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MTH120 College Algebra Chapter 7.4

 7.4 Partial Fractions

Partial fractions are a technique used in algebra to decompose a complex rational expression into simpler, more manageable fractions. This decomposition is particularly useful for integrating rational functions, simplifying equations, or solving equations. Partial fractions work by breaking down a rational expression into a sum of simpler fractions.

Here are the key steps and concepts related to partial fractions:

Partial Fraction Decomposition:

  1. Basic Idea: Given a complex rational expression, we aim to represent it as a sum of simpler fractions with denominators that are easier to work with.

  2. Types of Partial Fractions:

    • Proper Rational Functions: The degree of the numerator is less than the degree of the denominator.
    • Improper Rational Functions: The degree of the numerator is equal to or greater than the degree of the denominator.
    • Complex Rational Functions: Rational functions with quadratic factors in the denominator.

Partial Fraction Decomposition Process:

  1. Factor the Denominator: Factor the denominator of the rational expression completely. This includes factoring over the real numbers or complex numbers, depending on the context.

  2. Write the Decomposition: Write the rational expression as a sum of simpler fractions with each fraction having a factor of the denominator.

  3. Assign Variables: Assign variables (usually A, B, C, etc.) to the numerators of the simpler fractions.

  4. Find the Constants: Determine the values of the constants (A, B, C, etc.) by finding a common denominator, equating the numerators, and solving for the constants.

  5. Combine and Simplify: Combine the fractions and simplify the result as much as possible.

Types of Partial Fractions:

  1. Linear Factors: When the denominator consists of linear factors (e.g., 2�+3), the partial fraction decomposition is relatively straightforward.

  2. Repeated Linear Factors: When a linear factor appears multiple times in the denominator, each instance requires a different constant in the decomposition.

  3. Quadratic Factors: When the denominator includes quadratic factors (e.g., �2+4), the decomposition involves linear numerators.

  4. Irreducible Quadratic Factors: Quadratic factors that cannot be factored further require irreducible quadratic factors in the decomposition.

Partial fractions are especially valuable in calculus when integrating rational functions. They allow for simplifying complex integrals into easier-to-handle components. The technique is also used in solving differential equations and simplifying algebraic expressions.


Decomposing a rational function �(�)/�(�) where �(�) has only non repeated linear factors is relatively straightforward. The general idea is to express �(�)/�(�) as a sum of partial fractions, where each partial fraction has a simple linear factor in the denominator. Here are the steps:

Step 1: Factor the Denominator �(�): Factor the denominator �(�) into its non repeated linear factors. For example, if �(�) is a product of linear factors, you might have something like:

�(�)=(�−�)(�−�)(�−�)⋯

Step 2: Set Up the Partial Fractions: Write the rational function �(�)/�(�) as a sum of partial fractions. Each partial fraction will have a numerator that is a constant, and the denominator will be one of the linear factors from the factored �(�). Here's the setup:

�(�)�(�)=��−�+��−�+��−�+⋯

Step 3: Find the Constants (A, B, C, etc.): To find the constants (A, B, C, etc.), you'll need to clear the denominators and equate the numerators. Multiply both sides of the equation by the common denominator, which is the product of all the linear factors, and then equate the numerators.

For example, if you have the setup:

�(�)�(�)=��−�+��−�+��−�

You would clear the denominators by multiplying both sides by (�−�)(�−�)(�−�):

�(�)=�(�−�)(�−�)+�(�−�)(�−�)+�(�−�)(�−�)

Now, equate the numerators:

�(�)=�(�−�)(�−�)+�(�−�)(�−�)+�(�−�)(�−�)

Step 4: Solve for the Constants: To find the constants (A, B, C, etc.), you can choose specific values of � that make some of the terms zero, which will help you solve for the constants. For example, you can choose �=�, �=�, and �=� to eliminate the other terms and solve for each constant.

Step 5: Combine and Simplify: Once you've found the constants (A, B, C, etc.), rewrite the partial fraction decomposition as a single expression, and simplify if necessary.

This process allows you to express a rational function as a sum of simpler fractions, making it easier to integrate, simplify, or work with in various mathematical contexts.


Partial fraction decomposition of a rational function �(�)�(�) when �(�) has non repeated linear factors involves expressing the rational function as a sum of simpler fractions, where each simpler fraction has a linear factor in the denominator. Here are the steps for decomposition and examples:

Step 1: Factor the Denominator �(�): Factor the denominator �(�) into its non repeated linear factors. Each linear factor should appear only once. For example, if �(�) is given as:

�(�)=(�−�)(�−�)(�−�)⋯

Step 2: Set Up the Partial Fractions: Write the rational function �(�)�(�) as a sum of partial fractions. Each partial fraction will have a constant in the numerator and one of the linear factors from the factored �(�) in the denominator. The general form is:

�(�)�(�)=�1�−�+�2�−�+�3�−�+⋯

Here, �1,�2,�3,… are constants to be determined.

Step 3: Find the Constants �1,�2,�3,…: To find the constants, you'll need to clear the denominators by multiplying both sides by the common denominator, which is the product of all the linear factors. Then, equate the numerators.

Step 4: Solve for the Constants: Choose specific values of � to eliminate all but one of the fractions on the right side of the equation and solve for each constant. Typically, choosing � values that make some of the fractions zero is a helpful approach.

Step 5: Combine and Simplify: Once you've found the constants, rewrite the partial fraction decomposition as a single expression, and simplify as necessary.

Here are a couple of examples:

Example 1:

Decompose 2�(�−1)(�+2) into partial fractions.

Step 1: Factor the denominator: �(�)=(�−1)(�+2)

Step 2: Set up the partial fractions: 2�(�−1)(�+2)=��−1+��+2

Step 3: Find the constants: Multiply both sides by the common denominator (�−1)(�+2): 2�=�(�+2)+�(�−1)

Step 4: Solve for the constants: Let's choose values of � to eliminate one of the fractions. For example, if we set �=1, the �(�−1) term becomes zero, allowing us to solve for �.

2(1)=�(1+2)+�(1−1) 2=3� �=23

Similarly, we can choose �=−2 to eliminate the �(�+2) term and solve for �:

2(−2)=�(−2+2)+�(−2−1) −4=−3� �=43

Step 5: Combine and simplify: The partial fraction decomposition is: 2�(�−1)(�+2)=23(�−1)+43(�+2)

Example 2:

Decompose 3�2−2�+1(�−1)(�+2)(�+3) into partial fractions.

Step 1: Factor the denominator: �(�)=(�−1)(�+2)(�+3)

Step 2: Set up the partial fractions: 3�2−2�+1(�−1)(�+2)(�+3)=��−1+��+2+��+3

Step 3: Find the constants: Multiply both sides by the common denominator (�−1)(�+2)(�+3): 3�2−2�+1=�(�+2)(�+3)+�(�−1)(�+3)+�(�−1)(�+2)

This equation will lead to a system of equations involving �, �, and �, and you can solve for these constants.

By following these steps, you can decompose rational functions with non repeated linear factors into partial fractions, which can be useful for integration and simplification.


Decomposing a rational function �(�)�(�) where �(�) has repeated linear factors involves expressing the rational function as a sum of partial fractions, where each simpler fraction has a linear factor from �(�) in the denominator. Here are the steps for decomposition with examples:

Step 1: Factor the Denominator �(�): Factor the denominator �(�) completely, including the repeated linear factors. For example, if �(�) is given as:

�(�)=(�−�)�(�−�)�⋯

where (�−�)� and (�−�)� are repeated linear factors, with � and � greater than 1.

Step 2: Set Up the Partial Fractions: Write the rational function �(�)�(�) as a sum of partial fractions. Each partial fraction will have a constant in the numerator and one of the linear factors from the factored �(�) in the denominator. The general form is:

�(�)�(�)=�1�−�+�2(�−�)2+⋯+�1�−�+�2(�−�)2+⋯

Here, �1,�2,�1,�2,… are constants to be determined.

Step 3: Find the Constants �1,�2,�1,�2,…: To find the constants, you'll need to clear the denominators by multiplying both sides by the common denominator, which is the product of all the linear factors. Then, equate the numerators.

Step 4: Solve for the Constants: Choose specific values of � to eliminate all but one of the fractions on the right side of the equation and solve for each constant. The values of � chosen should match the repeated linear factors and their powers in the denominator. For example, choose �=� for the (�−�)� terms.

Step 5: Combine and Simplify: Once you've found the constants, rewrite the partial fraction decomposition as a single expression and simplify as necessary.

Here are some examples:

Example 1:

Decompose 3�2+7�+5(�−2)2(�+1) into partial fractions.

Step 1: Factor the denominator: �(�)=(�−2)2(�+1)

Step 2: Set up the partial fractions: 3�2+7�+5(�−2)2(�+1)=��−2+�(�−2)2+��+1

Step 3: Find the constants: Multiply both sides by the common denominator (�−2)2(�+1): 3�2+7�+5=�(�+1)+�+�(�−2)2

Step 4: Solve for the constants: You will need to choose values of � to eliminate specific terms. For �, choose �=2:

3(22)+7(2)+5=�(2+1)+�+�(2−2)2

This leads to �=5.

To solve for �, you can choose �=−1:

3(−1)2+7(−1)+5=5(−1+1)+�+�(−1−2)2

This leads to �=−2.

Finally, to solve for �, choose �=2 again:

3(22)+7(2)+5=5(2+1)+(−2)+�(2−2)2

This also leads to �=1.

Step 5: Combine and simplify: The partial fraction decomposition is: 3�2+7�+5(�−2)2(�+1)=5�−2−2(�−2)2+1�+1

Example 2:

Decompose �3−5�2+8�−4�2(�−2)3 into partial fractions.

Step 1: Factor the denominator: �(�)=�2(�−2)3

Step 2: Set up the partial fractions: �3−5�2+8�−4�2(�−2)3=��+��2+��−2+�(�−2)2+�(�−2)3

Step 3: Find the constants: Multiply both sides by the common denominator �2(�−2)3, and you will get a complicated equation that can be solved for each constant by selecting appropriate values of �.

By following these steps, you can decompose rational functions with repeated linear factors into partial fractions, which can be useful for integration, simplification, and solving equations.


Decomposing a rational function �(�)�(�) where �(�) has a non repeated irreducible quadratic factor involves expressing the rational function as a sum of partial fractions, where each simpler fraction has the irreducible quadratic factor in the denominator. Here are the steps for decomposition with an example:

Step 1: Factor the Denominator �(�): Factor the denominator �(�) completely, including the irreducible quadratic factor. For example, if �(�) is given as:

�(�)=(�−�)(�2+��+�)

where (�2+��+�) is the irreducible quadratic factor.

Step 2: Set Up the Partial Fractions: Write the rational function �(�)�(�) as a sum of partial fractions. Each partial fraction will have a constant in the numerator and the irreducible quadratic factor in the denominator. The general form is:

�(�)�(�)=��−�+��+��2+��+�

Here, �, �, and � are constants to be determined.

Step 3: Find the Constants �, �, and �: To find the constants, you'll need to clear the denominators by multiplying both sides by the common denominator. In this case, the common denominator is (�−�)(�2+��+�).

Step 4: Solve for the Constants: Choose specific values of � to eliminate specific terms. For the constants �, �, and �, you will need to choose values of � that make certain terms on the right side of the equation zero.

Step 5: Combine and Simplify: Once you've found the constants, rewrite the partial fraction decomposition as a single expression, and simplify as necessary.

Here's an example:

Example:

Decompose 3�2+2�−5(�−3)(�2+2�+5) into partial fractions.

Step 1: Factor the denominator: �(�)=(�−3)(�2+2�+5)

Step 2: Set up the partial fractions: 3�2+2�−5(�−3)(�2+2�+5)=��−3+��+��2+2�+5

Step 3: Find the constants: Multiply both sides by the common denominator (�−3)(�2+2�+5): 3�2+2�−5=�(�2+2�+5)+(��+�)(�−3)

Step 4: Solve for the constants: To find �, choose a value of � that makes the quadratic factor zero, such as �=−1:

3(−1)2+2(−1)−5=�((−1)2+2(−1)+5) 3−2−5=�(1−2+5) �=−4

To find � and �, you can use different values of � to eliminate terms. For example, choose �=3 to eliminate the � term:

3(3)2+2(3)−5=(�(3)+�)(3−3) 27+6−5=0 �(3)+�=−28

Now, choose a different value of �, like �=0, to eliminate the � and � terms:

3(0)2+2(0)−5=(−4(02+2(0)+5)+�(0−3) −5=−20+(−3�) −5+20=−3� 15=3� �=5

Step 5: Combine and simplify: The partial fraction decomposition is: 3�2+2�−5(�−3)(�2+2�+5)=−4�−3+5�+5�2+2�+5

This is how you can decompose a rational function when the denominator has a non repeated irreducible quadratic factor.


Decomposing a rational function �(�)�(�) where �(�) has a repeated irreducible quadratic factor involves expressing the rational function as a sum of partial fractions, where each simpler fraction has the irreducible quadratic factor in the denominator. Here are the steps for decomposition with an example:

Step 1: Factor the Denominator �(�): Factor the denominator �(�) completely, including the repeated irreducible quadratic factor. For example, if �(�) is given as:

�(�)=(�−�)(�2+��+�)�

where (�2+��+�) is the repeated irreducible quadratic factor, and � is the number of times it's repeated.

Step 2: Set Up the Partial Fractions: Write the rational function �(�)�(�) as a sum of partial fractions. Each partial fraction will have a constant in the numerator and the irreducible quadratic factor raised to different powers in the denominator. The general form is:

�(�)�(�)=�1�−�+�1�+�1�2+��+�+�2(�−�)2+�2�+�2(�2+��+�)2+⋯

Here, �1, �1, �1, �2, �2, �2, etc., are constants to be determined.

Step 3: Find the Constants �1, �1, �1, �2, �2, �2, etc.: To find the constants, you'll need to clear the denominators by multiplying both sides by the common denominator, which includes all the terms in the partial fractions.

Step 4: Solve for the Constants: Choose specific values of � to eliminate specific terms. For each constant, choose values of � that make certain terms on the right side of the equation zero.

Step 5: Combine and Simplify: Once you've found the constants, rewrite the partial fraction decomposition as a single expression, and simplify as necessary.

Here's an example:

Example:

Decompose 3�2−2�+5(�−2)(�2+2�+5)2 into partial fractions.

Step 1: Factor the denominator: �(�)=(�−2)(�2+2�+5)2

Step 2: Set up the partial fractions: 3�2−2�+5(�−2)(�2+2�+5)2=��−2+��+��2+2�+5+��+�(�2+2�+5)2

Step 3: Find the constants: Multiply both sides by the common denominator, which includes all the terms in the partial fractions.

Step 4: Solve for the constants: Choose specific values of � to eliminate specific terms. You may need to solve a system of equations to find all the constants.

Step 5: Combine and simplify: The partial fraction decomposition will include terms with various powers of (�2+2�+5) in the denominator. Simplify the expression by combining like terms.

In this example, the process can be more complex due to the repeated quadratic factor, but the method remains the same: clear the denominators, choose values of � to eliminate terms, and solve for the constants.


1. Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction 2. Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.) 3. Can you explain how to verify a partial fraction decomposition graphically? 4. You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer. 5. Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had 7x+133x2+8x+15=Ax+1+B3x+5, we eventually simplify to 7x+13=A(3x+5)+B(x+1). Explain how you could intelligently choose an x-value that will eliminate either A or B and solve for A and B.


  1. Not All Quotients of Polynomials Can Be Decomposed into Partial Fractions: No, not all quotients of polynomials can be decomposed into at least two partial fractions. In order to decompose a rational function into partial fractions, the denominator must be factorable into linear and irreducible quadratic factors. If the denominator has factors that cannot be factored further or has irreducible quadratic factors that are repeated, it may not be possible to decompose it into separate partial fractions. Here's an example:

    Consider the rational function 1(�2+1)2. The denominator (�2+1)2 has a repeated irreducible quadratic factor, but there are no linear factors. In this case, it cannot be decomposed into separate partial fractions.

  2. Uniqueness of Partial Fraction Decomposition: A partial fraction decomposition is unique for a given rational function. This uniqueness is a result of the Fundamental Theorem of Algebra. When you decompose a rational function into partial fractions, you are essentially reversing the process of combining fractions into a single rational function. Because each constant in the numerators corresponds to a unique term in the original rational function, there is only one way to express it as a sum of partial fractions. Any variation in the constants would result in different rational functions.

  3. Verifying Partial Fraction Decomposition Graphically: To verify a partial fraction decomposition graphically, you can compare the original rational function's graph with the sum of the graphs of the partial fractions. Here's how you can do it:

    • Graph the original rational function.
    • Graph each of the partial fractions separately.
    • The sum of the graphs of the partial fractions should match the graph of the original rational function.

    If the graphs do not match, it indicates an error in the partial fraction decomposition.

  4. Double-Checking a Partial Fraction Decomposition: If you're unsure about the correctness of your partial fraction decomposition, you can double-check it by:

    • Repeating the decomposition process to confirm the constants and denominators.
    • Using common denominators to recombine the partial fractions back into the original rational function and verifying that it matches the original expression.

    Additionally, you can perform algebraic operations on the decomposed fractions to ensure they are equivalent to the original expression. This includes verifying that the numerators of the partial fractions add up to the numerator of the original rational function and that the denominators match.

  5. Intelligently Choosing Values to Eliminate Constants (A and B): To solve for the constants (A and B) in the system of equations generated by the partial fraction decomposition, you can choose values of x that will eliminate one of the constants. For example, in the equation 7�+13=�(3�+5)+�(�+1), you can choose values of x to make one of the terms zero:

    • If you choose �=−1, the term with �(�+1) becomes zero, allowing you to solve for A.
    • If you choose �=−5/3, the term with �(3�+5) becomes zero, allowing you to solve for B.

    By intelligently selecting values of x, you can simplify the equations and isolate the constants A and B for easier calculation.

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