot-check.
Don't drop the question mark over the equals sign until you're sure that both sides are equal!
$$ \begin{gather} \cssId{s65}{\frac{5\cdot 0-2}{2\cdot 0^2 - 0 - 1} \overset{?}{=} \frac 3{2\cdot 0+1} + \frac 1{0 - 1}}\cr\cr \cssId{s66}{\frac{-2}{-1} \overset{?}{=} 3 - 1}\cr\cr \cssId{s67}{2 = 2}\cr \cssId{s68}{\text{Check!}} \end{gather} $$