Algebra:
Partial Fractions

1. 
\(\frac{8x+22}{{x}^{2}+4x-5}\)
 Solution
2. 
\(\frac{2x-3}{{x}^{3}+x}\)
 Solution
3. 
\(\frac{8x+22}{(x+1)(2+x)(x+3)}\)
 Solution
4. 
\(\frac{{x}^{2}+x+2}{{({x}^{2}+2)}^{2}}\)
 Solution
5. 
\(\frac{u}{{(1+u)}^{2}}\)
 Solution
6. 
\(\frac{x}{(2x+1)(2x+3)}\)
 Solution
7. 
\(\frac{1}{(2x+1)(2x+3)}\)
 Solution

Partial Fraction - Introduction

You have probably learned how to combine or simplify fractions that contain polynomials. These are fractions with rational expressions with variables in the numerator, denominator, or both. Naturally, it is also possible to reverse the process and discover the original set of polynomial fractions. This process is called partial fraction decomposition. Let's try to decompose some fractions in the next section!

Decomposing Polynomial Fractions

Let’s start with the simplified fraction
\(\frac{8x+22}{{x}^{2}+4x-5}\)
. How do we backtrack and discover the original set of fractions? Let's begin.
Step 1: Use polynomial factoring techniques to simplify the denominator. Let's rewrite the second term in the denominator
\({x}^{2}+4x-5\)
as a sum of two terms. You might ask, which two? First, let's find two numbers that sum up to
\(4\)
and multiply to
\(-5\)
. After trying out the factors of
\(-5\)
, you will find that these two numbers are
\(-1\)
and
\(5\)
. This allows us to simplify the denominator as
\((x-1)(x+5)\)
.
Step 2: Create a new set of fractions, each using one factor from the original denominator as their denominator. Since we do not know yet what the numerators are, let's assign them values of
\(A\)
and
\(B\)
. This gives the equation:
\(\frac{8x+22}{(x-1)(x+5)}=\frac{A}{x+5}+\frac{B}{x-1}\)
Step 3: Now, we have an equation with two unknowns. Our goal is to solve the equation and find their values. First, let's cross multiply, so that the right hand side has a single denominator.
\(\frac{8x+22}{(x-1)(x+5)}=\frac{A(x+5) + B(x-1)}{(x-1)(x+5)}\)
Step 4: We are now ready to cancel out the denominators. How do we do that? Let's multiply every term in the equation by
\(8x+22=A(x+5)+B(x-1)\)
Step 5: Expand the right side of the equation:
\(8x+22=Ax+5A+Bx-B\)
Step 6: Group terms by degree of
\(x\)
:
\(8x+22=(A+B)x+5A-B\)
Step 7: Create an equation system for
\(x\)
:
1) Since
\(8x\)
must equal
\((A+B)x\)
, this means
\(A+B\)
equals
\(8\)
.
2) Meanwhile,
\(5A-B\)
must equal
\(22\)
.
Therefore, our equation system is:
\(\begin{array}{lcl}{A+B=8}\\{5A-B=22}\end{array}\)
Step 8: Solve for
\(A\)
and
\(B\)
:
\(\begin{array}{lcl}{A=5}\\{B=3}\end{array}\)
There are quite many steps involved here. See here on how A and B are solved.
Step 9: Substitute the new values of
\(A\)
and
\(B\)
into the original equation to give:
\(\frac{8x+22}{(x-1)(x+5)}=\frac{5}{x+5}+\frac{3}{x-1}\)
We are done! Using partial fraction decomposition, we have successfully separated the fraction back to its original components, which are the two fractions above.

The Cymath Benefit

Want to get better at handling partial fractions? Try some of our partial fraction practice problems at the top of this page. Using the basic steps outlined above, challenge yourself to solve some of the questions without help. Stuck? Click on "Solution" to see how the partial fraction is decomposed, or sign up for Cymath Plus for additional help.