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Partial Fractions

Q: What does 'irreducible quadratic' mean again?

An irreducible quadratic is a quadratic expression, like $x^2+9$ or $x^2+x+1$, that cannot be factored into two linear terms using only real numbers. You can test this with the discriminant $b^2-4ac$; if it's negative, the quadratic is irreducible.

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Ever looked at a giant, messy fraction and wished you could split it into smaller, more manageable pieces? That's precisely what partial fraction decomposition allows you to do. It's a powerful algebraic technique for simplifying complex rational expressions, paving the way for more advanced math.

Partial Fractions — an original Algebra911 reference diagram defining partial fractions and a worked example. — A Beginner's Guide to Partial Fractions

What Are Partial Fractions?

Partial fraction decomposition is a method used in algebra to break down a single, complex rational expression into a sum of simpler, more manageable fractions. Think of it as the reverse of adding fractions. You already know how to take two simple fractions and combine them into one by finding a common denominator.

For instance, if we wanted to add $\frac{2}{x - 1}$ and $\frac{3}{x + 2}$ , we would do the following:

\frac{2}{x - 1} + \frac{3}{x + 2} = \frac{2 (x + 2)}{(x - 1) (x + 2)} + \frac{3 (x - 1)}{(x - 1) (x + 2)}

= \frac{2 x + 4 + 3 x - 3}{(x - 1) (x + 2)}

= \frac{5 x + 1}{x^{2} + x - 2}

The process of partial fractions takes us in the opposite direction. We start with the complicated fraction, $\frac{5 x + 1}{x^{2} + x - 2}$ , and figure out how to break it back down into the simpler 'partial' fractions, $\frac{2}{x - 1}$ and $\frac{3}{x + 2}$ . This skill is incredibly useful in higher-level mathematics, especially in calculus, where working with the simpler fractions is much easier than dealing with the combined one.

When Can You Use Partial Fractions?

Before you can start breaking down a fraction, you need to make sure it meets two important conditions. If it doesn't, the method won't work correctly.

The fraction must be a proper rational expression. A proper fraction is one where the degree (the highest exponent) of the numerator's polynomial is less than the degree of the denominator's polynomial.
- Proper: $\frac{x + 5}{x^{2} - 4}$ (Degree 1 is less than Degree 2)
- Improper: $\frac{x^{3} + 2 x}{x^{2} - 4}$ (Degree 3 is greater than Degree 2)
If you have an improper fraction, you must first perform polynomial long division. This will give you a polynomial plus a proper fraction, and you can then use partial fractions on the proper fraction part.
The denominator must be factorable. The entire process relies on your ability to factor the polynomial in the denominator into linear factors (like $x - 3$ ) or irreducible quadratic factors (like $x^{2} + 4$ ). If you can't factor the denominator, you can't use this method.

How Do You Handle Distinct Linear Factors?

This is the most straightforward case and the best place to start. "Distinct linear factors" means that when you factor the denominator, you get a set of unique factors of the form $(a x + b)$ , with none of them repeating.

For each distinct linear factor in the denominator, you create one partial fraction with a constant numerator (which we'll call A, B, C, etc.).

If the denominator is

(a x + b) (c x + d)

, the setup is:

\frac{A}{a x + b} + \frac{B}{c x + d}

Example 1

Decompose the fraction $\frac{x + 8}{x^{2} + x - 6}$ .

Step 1: Factor the denominator.
The denominator $x^{2} + x - 6$ factors into $(x + 3) (x - 2)$ . These are distinct linear factors.

Step 2: Set up the partial fractions.
Since we have two distinct linear factors, we'll need two partial fractions with unknown constants, A and B.

\frac{x + 8}{(x + 3) (x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2}

Step 3: Clear the denominators.
Multiply both sides of the equation by the original denominator, $(x + 3) (x - 2)$ , to get rid of the fractions.

x + 8 = A (x - 2) + B (x + 3)

Step 4: Solve for the constants A and B.
The easiest way to solve for A and B is to pick 'convenient' values for $x$ that make one of the terms zero. These values are the roots of the factors.

To find B, let's choose $x = 2$ . This will make the $A (x - 2)$ term become zero.
$(2) + 8 = A (2 - 2) + B (2 + 3)$
$10 = A (0) + B (5)$
$10 = 5 B$
$B = 2$
To find A, let's choose $x = - 3$ . This will make the $B (x + 3)$ term become zero.
$(- 3) + 8 = A (- 3 - 2) + B (- 3 + 3)$
$5 = A (- 5) + B (0)$
$5 = - 5 A$
$A = - 1$

Step 5: Write the final answer.
Now that we have $A = - 1$ and $B = 2$ , we substitute them back into our setup from Step 2.

\frac{x + 8}{x^{2} + x - 6} = \frac{- 1}{x + 3} + \frac{2}{x - 2}

What About Repeated Linear Factors?

Sometimes when you factor a denominator, you get the same linear factor multiple times, like $(x - 4)^{2}$ or $(2 x + 1)^{3}$ . This is called a repeated linear factor.

For a repeated linear factor $(a x + b)^{n}$ , you must create $n$ partial fractions: one for each power of the factor from 1 up to $n$ .

If the denominator has

(a x + b)^{2}

, the setup is:

\frac{A}{a x + b} + \frac{B}{(a x + b)^{2}}

Example 2

Decompose the fraction $\frac{3 x - 1}{(x - 2)^{2}}$ .

Step 1: Identify the factors.
The denominator is already factored as $(x - 2)^{2}$ . This is the linear factor $(x - 2)$ repeated twice.

Step 2: Set up the partial fractions.
We need a term for $(x - 2)$ and a term for $(x - 2)^{2}$ .

\frac{3 x - 1}{(x - 2)^{2}} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}}

Step 3: Clear the denominators.
Multiply both sides by $(x - 2)^{2}$ .

3 x - 1 = A (x - 2) + B

Step 4: Solve for the constants A and B.
We can still use the convenient value trick for one of the constants.

Let $x = 2$ to find B.
$3 (2) - 1 = A (2 - 2) + B$
$5 = A (0) + B$
$B = 5$

Now we have no other convenient values to plug in. To find A, we can substitute B back into the equation and use another method: equating coefficients. First, expand the equation:

3 x - 1 = A x - 2 A + B

Substitute $B = 5$ :

3 x - 1 = A x - 2 A + 5

Now, look at the terms with $x$ on both sides. The coefficient of $x$ on the left must equal the coefficient of $x$ on the right.

3 x = A x ⟹ A = 3

We can double-check this with the constant terms: $- 1 = - 2 A + 5$ . If $A = 3$ , then $- 1 = - 2 (3) + 5 = - 6 + 5 = - 1$ . It works!

Step 5: Write the final answer.
Substitute $A = 3$ and $B = 5$ back into the setup.

\frac{3 x - 1}{(x - 2)^{2}} = \frac{3}{x - 2} + \frac{5}{(x - 2)^{2}}

How Do You Decompose Irreducible Quadratic Factors?

The final common case involves quadratic factors that cannot be factored further using real numbers. These are called irreducible quadratic factors. A common example is $x^{2} + 1$ or $x^{2} + x + 5$ . You can test if a quadratic $a x^{2} + b x + c$ is irreducible by checking its discriminant, $b^{2} - 4 a c$ . If the discriminant is negative, the quadratic is irreducible.

When you have an irreducible quadratic factor in the denominator, its corresponding partial fraction numerator is not just a constant, but a linear term of the form $A x + B$ .

If the denominator has

(a x^{2} + b x + c)

, the setup is:

\frac{A x + B}{a x^{2} + b x + c}

Example 3

Decompose the fraction $\frac{5 x^{2} + 3 x + 4}{(x - 1) (x^{2} + x + 1)}$ .

Step 1: Identify the factors.
The denominator is factored into a linear factor $(x - 1)$ and a quadratic factor $(x^{2} + x + 1)$ . For $x^{2} + x + 1$ , the discriminant is $1^{2} - 4 (1) (1) = - 3$ , which is negative, so it's irreducible.

Step 2: Set up the partial fractions.
The linear factor gets a constant numerator (A), and the irreducible quadratic factor gets a linear numerator (Bx+C).

\frac{5 x^{2} + 3 x + 4}{(x - 1) (x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1}

Step 3: Clear the denominators.
Multiply both sides by $(x - 1) (x^{2} + x + 1)$ .

5 x^{2} + 3 x + 4 = A (x^{2} + x + 1) + (B x + C) (x - 1)

Step 4: Solve for the constants A, B, and C.
We'll use a combination of methods.

First, use the convenient value $x = 1$ to find A.
$5 (1)^{2} + 3 (1) + 4 = A (1^{2} + 1 + 1) + (B (1) + C) (1 - 1)$
$5 + 3 + 4 = A (3) + 0$
$12 = 3 A ⟹ A = 4$

Now, substitute $A = 4$ back into the equation and expand everything.

5 x^{2} + 3 x + 4 = 4 (x^{2} + x + 1) + (B x^{2} - B x + C x - C)

5 x^{2} + 3 x + 4 = 4 x^{2} + 4 x + 4 + B x^{2} - B x + C x - C

Group the terms on the right side by powers of $x$ .

5 x^{2} + 3 x + 4 = (4 + B) x^{2} + (4 - B + C) x + (4 - C)

Finally, equate the coefficients from the left and right sides.

Coefficients of $x^{2}$ : $5 = 4 + B ⟹ B = 1$
Constant terms: $4 = 4 - C ⟹ C = 0$
Check with coefficients of $x$ : $3 = 4 - B + C$ . Plugging in our values, $3 = 4 - 1 + 0$ , which is $3 = 3$ . It's correct!

Step 5: Write the final answer.
Substitute $A = 4$ , $B = 1$ , and $C = 0$ into the setup.

\frac{5 x^{2} + 3 x + 4}{(x - 1) (x^{2} + x + 1)} = \frac{4}{x - 1} + \frac{(1) x + 0}{x^{2} + x + 1} = \frac{4}{x - 1} + \frac{x}{x^{2} + x + 1}

What Are Common Mistakes to Avoid?

Partial fractions can be tricky, and small mistakes can lead to the wrong answer. Here are some common pitfalls to watch out for:

Forgetting to factor completely: Always make sure the denominator is factored as much as possible before you start. A factor like $x^{3} - x$ needs to be broken down into $x (x - 1) (x + 1)$ .
Incorrect setup for repeated factors: A common error is to forget a term. For a denominator with $(x - 5)^{3}$ , you need three fractions: one over $(x - 5)$ , one over $(x - 5)^{2}$ , and one over $(x - 5)^{3}$ .
Incorrect numerator for quadratic factors: Remember that irreducible quadratic factors get a linear numerator $(A x + B)$ , not just a constant $A$ .
Working with an improper fraction: If the degree of the numerator is greater than or equal to the degree of the denominator, you must perform polynomial long division first.
Simple algebra errors: Be very careful when expanding brackets and solving the system of equations for your constants. It's easy to make a sign error that throws off the entire result.

Quick Summary: The Four Cases

Here is a handy table to help you remember how to set up the partial fractions for different types of factors in the denominator.

Factor Type in Denominator	Example Factor	Partial Fraction Form
Distinct Linear	$(x - a)$	$\frac{A}{x - a}$
Repeated Linear	$(x - a)^{n}$	$\frac{A_{1}}{x - a} + \frac{A_{2}}{(x - a)^{2}} + \dots + \frac{A_{n}}{(x - a)^{n}}$
Irreducible Quadratic	$(a x^{2} + b x + c)$	$\frac{A x + B}{a x^{2} + b x + c}$
Repeated Irreducible Quadratic	$(a x^{2} + b x + c)^{n}$	$\frac{A_{1} x + B_{1}}{a x^{2} + b x + c} + \dots + \frac{A_{n} x + B_{n}}{(a x^{2} + b x + c)^{n}}$

Frequently Asked Questions

Why is it called 'partial' fractions?

It's called partial fractions because the method involves breaking a single, whole rational expression down into a sum of its simpler, component parts. Each of these simpler fractions is a 'part' of the original.

Do I always have to factor the denominator first?

Yes, absolutely. Factoring the denominator is the essential first step that tells you what kind of factors you have (linear, quadratic, repeated, etc.). This determines how you set up the partial fractions.

What if the degree of the numerator is bigger than the denominator?

If the degree of the numerator is greater than or equal to the degree of the denominator, you have an improper fraction. You must use polynomial long division first. The result will be a polynomial plus a proper fraction, which you can then decompose.

Is there more than one way to solve for the constants A, B, and C?

Yes, the two main methods are substituting 'convenient' values for x (the roots of the factors) and equating the coefficients of like terms. Often, the fastest approach is to use a combination of both methods.

Where is this technique used in higher math?

Partial fraction decomposition is a crucial skill in calculus, where it is used to simplify complex fractions before integrating them. It also appears in differential equations and various areas of engineering and physics.

What does 'irreducible quadratic' mean again?

An irreducible quadratic is a quadratic expression, like $x^{2} + 9$ or $x^{2} + x + 1$ , that cannot be factored into two linear terms using only real numbers. You can test this with the discriminant $b^{2} - 4 a c$ ; if it's negative, the quadratic is irreducible.

Can I use a calculator to do this?

While a calculator can help with solving the systems of equations for A, B, and C, the setup process requires algebraic understanding. Some advanced calculators can perform the entire decomposition, but it's essential to learn the manual process to understand the concepts.