Partial Fraction Decomposition - Beginner-Friendly Guide

Whether you're just starting to learn partial fraction decomposition in high school, prepping for a standardized exam, or looking for a refresher, you've come to the right place. 

In this guide, you’ll find simple definitions, step-by-step explanations, and solved examples of partial fraction decomposition. 

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What Is Partial Fraction Decomposition?  

Partial fraction decomposition is a technique in algebra that helps break down complex rational expressions into simpler, smaller fractions.  

Think of partial fraction decomposition like breaking down a complex recipe into its basic ingredients.  

Imagine you have a complicated recipe with lots of steps and ingredients all mixed together. It might be overwhelming to follow, right?  

But if you separate each ingredient and prepare it step by step, the whole recipe becomes easier to understand and manage.  

In partial fraction decomposition, you’re taking a complex rational expression and breaking it into simpler fractions, known as partial fractions.  

These smaller fractions are much easier to handle, and when you add them back together, they recreate the original complex expression — just like combining ingredients back to make the original recipe. 

Let’s Refresh Our Memory: What Are Rational Expressions?  

A quick detour before we proceed with partial fraction decomposition: To truly understand this technique, we first need to understand rational expressions. Rational expression is a fraction where both the numerator (the top part) and the denominator (the bottom part) are polynomials.  

A polynomial is an expression made of variables (like x) and constants (like 1, 2, 3), combined using addition, subtraction, and multiplication. 

For example: \( \Large \frac{2x+3}{x^2-1}\)

In this example, the numerator is 2x+3 (a polynomial), and the denominator is x²-1 (another polynomial).  

Rational expressions are really important in algebra because they come up in many problems involving fractions with variables.  

However, unlike regular fractions where the denominator is just a number, in rational expressions, we have to pay attention to what the variables are doing in the denominator.  

For instance, we always have to make sure the denominator isn't equal to zero because dividing by zero is undefined.  

Let’s look at a quick example: \( \Large \frac{5}{x-2}\)

This rational expression is fine as long as 𝑥≠2, because if 𝑥=2, the denominator becomes zero, and the expression doesn’t make sense. 

Here are a few more examples to help you understand:  

1. \( \Large \frac{3x+8}{(𝑥−4)(𝑥+1)}\)
2. \( \Large \frac{5}{(𝑥^2-5x+6)}\)
3. \( \Large \frac{7x-5}{x2-9}\)

Notice that in all the examples above, some rational expressions look simple, while others can seem more complex. A complex rational expression might be hard to integrate, solve, or analyze in its original form. By decomposing it into partial fractions, each smaller fraction becomes much easier to solve on its own. 

A Step-by-Step Guide to Decomposing Partial Fractions 

Now that we’re caught up with rational expressions, let’s get back to decomposing partial fractions. We’ll work out the steps using an example. 

Suppose we want to decompose the following rational function: 

\( \Large \frac{P(x)}{Q(x)}\)=\( \Large \frac{6x+7}{(x-1)(x+2)}\)

Here, 𝑄(𝑥)= (𝑥−1)(𝑥+2) consists of two linear factors (𝑥−1) and (𝑥+2).

Step 1: Ensure the Proper Fraction 

The first step is to check if the rational function is proper, meaning the degree of the numerator is less than the degree of the denominator. In this case: 

• The degree of the numerator 𝑃(𝑥)=6𝑥+7 is 1. 
• The degree of the denominator 𝑄(𝑥)=(𝑥−1)(𝑥+2)=𝑥²+𝑥−2 is 2.

Since the degree of the numerator is less than the degree of the denominator, the function is proper.  

If the function were improper (i.e., if the numerator had a higher degree), we would need to perform polynomial long division first.  

Step 2: Factor the Denominator  

The next step is to write the rational function as a sum of partial fractions, assigning a variable to each linear denominator factor.  

Since the factors in this example are linear, they each get a single constant numerator. 

So, we assume: 

\( \Large \frac{6x+7}{(x-1)(x-2)}\)=\( \Large \frac{A}{x-1}\)+\( \Large \frac{B}{x+2}\)

Here, A and B are unknowns we need to solve.  

Step 3: Combine the Right-Hand Side 

To combine the right-hand side of the equation, we need to find a common denominator. In this case, the denominators are (𝑥−1) and (𝑥+2), so we multiply the numerator and denominator of each fraction by the missing factor to make the denominators the same. This ensures that both fractions will have the same denominator of (𝑥−1)(𝑥+2).

We multiply the first fraction \( \Large \frac{𝐴}{(𝑥−1)}\) by \( \Large \frac{(𝑥−1)}{(𝑥−1)}\), and the second fraction \( \Large \frac{B}{x+2}\) by \( \Large \frac{(x+2)}{(x+2)}\). 

This gives us:  

(\( \Large \frac{A}{(x-1)}\) x \( \Large \frac{(x+2)}{(x+2)}\)) + (\( \Large \frac{B}{(x+2)}\) x \( \Large \frac{(x-1)}{(x-1)}\))

This step results in:  

\( \Large \frac{A(x+2)}{(x-1)(x+2)}\)+\( \Large \frac{B(x-1)}{(x-1)(x+2)}\)

Now that both fractions have the same denominator, we can combine the numerators, giving us: 

\( \Large \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}\)

This prepares the equation for the next step of simplifying and solving for the unknowns A and B. 

 \( \Large \frac{A}{x-1}\)+\( \Large \frac{B}{x+2}\)=\( \Large \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}\)

This gives us:

\( \Large \frac{6x+7}{(x-1)(x+2)}\)=\( \Large \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}\)

Step 4: Eliminate the Denominators 

Since the denominators are the same on both sides of the equation, we can eliminate them, leaving us with an equation in terms of X: 

6𝑥+7=𝐴(𝑥+2)+𝐵(𝑥−1)

Now, expand both sides of the equation: 

6𝑥+7=(𝐴 × 𝑥)+2𝐴+(𝐵 × 𝑥)−𝐵 

This simplifies to: 

6𝑥+7=(𝐴+𝐵)𝑥+(2𝐴−𝐵)

Step 5: Solve for the Unknowns  

To solve for A and B, we compare the coefficients of x and the constant terms on both sides of the equation. 

• The coefficient of 𝑥 on the left side is 6, so: 

 𝐴+𝐵=6 

• The constant term on the left side is 7, so: 

2𝐴−𝐵=7 

 Now solve this system of equations. Start with the first equation: 

𝐴+𝐵=6 (Equation 1) 

2𝐴−𝐵=7 (Equation 2) 

From Equation 1, express 𝐵 in terms of 𝐴:  

B=6−𝐴 

Substitute this into Equation 2: 

2𝐴−(6−𝐴)=7

Simplify and solve for 𝐴: 

2𝐴−6+𝐴=7 

3𝐴=13 

A=\( \Large \frac{13}{3}\)

Now substitute A into the expression for B: 

𝐵=6 −\( \Large \frac{13}{3}\)=\( \Large \frac{18}{3}\)-\( \Large \frac{13}{3}\)=\( \Large \frac{5}{3}\)

Step 6: Write the Decomposed Fractions

Now that we have A=\( \Large \frac{13}{3}\) and B=\( \Large \frac{5}{3}\) substitute them back into the partial fraction decomposition:  

\( \Large \frac{6x+7}{(x-1)(x+2)}\)=\(\Large \frac{\Large \frac{13}{3}}{(x-1)}\)+\(\Large \frac{\Large \frac{5}{3}}{(x-1)}\)

You can rewrite this as:

\( \Large \frac{6x+7}{(x-1)(x+2)}\)=\( \Large \frac{13}{3(x-1)}\)+\( \Large \frac{5}{3(x+2)}\)

And that's it! You've successfully decomposed the fraction!

Recap the steps: 

1. Check if the fraction is proper. 
2. Factor the denominator. 
3. Combine the right-hand side. 
4. Eliminate the denominators 
5. Solve for the unknowns. 
6. Write your final decomposed fractions. 

Solved Example of Partial Fraction Decomposition 

Let's go through another example of decomposing a rational function with linear factors in the denominator. 

\( \Large \frac{8x+5}{(x+1)(x-3)}\)

Remember your steps – first, you need to check if the fraction is proper.  

The given rational function is already proper because the degree of the numerator (1) is less than the degree of the denominator (2). So we can proceed directly to decomposition. 

For the next step, we need to write the decomposition for the function, assuming constants 𝐴 and 𝐵 in the numerators for each denominator factor: 

\( \Large \frac{8x+5}{(x+1)(x-3)}\) = \( \Large \frac{A}{x+1}\)+\( \Large \frac{B}{x-3}\)

To combine the fractions on the right-hand side, we first find a common denominator.  

We multiply the numerator and denominator of each fraction by the missing factor to make the denominators the same. This ensures that both fractions will have the same denominator of (𝑥+1)(𝑥+3).

This gives us: 

(\( \Large \frac{A}{(x+1)}\)x \( \Large \frac{(x-3)}{(x-3)}\))+(\( \Large \frac{B}{(x-3)}\)x\( \Large \frac{(x+1)}{(x+1)}\))

And the step results in:

\( \Large \frac{A(x-3)}{(x+1)(x-3)}\)+\( \Large \frac{B(x+1)}{(x+1)(x-3)}\)

Both fractions have the same denominator, so now we can combine the numerators, giving us:

\( \Large \frac{A(x-3)+B(x+1)}{(x+1)(x-3)}\)

This prepares the equation for the next step of simplifying and solving for the unknowns A and B. 

\( \Large \frac{A}{(x+1)}\)+\( \Large \frac{B}{(x-3)}\)=\( \Large \frac{A(x-3)+B(x+1)}{(x+1)(x-3)}\)

Now both sides of the equation have the same denominator, so we can equate the numerators and get: 

8𝑥+5=𝐴(𝑥−3)+𝐵(𝑥+1)

Then, we expand the terms on the right-hand side: 

𝐴(𝑥−3)=𝐴 × 𝑥−3𝐴

𝐵(𝑥+1)=𝐵 × 𝑥+𝐵 

So the equation becomes: 

8𝑥+5=(𝐴 × 𝑥)−3𝐴+(𝐵 × 𝑥)+𝐵

Now, group the x-terms and the constants together: 

8𝑥+5=(𝐴+𝐵)𝑥+(−3𝐴+𝐵)

The fifth step involves comparing the coefficients of x and the constant terms on both sides of the equation: 

• Coefficient of X: 𝐴+𝐵=8 
• Constant term: −3𝐴+𝐵=5

We now have a system of two equations: 

Equation 1: 𝐴+𝐵=8 

Equation 2: −3𝐴+𝐵=5 

Solve the system of equations: 

From Equation 1, solve for 𝐵 in terms of 𝐴: 

𝐵=8−𝐴 

Substitute this into Equation 2: 

−3𝐴+(8−𝐴)=5

Simplify: 

−3𝐴+8−𝐴=5 

−4𝐴+8=5 

−4𝐴=−3  

 𝐴=34

Now substitute A back in the expression for B: 

𝐵=8−\( \Large \frac{3}{4}\)=\( \Large \frac{32}{4}\)-\( \Large \frac{3}{4}\)=\( \Large \frac{29}{4}\)

And for the final step, write down the final decomposition.

Substitute the values we have for A and B back into the partial fraction decomposition: 

\( \Large \frac{8x+5)}{(x-1)(x+3)}\)=\(\Large \frac{\Large \frac{3}{4}}{(x-1)}\)+\(\Large \frac{\Large \frac{29}{4}}{(x+3)}\)

You can rewrite it like this: 

\( \Large \frac{8x+5}{(x-1)(x+3)}\)=\( \Large \frac{3}{4(x-1)}\)+\( \Large \frac{29}{4(x+3)}\)

Practice Partial Fractions with Mathnasium La Jolla

Quiz! Test Your Knowledge of Partial Fraction Decomposition 

To solidify your knowledge of partial fraction decomposition, try and solve the examples below: 

1. \( \Large \frac{4x+1}{(x+2)(x-3)}\)
2. \( \Large \frac{x+7}{(x+1)(x-2)}\)
3. \( \Large \frac{6x+3}{(x+2)(x-3)}\)
4. \( \Large \frac{9}{(x+7)(x-1)}\)
5. \( \Large \frac{6x-1}{(x+3)(x-5)}\)

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FAQs About Partial Fraction Decomposition 

1. When do I use partial fraction decomposition? 

You use partial fraction decomposition when you have a rational expression with a more complicated denominator (bottom of the fraction) that can be factored.  

It’s often used in calculus for integration but can also be helpful in solving algebraic equations involving fractions. 

2. How do I know if a fraction can be decomposed? 

To see if a fraction can be decomposed, check if the denominator can be factored into simpler expressions. If it can, you can use partial fraction decomposition.  

For example, if the denominator is 𝑥²−1 you can factor it into (𝑥+1)(𝑥+1) which means it’s possible to decompose it. 

3. Can partial fraction decomposition be used for any rational expression? 

Partial fraction decomposition works best with proper rational expressions, where the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is higher or the same as the denominator, you’ll need to perform polynomial division first. 

4. What should I do if I get stuck on a partial fraction decomposition problem? 

If you’re stuck, go back and check each step carefully: 

• Make sure the denominator is factored correctly. 
• Double-check that each partial fraction is set up for each factor in the denominator. 
• Review your steps for solving the variables in the numerators.  
• If all else fails, ask for help from your Mathnasium La Jolla tutor.  

PSST! Check Your Quiz Answers Here 

Here are the answers to the 6 exercise problems, let’s see how you did! 

1. \( \Large \frac{7}{5(x+13)}\)+\( \Large \frac{13}{5(x-3)}\)
2. \( \Large \frac{-2}{x+1}\)+\( \Large \frac{3}{x-2}\)
3. \( \Large \frac{9}{5(x+2)}\)+\( \Large \frac{21}{5(x-3)}\)
4. -\( \Large \frac{9}{8(x+7)}\)+\( \Large \frac{9}{8(x-1)}\)
5. \( \Large \frac{19}{8(x+3)}\)+\( \Large \frac{29}{8(x-5)}\)
6. \( \Large \frac{-1}{x+2}\)+\( \Large \frac{3}{x-2}\)