Partial Fraction Decomposition Linear Factors Lesson

In this lesson, we will learn how to find the partial fraction decomposition with distinct linear factors and repeated linear factors. We previously studied how to add and subtract rational expressions. With partial fraction decomposition, we will reverse this process. In other words, if we are now given a single rational expression, we will write it as the sum of two or more rational expressions. Each term in this sum is known as a partial fraction. Before we get into our procedure, let's list a few definitions here for reference.

• Distinct Linear Factors: refers to each linear factor being different or unique
• Linear Factor: refers to a factor such as (ax + b), a ≠ 0
  • a and b represent real numbers
  • The factor contains a variable raised to the first power
  • The factor does not contain a variable raised to any higher power
  • Ex: (3x + 2) is a linear factor (highest exponent is 1)
  • Ex: (3x² + x + 2) is not a linear factor (highest exponent is 2)

Partial Fraction Decomposition of f(x)/g(x):

1. If f(x)/g(x) is not a proper fraction, perform polynomial long division and apply the steps to the remainder
   • A proper fraction means the degree of the numerator is strictly less than the degree of the denominator
   • $$\text{Proper:} \, \frac{x}{x^2 + x + 5}$$
   • $$\text{Improper:} \, \frac{2x^2}{x + 3}$$
2. Factor the denominator g(x) completely into linear factors of the form:
   • (ax + b)ⁿ, where n is a positive integer (1, 2, 3,...)
   • When n > 1, we have a repeated linear factor
3. For each distinct linear factor (ax + b), we produce a partial fraction:
   • $$\frac{A}{ax + b}$$
   • The capital letter "A" represents an unknown constant
4. For each repeated linear factor (ax + b)ⁿ, n > 1:
   • $$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$$
5. Set the original rational expression equal to the sum of the partial fractions
6. Set up a system of linear equations and solve for A, B, C,...
   • Clear all denominators (multiply both sides by the LCD)
   • Write both sides in descending powers of x
   • Equate the coefficients of like powers of x
   • Equate constant terms
   • Solve the given system using algebraic methods

Partial Fraction Decomposition with Distinct Linear Factors in the Denominator

We will begin with the easiest case, which is when we have distinct linear factors. Let's look at 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. In this case, the degree of the numerator is 1 and the degree of the denominator is 2, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^2 + 3x=x(x + 3)$$ 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, set up a linear system, and solve the system. $$\frac{9x + 12}{x^2 + 3x}=\frac{A}{x}+ \frac{B}{x + 3}$$ 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. $$\frac{9x + 12}{x(x + 3)} \cdot x(x + 3) = x(x + 3) \left(\frac{A}{x}+ \frac{B}{x + 3}\right)$$ $$\require{cancel}\frac{9x + 12}{\cancel{x(x + 3)}} \cdot \cancel{x(x + 3)} = \cancel{x}(x + 3) \cdot \frac{A}{\cancel{x}} + x\cancel{(x + 3)} \cdot \frac{B}{\cancel{(x + 3)}}$$ $$9x+12=A(x+3) + Bx$$ 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. $$9x + 12=Ax + 3A + Bx$$ $$9x + 12 = Ax + Bx + 3A$$ $$9x + 12=(A + B)x + 3A$$ These two expressions can only be equal if: $$A + B = 9$$ $$3A=12$$ We can easily solve this with substitution. Let's first solve for A: $$\frac{3}{3}A = \frac{12}{3}$$ $$A=4$$ Substitute in for A to find B: $$A + B = 9$$ $$4 + B = 9$$ $$B=5$$ We just replace A with 4 and B with 5 and we are done: $$\frac{9x + 12}{x^2 + 3x}=\frac{4}{x}+ \frac{5}{x + 3}$$

Partial Fraction Decomposition with Repeated Linear Factors in the Denominator

In some cases, you will have a repeated linear factor to deal with. If you had (ax + b)ⁿ, n > 1, then the partial fraction decomposition will contain a sum of n fractions for this factor of the denominator. In other words, you would need to include one fraction with a constant numerator for each power of (ax + b). For example, if you have a linear factor raised to the 2^nd power, you would include a fraction with the linear factor raised to the first power and also a fraction with the linear factor raised to the second power. $$\frac{P(x)}{(ax + b)^n} = \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$$ Let's look at some examples.
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. In this case, the degree of the numerator is 1 and the degree of the denominator is 2, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^2 - 4x + 4=(x - 2)^2$$ Step 3) For each distinct linear factor, we set up a fraction with the factor as the denominator and a variable as the numerator.
Here we have a repeated linear factor, so we have to include one fraction with a constant numerator for each power of (x - 2). $$\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ Notice how we included a fraction for each power of (x - 2): the first power (x - 2) and the second power (x - 2)².
Step 4) Set this equal to the original rational expression, set up a linear system, and solve the system. $$\frac{-5x + 14}{x^2 - 4x + 4}=\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}$$ 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. $$\frac{-5x + 14}{(x - 2)^2} \cdot (x - 2)^2 = (x - 2)^2 \left(\frac{A}{x - 2}+ \frac{B}{(x - 2)^2}\right)$$ $$\frac{-5x + 14}{\cancel{(x-2)^2}} \cdot \cancel{(x-2)^2} = (x-2)^{1\cancel{2}} \cdot \frac{A}{\cancel{(x - 2)}}+ \cancel{(x-2)^2} \cdot \frac{B}{\cancel{(x - 2)^2}}$$ $$-5x + 14=A(x-2) + B$$ 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. $$A=-5$$ $$B=14 + 2A$$ Substitute in for A to find B: $$B=14 + 2(-5)$$ $$B=14 - 10$$ $$B=4$$ We just replace A with -5 and B with 4 and we are done: $$\frac{-5x + 14}{x^2 - 4x + 4}=-\frac{5}{x - 2} + \frac{4}{(x - 2)^2}$$ Example #3: Find the partial fraction decomposition. $$\frac{-9x^2 + 38x - 27}{x^3 - 6x^2 + 9x}$$ 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. In this case, the degree of the numerator is 2 and the degree of the denominator is 3, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2$$ Step 3) For each distinct linear factor, we set up a fraction with the factor as the denominator and a variable as the numerator.
Here we have a repeated linear factor, so we have to include one fraction with a constant numerator for each power of (x - 3). $$\frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}$$ Notice how we included a fraction for each power of (x - 3): the first power (x - 3) and the second power (x - 3)².
Step 4) Set this equal to the original rational expression, set up a linear system, and solve the system. $$\frac{-9x^2 + 38x - 27}{x^3 - 6x^2 + 9x} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}$$ 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. $$\frac{-9x^2 + 38x - 27}{x(x - 3)^2} \cdot x(x - 3)^2 = x(x - 3)^2\left(\frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}\right)$$ $$\frac{-9x^2 + 38x - 27}{\cancel{x(x - 3)^2}} \cdot \cancel{x(x - 3)^2} = \cancel{x}(x - 3)^2 \cdot \frac{A}{\cancel{x}} + x{(x - 3)^{1 \cancel{2}}} \cdot \frac{B}{\cancel{(x - 3)}} + x\cancel{(x - 3)^2} \cdot \frac{C}{\cancel{(x - 3)^2}}$$ $$-9x^2 + 38x - 27 = A(x - 3)^2 + Bx(x - 3) + Cx$$ 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. $$-9x^2 + 38x - 27 = A(x^2 - 6x + 9) + Bx(x - 3) + Cx$$ $$-9x^2 + 38x - 27 = Ax^2 - 6Ax + 9A + Bx^2 - 3Bx + Cx$$ $$-9x^2 + 38x - 27 = Ax^2 + Bx^2 - 6Ax - 3Bx + Cx + 9A$$ $$-9x^2 + 38x - 27 = (A + B)x^2 + (-6A - 3B + C)x + 9A$$ These two expressions can only be equal if: $$A + B = -9$$ $$-6A - 3B + C = 38$$ $$9A = -27$$ We can easily solve with substitution. Let's first solve for A: $$\frac{9}{9}A = \frac{-27}{9}$$ $$A = -3$$ Substitute in for A to find B: $$A + B = -9$$ $$-3 + B = -9$$ $$B = -6$$ Substitute in for A and B to find C: $$-6A - 3B + C = 38$$ $$-6(-3) - 3(-6) + C = 38$$ $$18 + 18 + C = 38$$ $$C + 36 = 38$$ $$C = 2$$ Just replace A with -3, B with -6, and C with 2, and we are done. $$\frac{-9x^2 + 38x - 27}{x^3 - 6x^2 + 9x} = -\frac{3}{x} - \frac{6}{x - 3} + \frac{2}{(x - 3)^2}$$

Partial Fraction Decomposition with an Improper Fraction

As mentioned previously, we may be given a situation in which our rational expression is not a proper fraction. An improper fraction will occur when the degree of the numerator is greater than or equal to the degree of the denominator. When this happens, we will perform a polynomial long division and then apply the steps to the remainder. Let's look at an example.
Example #4: Find the partial fraction decomposition. $$\frac{-2x^2 + 12x - 26}{x^2 - 4x + 3}$$ 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. In this case, the degree of the numerator is 2 and the degree of the denominator is also 2, so we do not have a proper fraction.
When this occurs, we need to perform a polynomial long division. $$\frac{-2x^2 + 12x - 26}{x^2 - 4x + 3} = -2 + \frac{4x - 20}{x^2 - 4x + 3}$$ For now, we will just work with the remainder. When we write our final answer, the quotient of -2 will need to be included. $$\frac{4x - 20}{x^2 - 4x + 3}$$ Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^2 - 4x + 3 = (x - 1)(x - 3)$$ Step 3) For each distinct linear factor, we set up a fraction with the factor as the denominator and a variable as the numerator. $$\frac{A}{x - 1} + \frac{B}{x - 3}$$ Step 4) Set this equal to the original rational expression, set up a linear system, and solve the system. Here, let's make a note that in this case, the "original rational expression" will refer to our rational expression produced from the remainder of our long division. $$\frac{4x - 20}{x^2 - 4x + 3} = \frac{A}{x - 1} + \frac{B}{x - 3}$$ 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. $$\frac{4x - 20}{(x - 1)(x - 3)} \cdot (x - 1)(x - 3) = (x - 1)(x - 3)\left(\frac{A}{x - 1} + \frac{B}{x - 3}\right)$$ $$\frac{4x - 20}{\cancel{(x - 1)(x - 3)}} \cdot \cancel{(x - 1)(x - 3)} = \cancel{(x - 1)}(x - 3) \cdot \frac{A}{\cancel{(x - 1)}} + (x - 1)\cancel{(x - 3)} \cdot \frac{B}{\cancel{(x - 3)}}$$ $$4x - 20 = A(x - 3) + B(x - 1)$$ 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. $$4x - 20 = Ax - 3A + Bx - B$$ $$4x - 20 = Ax + Bx - 3A - B$$ $$4x - 20 = (A + B)x - (3A + B)$$ These two expressions can only be equal if: $$4 = A + B$$ $$20 = 3A + B$$ Note: Be very careful when equating the constant terms, it's really easy to make a sign mistake. Notice that we matched up a minus sign on the left with a minus sign on the right. We also could have changed the form: $$4x - 20 = (A + B)x - (3A + B)$$ $$4x + (-20) = (A + B)x + (-3A - B)$$ Which would now produce a slightly different system with the same answer. $$4 = A + B$$ $$-20 = -3A - B$$ If you divide both sides of the second equation by -1 you will obtain the same system as above. We will work with the original system below.
Multiply the top equation by -3: $$-12 = -3A -3B$$ $$20 = 3A + B$$ Add the two equations: $$8 = -2B$$ $$B = -4$$ Substitute in for B in the original top equation to find A: $$4 = A + B$$ $$4 = A - 4$$ $$A = 8$$ First, we will replace A with 8 and B with -4: $$\frac{4x - 20}{x^2 - 4x + 3} = \frac{8}{x - 1} - \frac{4}{x - 3}$$ Since the rational expression on the left is not the original, we will consider our results from the long division. $$\frac{-2x^2 + 12x - 26}{x^2 - 4x + 3} = -2 + \frac{4x - 20}{x^2 - 4x + 3}$$ Now we can just replace what we found to obtain our final answer: $$\frac{-2x^2 + 12x - 26}{x^2 - 4x + 3} = -2 + \frac{8}{x - 1} - \frac{4}{x - 3}$$

Choosing Suitable Values for x

Up to this point, we have found our unknown constants by creating and solving a linear system. In most cases, a faster approach is to find suitable choices for x that allow us to quickly solve for the unknown constants. This method is less straightforward and sometimes a linear system still needs to be created. Let's look at an example.
Example #5: Find the partial fraction decomposition. $$\frac{x - 33}{x^2 - 9}$$ 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. In this case, the degree of the numerator is 1 and the degree of the denominator is 2, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^2 - 9 = (x + 3)(x - 3)$$ 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,...}. $$\frac{A}{x + 3}+ \frac{B}{x - 3}$$ Step 4) Set this equal to the original rational expression. $$\frac{x - 33}{x^2 - 9} = \frac{A}{x + 3}+ \frac{B}{x - 3}$$ Instead of following the normal approach of creating and solving a linear system, we will clear the denominators and look for suitable choices for x to solve for our unknown constants A and B.
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. $$\frac{x - 33}{(x + 3)(x - 3)} \cdot (x + 3)(x - 3) = (x + 3)(x - 3)\left(\frac{A}{x + 3}+ \frac{B}{x - 3}\right)$$ $$\frac{x - 33}{\cancel{(x + 3)(x - 3)}} \cdot \cancel{(x + 3)(x - 3)} = \cancel{(x + 3)}(x - 3) \cdot \frac{A}{\cancel{(x + 3)}} + (x + 3)\cancel{(x - 3)} \cdot \frac{B}{\cancel{(x - 3)}}$$ $$x - 33 = A(x - 3) + B(x + 3)$$ Looking at our equation above, think about what would happen if we let x = 3. $$3 - 33 = A(3 - 3) + B(3 + 3)$$ $$-30 = A(0) + B(6)$$ $$-30 = 6B$$ $$B = \frac{-30}{6} = -5$$ Notice that substituting a value of 3 in for x eliminated A, which allowed us to solve for B. Since we still need A, let's repeat the process and let x = -3. $$-3 - 33 = A(-3 - 3) + B(-3 + 3)$$ $$-36 = -6A + B(0)$$ $$-36 = -6A$$ $$A = \frac{-36}{-6} = 6$$ Just replace A with 6 and B with -5, and we are done. $$\frac{x - 33}{x^2 - 9} = \frac{6}{x + 3} - \frac{5}{x - 3}$$ In the previous example, picking a suitable number to eliminate A so that we could solve for B (and vice versa) worked out nicely, but this won't always be the case. Let's look at an example where this method wouldn't be as straightforward.
Example #6: Find the partial fraction decomposition. $$\frac{-3x - 8}{9x^2 + 6x + 1}$$ 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. In this case, the degree of the numerator is 1 and the degree of the denominator is 2, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$9x^2 + 6x + 1 = (3x + 1)^2$$ 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 include one fraction with a constant numerator for each power of (3x + 1). $$\frac{A}{3x + 1} + \frac{B}{(3x + 1)^2 }$$ Notice how we included a fraction for each power of (3x + 1): the first power (3x + 1) and the second power (3x + 1)².
Step 4) Set this equal to the original rational expression. $$\frac{-3x - 8}{9x^2 + 6x + 1} = \frac{A}{3x + 1} + \frac{B}{(3x + 1)^2 }$$ Instead of following the normal approach of creating and solving a linear system, we will clear the denominators and look for suitable choices for x to solve for our unknown constants A and B.
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. $$\frac{-3x - 8}{(3x + 1)^2} \cdot (3x + 1)^2 = (3x + 1)^2 \left(\frac{A}{3x + 1} + \frac{B}{(3x + 1)^2 }\right)$$ $$\frac{-3x - 8}{\cancel{(3x + 1)^2}} \cdot \cancel{(3x + 1)^2} = (3x + 1)^{1 \cancel{2}} \cdot \frac{A}{\cancel{(3x + 1)}} + \cancel{(3x + 1)^2} \cdot \frac{B}{\cancel{(3x + 1)^2}}$$ $$-3x - 8 = A(3x + 1) + B$$ Looking at our equation above, think about what would happen if we let x = -1/3. $$-3\left(-\frac{1}{3}\right) - 8 = A\left(3 \cdot -\frac{1}{3} + 1\right) + B$$ $$1 - 8 = A(0) + B$$ $$B = -7$$ We have found that B is -7. But how can we find A using the same approach? If we inspect our equation, we see that there is no variable associated with B. So, how can we eliminate B? $$-3x - 8 = A(3x + 1) + B$$ The answer here is to substitute in for B in our equation. $$-3x - 8 = A(3x + 1) - 7$$ Now we can solve for A but there is still an issue with the process. $$-3x - 8 = A(3x + 1) - 7$$ What can we substitute in for x here so that we can solve for A? We can pick any number we want as long as it's not -1/3 since that eliminates A as we saw before. To make things simple, let's just pick an x-value of 0. $$-3(0) - 8 = A(3(0) + 1) - 7$$ $$-8 = A - 7$$ $$A = -8 + 7 = -1$$ Just replace A with -1 and B with -7, and we are done. $$\frac{-3x - 8}{9x^2 + 6x + 1} = -\frac{1}{3x + 1} - \frac{7}{(3x + 1)^2 }$$

Heaviside Cover-Up Method

When you have distinct linear factors, clearing the denominators is unnecessary, and the unknown constants can be found with immediate substitution. Let's look at an example.
Example #7: Find the partial fraction decomposition. $$\frac{x^2 + 20x + 60}{x^3 + 9x^2 + 20x}$$ 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. In this case, the degree of the numerator is 2 and the degree of the denominator is 3, so we do have a proper fraction.
Step 2) Factor the denominator completely into linear factors of the form: (ax + b)ⁿ. $$x^3 + 9x^2 + 20x = x(x + 4)(x + 5)$$ 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,...}. $$\frac{A}{x} + \frac{B}{x + 4} + \frac{C}{x + 5}$$ Step 4) Set this equal to the original rational expression. $$\frac{x^2 + 20x + 60}{x(x + 4)(x + 5)} = \frac{A}{x} + \frac{B}{x + 4} + \frac{C}{x + 5}$$ At this point, we normally clear our denominators but this is unnecessary. We start with our first fraction on the right side of the equation: $$\frac{A}{x}$$ How can we make the denominator 0? Since we only have x in the denominator, we can just replace x with 0. So on the left side, we will remove x from the denominator and plug in a 0 everywhere else to find A. $$A = \frac{(0)^2 + 20(0) + 60}{(0 + 4)(0 + 5)} = \frac{60}{20} = 3$$ Before we continue, why does this work? $$\frac{x^2 + 20x + 60}{x(x + 4)(x + 5)} = \frac{A}{x} + \frac{B}{x + 4} + \frac{C}{x + 5}$$ Multiply both sides by x: $$\frac{x^2 + 20x + 60}{\cancel{x}(x + 4)(x + 5)} \cdot \cancel{x} = \cancel{x} \cdot \frac{A}{\cancel{x}} + x \cdot \frac{B}{x + 4} + x \cdot \frac{C}{x + 5}$$ $$\frac{x^2 + 20x + 60}{(x + 4)(x + 5)} = A + \frac{Bx}{x + 4} + \frac{Cx}{x + 5}$$ Let's substitute in an x-value of 0: $$\frac{(0)^2 + 20(0) + 60}{(0 + 4)(0 + 5)} = A + \frac{B(0)}{0 + 4} + \frac{C(0)}{0 + 5}$$ $$\frac{60}{20} = A$$ $$A = 3$$ Let's move on and figure out B and C using the same process. To find B: $$\frac{B}{x + 4}$$ How can we make the denominator 0? $$x + 4 = 0$$ $$x = -4$$ We plug in a -4 for x on the left side after removing the (x + 4) factor in the denominator. $$B = \frac{(-4)^2 + 20(-4) + 60}{-4(-4 + 5)} = \frac{16 - 80 + 60}{-4} = \frac{-4}{-4} = 1$$ To find C: $$\frac{C}{x + 5}$$ How can we make the denominator 0? $$x + 5 = 0$$ $$x = -5$$ We plug in a -5 for x on the left side after removing the (x + 5) factor in the denominator. $$C = \frac{(-5)^2 + 20(-5) + 60}{-5(-5 + 4)} = \frac{25 - 100 + 60}{5} = \frac{-15}{5} = -3$$ Just replace A with 3, B with 1, and C with -3 and we are done. $$\frac{x^2 + 20x + 60}{x(x + 4)(x + 5)} = \frac{3}{x} + \frac{1}{x + 4} - \frac{3}{x + 5}$$