Lesson Objectives

- Learn how to find the partial fraction decomposition with quadratic factors

## How to Find the Partial Fraction Decomposition with Quadratic Factors

In this lesson, we will learn how to find the partial fraction decomposition with quadratic factors. At this point in our course, one should be fully comfortable with the process of combining two or more rational expressions into a single rational expression. Here, we will consider how to reverse this process and turn a single rational expression into the sum of two or more rational expressions. Let's look at this process with an example.

Example #1: Find the partial fraction decomposition. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. If you do not have a proper fraction, you must do long division and then apply the following steps to the remainder.

Step 2) Factor the denominator completely. $$x^4 - 3x^2 - 10=(x^2 + 2)(x^2 - 5)$$ Step 3) For each distinct linear or quadratic factor, we set up a fraction with the factor as the denominator and a variable as the numerator. It is typical to see capital letters as the variables:{A, B, C, D, E,...}$$\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ Step 4) Set this equal to the original rational expression. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression. $$5x^2 - 4=A(x^2 - 5) + B(x^2 + 2)$$ Match up the two sides. To do this, we will simplify the right side and change it into the same form as the left side. $$5x^2 - 4=Ax^2 - 5A + Bx^2 + 2B$$ $$5x^2 - 4=(A + B)x^2 + (2B - 5A)$$ These two expressions can only be equal if: $$A + B=5$$ $$2B - 5A=-4$$ We can easily solve this with substitution. $$A=5 - B$$ Plug into the second equation: $$2B - 5(5 - B)=-4$$ $$2B - 25 + 5B=-4$$ $$7B=21$$ $$B=3$$ Plug into the first equation: $$A + 3=5$$ $$A=2$$ We just replace A with 2 and B with 3 and we are done: $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{2}{x^2 + 2}+ \frac{3}{x^2 - 5}$$

Example #1: Find the partial fraction decomposition. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. If you do not have a proper fraction, you must do long division and then apply the following steps to the remainder.

Step 2) Factor the denominator completely. $$x^4 - 3x^2 - 10=(x^2 + 2)(x^2 - 5)$$ Step 3) For each distinct linear or quadratic factor, we set up a fraction with the factor as the denominator and a variable as the numerator. It is typical to see capital letters as the variables:{A, B, C, D, E,...}$$\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ Step 4) Set this equal to the original rational expression. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression. $$5x^2 - 4=A(x^2 - 5) + B(x^2 + 2)$$ Match up the two sides. To do this, we will simplify the right side and change it into the same form as the left side. $$5x^2 - 4=Ax^2 - 5A + Bx^2 + 2B$$ $$5x^2 - 4=(A + B)x^2 + (2B - 5A)$$ These two expressions can only be equal if: $$A + B=5$$ $$2B - 5A=-4$$ We can easily solve this with substitution. $$A=5 - B$$ Plug into the second equation: $$2B - 5(5 - B)=-4$$ $$2B - 25 + 5B=-4$$ $$7B=21$$ $$B=3$$ Plug into the first equation: $$A + 3=5$$ $$A=2$$ We just replace A with 2 and B with 3 and we are done: $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{A}{x^2 + 2}+ \frac{B}{x^2 - 5}$$ $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}=\frac{2}{x^2 + 2}+ \frac{3}{x^2 - 5}$$

#### Skills Check:

Example #1

Find the partial fraction decomposition.

Please choose the best answer. $$\frac{-7x^2 + 1}{x^4 - x^2 - 6}$$

A

$$\frac{2}{x^2 - 3}+ \frac{4}{x^2 + 2}$$

B

$$\frac{2}{x^2 - 3}- \frac{3}{x^2 + 2}$$

C

$$-\frac{4}{x^2 - 3}- \frac{3}{x^2 + 2}$$

D

$$-\frac{6}{x^2 - 3}- \frac{3}{x^2 + 2}$$

E

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

Example #2

Find the partial fraction decomposition. $$\frac{3x^2 + x - 14}{x^3 - 5x + x^2 - 5}$$

Please choose the best answer.

A

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

B

$$-\frac{6}{x + 1}+ \frac{3}{x^2 - 5}$$

C

$$\frac{3}{x + 1}+ \frac{1}{x^2 - 5}$$

D

$$\frac{4}{x + 1}- \frac{7}{x^2 - 5}$$

E

$$\frac{3}{x - 1}+ \frac{2}{x^2 - 3}$$

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