Lesson Objectives

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

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

In this lesson, we will learn how to find the partial fraction decomposition with linear 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. $$\frac{9x + 12}{x^2 + 3x}$$ 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. $$x^2 + 3x=x(x + 3)$$ Step 3) For each distinct linear 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,...}$$\frac{A}{x}+ \frac{B}{x + 3}$$ Step 4) Set this equal to the original rational expression. $$\frac{9x + 12}{x^2 + 3x}=\frac{A}{x}+ \frac{B}{x + 3}$$ 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. $$9x+12=A(x+3) + Bx$$ 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. $$9x + 12=Ax + 3A + Bx$$ $$9x + 12=(A + B)x + 3A$$ These two expressions can only be equal if: $$9=A + B$$ $$3A=12$$ We can easily solve this with substitution. $$A=4$$ $$9=4 + B$$ $$B=5$$ We just replace A with 4 and B with 5 and we are done: $$\frac{9x + 12}{x^2 + 3x}=\frac{A}{x}+ \frac{B}{x + 3}$$ $$\frac{9x + 12}{x^2 + 3x}=\frac{4}{x}+ \frac{5}{x + 3}$$ In some cases, you will have a repeated linear factor to deal with. For this scenario, you must ensure that you build up the power. This means if you have a factor raised to the 2

Example #2: Find the partial fraction decomposition. $$\frac{-5x + 14}{x^2 - 4x + 4}$$ 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. $$x^2 - 4x + 4=(x - 2)^2$$ Step 3) For each distinct linear 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,...}

Here we have a repeated linear factor, so we have to build up the power. $$\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ Notice how we included the first power and the second power. If this had been a factor raised to the third power, we would include the first, second, and third power.

Step 4) Set this equal to the original rational expression. $$\frac{-5x + 14}{x^2 - 4x + 4}=\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ 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. $$-5x + 14=A(x-2) + B$$ 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. $$-5x + 14=A(x-2) + B$$ $$-5x + 14=Ax - 2A + B$$ $$-5x + 14=Ax + (B - 2A)$$ These two expressions can only be equal if: $$-5=A$$ $$14=B - 2A$$ We can easily solve this with substitution. $$A=-5$$ $$B=14 + 2A$$ $$B=14 + 2(-5)$$ $$B=14 - 10$$ $$B=4$$ We just replace A with -5 and B with 4 and we are done: $$\frac{-5x + 14}{x^2 - 4x + 4}=\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ $$\frac{-5x + 14}{x^2 - 4x + 4}=-\frac{5}{x - 2}+ \frac{4}{(x - 2)^2}$$

Example #1: Find the partial fraction decomposition. $$\frac{9x + 12}{x^2 + 3x}$$ 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. $$x^2 + 3x=x(x + 3)$$ Step 3) For each distinct linear 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,...}$$\frac{A}{x}+ \frac{B}{x + 3}$$ Step 4) Set this equal to the original rational expression. $$\frac{9x + 12}{x^2 + 3x}=\frac{A}{x}+ \frac{B}{x + 3}$$ 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. $$9x+12=A(x+3) + Bx$$ 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. $$9x + 12=Ax + 3A + Bx$$ $$9x + 12=(A + B)x + 3A$$ These two expressions can only be equal if: $$9=A + B$$ $$3A=12$$ We can easily solve this with substitution. $$A=4$$ $$9=4 + B$$ $$B=5$$ We just replace A with 4 and B with 5 and we are done: $$\frac{9x + 12}{x^2 + 3x}=\frac{A}{x}+ \frac{B}{x + 3}$$ $$\frac{9x + 12}{x^2 + 3x}=\frac{4}{x}+ \frac{5}{x + 3}$$ In some cases, you will have a repeated linear factor to deal with. For this scenario, you must ensure that you build up the power. This means if you have a factor raised to the 2

^{nd}power, you include a fraction with the first power and also a fraction with the second power. Each power gets a fraction starting with the first and building up to the final power. Let's see an example.Example #2: Find the partial fraction decomposition. $$\frac{-5x + 14}{x^2 - 4x + 4}$$ 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. $$x^2 - 4x + 4=(x - 2)^2$$ Step 3) For each distinct linear 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,...}

Here we have a repeated linear factor, so we have to build up the power. $$\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ Notice how we included the first power and the second power. If this had been a factor raised to the third power, we would include the first, second, and third power.

Step 4) Set this equal to the original rational expression. $$\frac{-5x + 14}{x^2 - 4x + 4}=\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ 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. $$-5x + 14=A(x-2) + B$$ 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. $$-5x + 14=A(x-2) + B$$ $$-5x + 14=Ax - 2A + B$$ $$-5x + 14=Ax + (B - 2A)$$ These two expressions can only be equal if: $$-5=A$$ $$14=B - 2A$$ We can easily solve this with substitution. $$A=-5$$ $$B=14 + 2A$$ $$B=14 + 2(-5)$$ $$B=14 - 10$$ $$B=4$$ We just replace A with -5 and B with 4 and we are done: $$\frac{-5x + 14}{x^2 - 4x + 4}=\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ $$\frac{-5x + 14}{x^2 - 4x + 4}=-\frac{5}{x - 2}+ \frac{4}{(x - 2)^2}$$

#### Skills Check:

Example #1

Find the partial fraction decomposition. $$\frac{5x + 11}{x^2 + 2x + 1}$$

Please choose the best answer.

A

$$\frac{5}{x + 1}+ \frac{6}{(x + 1)^2}$$

B

$$\frac{1}{x + 1}+ \frac{3}{(x + 3)}$$

C

$$\frac{5}{x + 1}- \frac{6}{(x + 2)}$$

D

$$\frac{7}{x - 1}+ \frac{3}{(x - 1)^2}$$

E

$$\frac{1}{x + 1}+ \frac{3}{(x + 1)^2}$$

Example #2

Find the partial fraction decomposition. $$\frac{-5x + 4}{x^2 - 2x + 1}$$

Please choose the best answer.

A

$$\frac{9}{x - 1}+ \frac{4}{x + 3}$$

B

$$\frac{3}{x + 2}+ \frac{5}{x + 7}$$

C

$$-\frac{5}{x - 1}- \frac{4}{(x - 1)^2}$$

D

$$-\frac{5}{x - 1}- \frac{1}{(x - 1)^2}$$

E

$$-\frac{1}{x - 1}- \frac{1}{(x - 1)^2}$$

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