Partial Fraction Decomposition
UVU Math Lab
Using Systems of Equations:
Find the partial fraction decomposition for the following rational expression:
2
𝑥𝑥 5 +𝑥𝑥 3 −𝑥𝑥 4 −𝑥𝑥 2
STEP 1: Factor the denominator and set expression equal to the form of partial fraction decomposition
with the unknown constants (A, B, C,…) in the numerators of the decomposition.
2
𝒙𝒙𝟐𝟐 (𝒙𝒙−𝟏𝟏)(𝒙𝒙𝟐𝟐 +𝟏𝟏)
𝐴𝐴
𝐵𝐵
𝐶𝐶
= 𝑥𝑥 + 𝑥𝑥 2 + 𝑥𝑥−1 +
𝐷𝐷𝐷𝐷+𝐸𝐸
𝑥𝑥 2 +1
STEP 2: Multiply both sides by the LCD to eliminate denominators and multiply through.
𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) ∙
2
𝐴𝐴
𝐵𝐵
𝐶𝐶
𝐷𝐷𝐷𝐷 + 𝐸𝐸 2
= ∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 2 ∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) +
∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 2
∙ 𝑥𝑥 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1)
𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) 𝑥𝑥
𝑥𝑥
𝑥𝑥 − 1
𝑥𝑥 + 1
2 = 𝐴𝐴𝐴𝐴(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 𝐵𝐵(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 𝐶𝐶𝑥𝑥 2 (𝑥𝑥 2 + 1) + (𝐷𝐷𝐷𝐷 + 𝐸𝐸)𝑥𝑥 2 (𝑥𝑥 − 1)
2 = 𝐴𝐴𝑥𝑥 4 +𝐴𝐴𝐴𝐴 2 −𝐴𝐴𝐴𝐴 3 − 𝐴𝐴𝐴𝐴+𝐵𝐵𝐵𝐵 3 + 𝐵𝐵𝐵𝐵−𝐵𝐵𝐵𝐵 2 − 𝐵𝐵+𝐶𝐶𝐶𝐶 4 +𝐶𝐶𝐶𝐶 2 +𝐷𝐷𝐷𝐷 4 −𝐷𝐷𝐷𝐷 3 +𝐸𝐸𝑥𝑥 3 −𝐸𝐸𝐸𝐸 2
STEP 3: Group like terms and write both sides of the equation in descending powers of 𝑥𝑥.
0𝑥𝑥 4 + 0𝑥𝑥 3 + 0𝑥𝑥 2 + 0𝑥𝑥1 +2𝑥𝑥 0
= 𝑥𝑥 4 (𝐴𝐴 + 𝐶𝐶 + 𝐷𝐷) + 𝑥𝑥 3 (−𝐴𝐴 + 𝐵𝐵 − 𝐷𝐷 + 𝐸𝐸) + 𝑥𝑥 2 (𝐴𝐴 − 𝐵𝐵 + 𝐶𝐶 − 𝐸𝐸) + 𝑥𝑥(−𝐴𝐴 + 𝐵𝐵) − 𝐵𝐵
STEP 4: Equate the coefficients of like powers and solve the resulting system of equations.
4
(4) 𝑥𝑥 : 0 = 𝐴𝐴 + 𝐶𝐶 + 𝐷𝐷
(3) 𝑥𝑥 3 : 0 = −𝐴𝐴 + 𝐵𝐵 − 𝐷𝐷 + 𝐸𝐸
(2) 𝑥𝑥 2 : 0 = 𝐴𝐴 − 𝐵𝐵 + 𝐶𝐶 − 𝐸𝐸
(1) 𝑥𝑥1 : 0 = −𝐴𝐴 + 𝐵𝐵
(0) 𝑥𝑥 0 : 2 = −𝐵𝐵
(0) 2 = −𝐵𝐵
(1) 0 = −𝐴𝐴 + (−2)
→ 𝑩𝑩 = −𝟐𝟐
→ 𝑨𝑨 = −𝟐𝟐
(3) 0 = −(−2) + (−2) − 𝐷𝐷 + 𝐸𝐸
(2) 0 = (−2) − (−2) + 𝐶𝐶 − 𝐸𝐸
(5) 0 = 𝐶𝐶 −D
(4) 0 = (−2) + 𝐶𝐶 + 𝐷𝐷
(5) 0 =
𝐶𝐶 − 𝐷𝐷
(6) 0 = 2 + 2𝐶𝐶
(2) 0 = −2 + 2 + 1 − 𝐸𝐸
(3) 0 = 2 − 2 − 𝐷𝐷 + 1
STEP 5: Substitute the values found for the unknown constants back into the decomposition.
𝑥𝑥 2 (𝑥𝑥
2
2 2
1
𝑥𝑥 + 1
=− − 2+
+ 2
2
− 1)(𝑥𝑥 + 1)
𝑥𝑥 𝑥𝑥
𝑥𝑥 − 1 𝑥𝑥 + 1
More handouts like this are available at: www.uvu.edu/mathlab/mathresources/
→ 𝑪𝑪 = 𝟏𝟏
→ 𝑬𝑬 = 𝟏𝟏
→ 𝑫𝑫 = 𝟏𝟏
Partial Fraction Decomposition
UVU Math Lab
Using the Zero-Out Method:
(The Short Cut!)
Find the partial fraction decomposition for the following rational expression:
2
𝑥𝑥 5 +𝑥𝑥 3 −𝑥𝑥 4 −𝑥𝑥 2
STEP 1: Factor the denominator and set expression equal to the form of partial fraction decomposition
with the unknown constants (A, B, C,…) in the numerators of the decomposition.
2
𝒙𝒙𝟐𝟐 (𝒙𝒙−𝟏𝟏)(𝒙𝒙𝟐𝟐 +𝟏𝟏)
𝐴𝐴
𝐵𝐵
𝐶𝐶
= 𝑥𝑥 + 𝑥𝑥 2 + 𝑥𝑥−1 +
𝐷𝐷𝐷𝐷+𝐸𝐸
𝑥𝑥 2 +1
STEP 2: Multiply both sides by the LCD to eliminate denominators.
𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) ∙
2
𝐴𝐴
𝐵𝐵
𝐶𝐶
𝐷𝐷𝐷𝐷 + 𝐸𝐸 2
= ∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 2 ∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) +
∙ 𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 2
∙ 𝑥𝑥 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1)
𝑥𝑥 2 (𝑥𝑥 − 1)(𝑥𝑥 2 + 1) 𝑥𝑥
𝑥𝑥
𝑥𝑥 − 1
𝑥𝑥 + 1
2 = 𝐴𝐴𝐴𝐴(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 𝐵𝐵(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 𝐶𝐶𝑥𝑥 2 (𝑥𝑥 2 + 1) + (𝐷𝐷𝐷𝐷 + 𝐸𝐸)𝑥𝑥 2 (𝑥𝑥 − 1)
STEP 3: Choose a value for 𝑥𝑥 that will result in zero factors and solve for the remaining letter.
Let 𝑥𝑥 = 0:
2 = 0 + 𝐵𝐵(−1)(1) + 0 + 0
2 = −𝐵𝐵
𝑩𝑩 = −𝟐𝟐
Let 𝑥𝑥 = 1:
2 = 0 + 0 + 𝐶𝐶(1)(2) + 0
2 = 2𝐶𝐶
𝐶𝐶 = 𝟏𝟏
(4) 𝑥𝑥 4 : − 1 = 𝐴𝐴 + 𝐷𝐷
(3) 𝑥𝑥 3 : 2 = −𝐴𝐴 − 𝐷𝐷 + 𝐸𝐸
(2) 𝑥𝑥 2 : − 3 = 𝐴𝐴 − 𝐸𝐸
(1) 𝑥𝑥1 : 2 = −𝐴𝐴
(1) 𝑨𝑨 = −𝟐𝟐
(2) −3 = −2 − 𝐸𝐸
(4) −1 = −2 + 𝐷𝐷
STEP 4: Substitute solved values into equation and then begin with STEP 3 on opposite side to solve any
remaining values using systems of equations.
2 = 𝐴𝐴𝐴𝐴(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) − 2(𝑥𝑥 − 1)(𝑥𝑥 2 + 1) + 1𝑥𝑥 2 (𝑥𝑥 2 + 1) + (𝐷𝐷𝐷𝐷 + 𝐸𝐸)𝑥𝑥 2 (𝑥𝑥 − 1)
2 = 𝐴𝐴𝑥𝑥 4 +𝐴𝐴𝐴𝐴 2 −𝐴𝐴𝐴𝐴 3 − 𝐴𝐴𝐴𝐴−2𝑥𝑥 3 − 2𝑥𝑥+2𝑥𝑥 2 + 2+1𝑥𝑥 4 +1𝑥𝑥 2 +𝐷𝐷𝐷𝐷 4 −𝐷𝐷𝐷𝐷 3 +𝐸𝐸𝐸𝐸 3 −𝐸𝐸𝐸𝐸 2
2 = 𝑥𝑥 4 (𝐴𝐴 + 1 + 𝐷𝐷) + 𝑥𝑥 3 (−𝐴𝐴 − 2 − 𝐷𝐷 + 𝐸𝐸) + 𝑥𝑥 2 (𝐴𝐴 + 2 + 1 − 𝐸𝐸) + 𝑥𝑥(−𝐴𝐴 − 2) + 2
→ 𝑬𝑬 = 𝟏𝟏
→ 𝑫𝑫 = 𝟏𝟏
STEP 5: Substitute the values found for the unknown constants back into the decomposition.
𝑥𝑥 2 (𝑥𝑥
2
2 2
1
𝑥𝑥 + 1
=− − 2+
+ 2
2
− 1)(𝑥𝑥 + 1)
𝑥𝑥 𝑥𝑥
𝑥𝑥 − 1 𝑥𝑥 + 1
More handouts like this are available at: www.uvu.edu/mathlab/mathresources/
Partial Fraction Decomposition
UVU Math Lab
The Four Types of Partial Fraction Decomposition:
Previously, we learned that in order to find the sum of rational expressions we had to find a common
denominator.
2
3
𝑥𝑥
+
−
𝑥𝑥 + 1 𝑥𝑥 − 1 𝑥𝑥
𝑥𝑥
𝑥𝑥(𝑥𝑥 − 1)
2
𝑥𝑥(𝑥𝑥 + 1) 3 (𝑥𝑥 + 1)(𝑥𝑥 − 1)
=
∙
+
∙
− ∙
𝑥𝑥 + 1 𝑥𝑥(𝑥𝑥 − 1) 𝑥𝑥 − 1 𝑥𝑥(𝑥𝑥 + 1) 𝑥𝑥 (𝑥𝑥 − 1)(𝑥𝑥 + 1)
𝑥𝑥 3 −2𝑥𝑥 2 + 2𝑥𝑥 + 3
=
𝑥𝑥(𝑥𝑥 2 − 1)
With partial fraction decomposition, we reverse that process and split the rational expression into partial
fractions, expressions with irreducible denominators.
𝑃𝑃(𝑥𝑥)
Given the rational expression 𝑄𝑄(𝑥𝑥), we factor the denominator, 𝑄𝑄(𝑥𝑥) and then set the expression equal to the
partial fraction decomposition according to the following forms.
Type I:
The denominator is the product of distinct, non-repeating linear factors, such as (𝑥𝑥 − 𝑎𝑎):
𝑃𝑃(𝑥𝑥)
𝐴𝐴
𝐵𝐵
𝐶𝐶
=
+
+
+⋯
𝑄𝑄(𝑥𝑥) 𝑥𝑥 − 𝑎𝑎 𝑥𝑥 − 𝑏𝑏 𝑥𝑥 − 𝑐𝑐
Type II:
The denominator is the product of repeating linear factors, such as (𝑥𝑥 − 𝑎𝑎)𝑛𝑛 :
𝐶𝐶1
𝐶𝐶2
𝐶𝐶𝑛𝑛
𝑃𝑃(𝑥𝑥)
=
+
+⋯+
2
𝑄𝑄(𝑥𝑥) 𝑥𝑥 − 𝑎𝑎 (𝑥𝑥 − 𝑎𝑎)
(𝑥𝑥 − 𝑎𝑎)𝑛𝑛
Type III:
The denominator is the product of distinct irreducible quadratic factors (cannot be factored further) of the form
(𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐):
𝑃𝑃(𝑥𝑥) 𝐴𝐴𝐴𝐴 + 𝐵𝐵 𝐶𝐶𝐶𝐶 + 𝐷𝐷 𝐷𝐷𝐷𝐷 + 𝐸𝐸
=
+
+ 2
+⋯
𝑄𝑄(𝑥𝑥) 𝑥𝑥 2 + 𝑎𝑎 𝑥𝑥 2 + 𝑏𝑏
𝑥𝑥 + 𝑐𝑐
Type IV:
The denominator is the product of repeating irreducible quadratic factors, such as (𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)𝑛𝑛 :
𝑃𝑃(𝑥𝑥) 𝐶𝐶1 𝑥𝑥 + 𝐶𝐶2
𝐶𝐶3 𝑥𝑥+𝐶𝐶4
𝐶𝐶𝑚𝑚−1 𝑥𝑥 + 𝐶𝐶𝑚𝑚
= 2
+ 2
+⋯+
2
𝑄𝑄(𝑥𝑥)
𝑥𝑥 + 𝑎𝑎
(𝑥𝑥 + 𝑎𝑎)
(𝑥𝑥 2 + 𝑎𝑎)𝑛𝑛
More handouts like this are available at: www.uvu.edu/mathlab/mathresources/