Partial Fraction Decomposition

Linear Factors in the Denominator

Denominators without Repeated Factors

A rational expression with only linear factors in the denominator and with no repeating factors can be decomposed into partial fractions.

A rational expression is an expression in the form PQ\frac{P}{Q}, where PP and QQ are polynomials and the value of QQ is not zero. The degree of a term in a polynomial is the exponent of the variable. The degree of a polynomial is the degree of the term with the greatest degree. A linear factor of a polynomial is a factor in the form (x+n)(x + n), where nn is a constant.

Sometimes it is helpful to rewrite a rational expression as a sum of simpler expressions. Partial fraction decomposition is a method of rewriting a rational expression PQ\frac{P}{Q} as a sum of partial fractions, where each partial fraction is a rational expression with a denominator that has a degree less than the degree of QQ.

To decompose a rational expression into a sum of partial fractions:

1. Factor the denominator of the rational expression.

2. Write an equation setting the original rational expression equal to a sum of partial fractions with unknown numerators. The numerator of each partial fraction must have a degree that is one less than the denominator.

3. Multiply both sides of the equation by the denominator of the original rational expression, which will eliminate the denominators.

4. Solve for the unknown values in the numerators.

5. Substitute the values for the numerators into the partial fractions.

Step-By-Step Example
Decomposing a Rational Expression with No Repeated Factors
Write the rational expression as a sum of partial fractions.
2x26x2+2x8\frac{-2x-26}{x^2+2x-8}
Step 1
Factor the denominator.
2x26x2+2x8=2x26(x+4)(x2)\frac{-2x-26}{x^2+2x-8} = \frac{-2x-26}{(x+4)(x-2)}
Step 2
Rewrite the fraction as a sum of two fractions. Each denominator is one of the factors of the original denominator, and the numerator of each fraction is unknown. In this case, the denominators are linear. Since the numerators must be one degree less, they are constants. Use AA and BB to represent the unknown constants.
2x26(x+4)(x2)=Ax+4+Bx2\frac{-2x-26}{(x+4)(x-2)}=\frac{A}{x+4}+\frac{B}{x-2}
Step 3

Clear the denominators on each side of the equation in Step 2.

Multiply both sides of the equation by (x+4)(x2)(x+4)(x-2). Then simplify the left side.
(x+4)(x2)2x26(x+4)(x2)=(x+4)(x2)(Ax+4+Bx2)(x+4)(x2)2x26(x+4)(x2)=(x+4)(x2)(Ax+4+Bx2)2x26=(x+4)(x2)(Ax+4+Bx2)\begin{aligned}(x+4)(x-2)\frac{-2x-26}{(x+4)(x-2)}&=(x+4)(x-2)\left(\frac{A}{x+4}+\frac{B}{x-2}\right)\\\cancel{(x+4)}\cancel{(x-2)}\frac{-2x-26}{\cancel{(x+4)}\cancel{(x-2)}}&=(x+4)(x-2)\left(\frac{A}{x+4}+\frac{B}{x-2}\right)\\-2x-26&=(x+4)(x-2)\left(\frac{A}{x+4}+\frac{B}{x-2}\right)\end{aligned}
Next simplify the right side.
2x26=(x+4)(x2)(Ax+4)+(x+4)(x2)(Bx2)2x26=(x+4)(x2)(Ax+4)+(x+4)(x2)(Bx2)2x26=A(x2)+B(x+4)\begin{aligned}-2x-26&=(x+4)(x-2)\left(\frac{A}{x+4}\right)+(x+4)(x-2)\left(\frac{B}{x-2}\right)\\-2x-26&=\cancel{(x+4)}(x-2)\left(\frac{A}{\cancel{x+4}}\right)+(x+4)\cancel{(x-2)}\left(\frac{B}{\cancel{x-2}}\right)\\-2x-26&=A(x-2)+B(x+4)\end{aligned}
Step 4
Write a system of linear equations to find AA and BB. Expand the right side of the equation from Step 3. Then combine like terms.
2x26=A(x2)+B(x+4)2x26=Ax2A+Bx+4B2x26=(A+B)x+(2A+4B)\begin{aligned}-2x - 26 &= A(x-2) + B(x+4)\\-2x-26 &=Ax-2A+Bx+4B\\-2x-26 &= (A+B)x+(-2A+4B)\end{aligned}
The coefficient of xx is –2 on the left side and (A+B)(A+B) on the right side. The coefficients must be equal, so:
A+B=2A+B=-2
Likewise, set the constant terms equal:
2A+4B=26-2A+4B=-26
Step 5
Solve the system by elimination.
A+B=22A+4B=26\begin{aligned}A+B&=-2\\-2A+4B&=-26\end{aligned}
Multiply the first equation by 2 so that the AA terms are opposites.
2A+2B=42A+4B=26\begin{aligned}2A+2B&=-4\\-2A+4B&=-26\end{aligned}
Add the equations together, and solve for BB.
6B=30B=5\begin{aligned}6B&=-30\\B&=-5\end{aligned}
Substitute the value of BB in the first equation of the system to find AA.
A+5=2A=3\begin{aligned}A+-5&=-2\\A&=3\end{aligned}
Solution
Substitute values for AA and BB into the partial fraction sum from Step 2.
2x26x2+2x8=Ax+4+Bx2=3x+45x2\begin{aligned}\frac{-2x-26}{x^2+2x-8}&=\frac A{x+4}+\frac B{x-2}\\&=\frac3{x+4}-\frac5{x-2}\end{aligned}

A rational expression is a proper fraction if the degree of the denominator is greater than the degree of the numerator. When the numerator is an expression with a degree greater than or equal to the denominator, it is an improper fraction. To decompose an improper fraction, first divide the numerator by the denominator to rewrite the expression as a sum of the quotient and a remainder written as a proper fraction.
Step-By-Step Example
Decomposing an Improper Rational Expression with No Repeated Factors
Write the rational expression as a sum of partial fractions.
3x2+4x8x24\frac{3x^2+4x-8}{x^2-4}
Step 1
Divide using long division.
x24)3x2+4x32x24)3x2+4x82x24)3x2+0x12x24)0x2+4x+42\begin{aligned}\phantom{x^{2}-4\overline{)3x^{2}+4x-}}3\phantom{2}\\x^{2}-4\overline{)3x^{2}+4x-8\phantom{2}}\\\phantom{x^{2}-4\overline{)}}\underline{3x^2+0x-12}\\\phantom{x^{2}-4\overline{)}}0x^{2}+4x+4\phantom{2}\end{aligned}
Write the expression as the sum of the quotient and the remainder as a proper fraction.
3x2+4x8x24=3+4x+4x24\frac{3x^2+4x-8}{x^2-4} = 3+\frac{4x+4}{x^2-4}
Step 2
Factor the denominator.
3+4x+4x24=3+4x+4(x+2)(x2)3+\frac{4x+4}{x^2-4} = 3+\frac{4x+4}{(x+2)(x-2)}
Step 3
Rewrite the remainder as a sum of two fractions. The numerator of each addend is an unknown constant, and each denominator is one of the factors.
4x+4(x+2)(x2)=Ax+2+Bx2\frac{4x+4}{\left(x+2\right)\left(x-2\right)}=\frac A{x+2}+\frac B{x-2}
Step 4
Multiply both sides by both factors and then clear the denominators on each side of the equation.
(x+2)(x2)(4x+4(x+2)(x2))=(x+2)(x2)(Ax+2+Bx2)4x+4=A(x2)+B(x+2)\begin{aligned} (x+2)(x-2)\left ( \frac{4x+4}{(x+2)(x-2)} \right )&=(x+2)(x-2)\left ( \frac{A}{x+2}+\frac{B}{x-2} \right )\\ 4x+4&=A(x-2)+B(x+2)\end{aligned}
Step 5
Write a system of linear equations to find AA and BB.
4x+4=A(x2)+B(x+2)4x+4=Ax2A+Bx+2B4x+4=(A+B)x+(2A+2B)\begin{aligned}4x+4&=A(x-2)+B(x+2)\\4x+4&=Ax-2A+Bx+2B\\4x+4&=(A+B)x+(-2A+2B)\end{aligned}
Set the coefficients of the xx-terms equal and the constants equal to get a linear system of equations.
A+B=42A+2B=4\begin{aligned}A+B&=4\\-2A+2B&=4\end{aligned}
Step 6
Solve the system by elimination. Multiply the first equation by 2 so that the AA terms are opposites.
2A+2B=82A+2B=4\begin{aligned}2A+2B&=8\\-2A+2B&=4\end{aligned}
Add the equations together. Solve for BB.
4B=12B=3\begin{aligned}4B &= 12\\B &=3\end{aligned}
Substitute the value of BB in the first equation of the original system to find AA.
A+3=4A=1\begin{aligned}A+3 &= 4\\A &=1\end{aligned}
Solution
Substitute values for AA and BB into the partial fraction sum from Step 3, and rewrite the equation from Step 1 with the partial fractions as the remainder.
3x2+4x8x24=3+Ax+2+Bx2=3+1x+2+3x2\begin{aligned}\frac{3x^2+4x-8}{x^2-4}&=3+\frac{A}{x+2}+\frac{B}{x-2}\\&=3+\frac{1}{x+2}+\frac{3}{x-2}\end{aligned}

Denominators with Repeated Factors

A rational expression with only linear factors in the denominator with repeating factors can be decomposed into partial fractions.
Sometimes, there are repeated factors in the denominator when decomposing a rational expression. In this case, the process is altered slightly to include additional addends with the repeated factors in the denominator.
Step-By-Step Example
Decompose a Rational Expression with Repeated Linear Factors
Write the rational expression as a sum of partial fractions.
5x22x+1(x+2)(x3)2\frac{5x^2-2x+1}{(x+2)(x-3)^2}
Step 1

Write the fraction as a sum with unknown numerators, using each linear factor once.

If a factor is repeated nn times, the sum will include nn partial fractions with a constant numerator and the linear factor as the denominator. For each of those partial fractions, the exponent on the denominator will increase by 1. In this example, the factor (x3)(x-3) occurs twice, so there are two partial fractions. The first has a denominator of (x3)(x-3), and the second has a denominator of (x3)2(x-3)^2.
5x22x+1(x+2)(x3)2=Ax+2+Bx3+C(x3)2\frac{5x^2-2x+1}{(x+2)(x-3)^2} = \frac{A}{x+2} + \frac{B}{x-3}+ \frac{C}{(x-3)^2}
Step 2
Add the sum using the original denominator. Each partial fraction is multiplied by a form of 1 to get the common denominator.
5x22x+1(x+2)(x3)2=A(x3)2(x+2)(x3)2+B(x+2)(x3)(x+2)(x3)2+C(x+2)(x+2)(x3)25x22x+1(x+2)(x3)2=A(x3)2+B(x+2)(x3)+C(x+2)(x+2)(x3)2\begin{aligned}\frac{5x^2-2x+1}{(x+2)(x-3)^2} &= \frac{A(x-3)^2}{(x+2)(x-3)^2}+\frac{B(x+2)(x-3)}{(x+2)(x-3)^2}+\frac{C(x+2)}{(x+2)(x-3)^2}\\\frac{5x^2-2x+1}{(x+2)(x-3)^2}&= \frac{A(x-3)^2+B(x+2)(x-3)+C(x+2)}{(x+2)(x-3)^2}\end{aligned}
Step 3
Multiply both sides of the equation by the denominator. Then, set the numerators equal to each other.
5x22x+1=A(x3)2+B(x+2)(x3)+C(x+2)5x^2-2x+1= A(x-3)^2+B(x+2)(x-3)+C(x+2)
Expand the right side, and combine like terms.
5x22x+1=A(x26x+9)+B(x2x6)+C(x+2)5x22x+1=Ax26Ax+9A+Bx2Bx6B+Cx+2C5x22x+1=(A+B)x2+(6AB+C)x+(9A6B+2C)\begin{aligned}5x^2-2x+1&=A(x^2-6x+9)+B(x^2-x-6)+C(x+2)\\5x^2-2x+1&=Ax^2-6Ax+9A+Bx^2-Bx-6B+Cx+2C\\5x^2-2x+1&=(A+B)x^2+(-6A-B+C)x+(9A-6B+2C)\end{aligned}
The like terms on either side of the equation must be equal. Use this fact to solve for AA, BB, and CC.
5x2Quadratic2xLinear+1Constant=(A+B)x2Quadratic+(6AB+C)xLinear+(9A6B+2C)Constant\overbrace{5x^2}^{\text{Quadratic}}-\overbrace{2x}^{\text{Linear}}+\overbrace{1}^{\text{Constant}}=\overbrace{(A+B)x^2}^{\text{Quadratic}}+\overbrace{(-6A-B+C)x}^{\text{Linear}}+\overbrace{(9A-6B+2C)}^{\text{Constant}}
Write a linear system in three variables using the coefficients of the quadratic, linear, and constant terms.
A+B=56AB+C=29A6B+2C=1\begin{aligned}A+B&=5\\-6A-B+C&=-2\\9A-6B+2C&=1\end{aligned}
Step 4
Solve the system by substitution and elimination. Solve the first equation for AA.
A=B+5A=-B+5
Substitute B+5-B+5 for AA into the second and third equations, and simplify. Second equation:
6(B+5)B+C=25B+C=28\begin{aligned}\\-6(B+5)-B+C&=-2\\5B+C&=28\end{aligned}
Third Equation:
9(B+5)6B+2C=115B+2C=44\begin{aligned}9(B+5)-6B+2C&=1\\-15B+2C&=-44\end{aligned}
Multiply 5B+C=285B+C=28 by 3 so that the BB-terms have opposite coefficients.
15B+3C=8415B+2C=44\begin{aligned}15B+3C&=84\\-15B+2C&=-44\end{aligned}
Add the two equations, and solve for CC.
5C=40C=8\begin{aligned}5C &= 40\\C &=8\end{aligned}
Substitute the value of CC to find BB.
5B+8=285B=20B=4\begin{aligned}5B+8 &= 28\\5B &= 20\\B &=4\end{aligned}
Substitute the value of BB to find AA.
A+4=5A=1\begin{aligned}A+4 &= 5\\A &=1\end{aligned}
Solution
Substitute values for AA, BB, and CC into the partial fraction sum.
5x22x+1x34x23x+18=1x+2+4x3+8(x3)2\frac{5x^2-2x+1}{x^3-4x^2-3x+18}=\frac{1}{x+2} + \frac{4}{x-3}+ \frac{8}{(x-3)^2}