Lesson 11.4: Partial Fractions

New Concept

Master the art of breaking down complex rational expressions! You'll learn to decompose $\dfrac{P(x)}{Q(x)}$ into a sum of simpler "partial fractions," a key skill for tackling different types of factors in the denominator.

What’s next

You've got the foundation. Now, you'll master this technique with worked examples, practice cards, and interactive challenges for each type of denominator factor.

Property

The partial fraction decomposition of $\frac{P(x)}{Q(x)}$ when $Q(x)$ has nonrepeated linear factors and the degree of $P(x)$ is less than the degree of $Q(x)$ is

To decompose a rational expression with distinct linear factors: First, assign a variable for each numerator over a single factor. Next, multiply the entire equation by the common denominator. Then, expand the right side, collect like terms, and set the coefficients of like terms equal to create a system of equations to solve for the numerators.

Examples

• To decompose $\frac{8x-1}{(x-2)(x+3)}$, we set it up as $\frac{A}{x-2} + \frac{B}{x+3}$. Multiplying by $(x-2)(x+3)$ gives $8x-1 = A(x+3) + B(x-2)$. Setting $x=2$ gives $15=5A$, so $A=3$. Setting $x=-3$ gives $-25=-5B$, so $B=5$. The decomposition is $\frac{3}{x-2} + \frac{5}{x+3}$.

• For $\frac{10x-2}{x^2-4}$, we first factor the denominator as $(x-2)(x+2)$. This gives $\frac{A}{x-2} + \frac{B}{x+2}$. Multiplying by the common denominator yields $10x-2 = A(x+2) + B(x-2)$. If $x=2$, then $18=4A$, so $A=\frac{9}{2}$. If $x=-2$, then $-22=-4B$, so $B=\frac{11}{2}$. The result is $\frac{\frac{9}{2}}{x-2} + \frac{\frac{11}{2}}{x+2}$.

Property

The partial fraction decomposition of $\frac{P(x)}{Q(x)}$, when $Q(x)$ has a repeated linear factor occurring $n$ times and the degree of $P(x)$ is less than the degree of $Q(x)$, is

Write the denominator powers in increasing order. Use a variable for each numerator and account for increasing powers of the denominators. Multiply by the common denominator, expand the right side, collect like terms, and set coefficients equal to create a system of equations.

Examples

• To decompose $\frac{2x+9}{(x+3)^2}$, we write $\frac{A}{x+3} + \frac{B}{(x+3)^2}$. This gives $2x+9 = A(x+3)+B$. Expanding gives $2x+9 = Ax+3A+B$. Comparing coefficients, $A=2$ and $3A+B=9$. Substituting $A=2$ gives $6+B=9$, so $B=3$. The result is $\frac{2}{x+3} + \frac{3}{(x+3)^2}$.

• Decompose $\frac{3x^2-2x+1}{x^2(x-1)}$. We set up $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}$. This leads to $3x^2-2x+1 = Ax(x-1) + B(x-1) + Cx^2$. Setting $x=1$ gives $C=2$. Setting $x=0$ gives $B=-1$. Comparing $x^2$ coefficients, $3=A+C$, so $3=A+2$, which means $A=1$. The decomposition is $\frac{1}{x} - \frac{1}{x^2} + \frac{2}{x-1}$.

Property

The partial fraction decomposition of $\frac{P(x)}{Q(x)}$ such that $Q(x)$ has a nonrepeated irreducible quadratic factor and the degree of $P(x)$ is less than the degree of $Q(x)$ is written as

The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: $A$, $B$, $C$, and so on. For numerators over quadratic factors, use linear expressions like $A_1x+B_1$. Then, multiply by the common denominator, expand, and set coefficients equal to create a system of equations.

Examples

• Decompose $\frac{3x^2+2x+8}{(x-2)(x^2+4)}$. We set this as $\frac{A}{x-2} + \frac{Bx+C}{x^2+4}$. This gives $3x^2+2x+8 = A(x^2+4) + (Bx+C)(x-2)$. For $x=2$, $24=8A$, so $A=3$. Expanding gives $3x^2+2x+8=(A+B)x^2+(-2B+C)x+(4A-2C)$. Comparing $x^2$ terms: $3=A+B → 3=3+B → B=0$. Comparing constants: $8=4A-2C → 8=12-2C → C=2$. The result is $\frac{3}{x-2} + \frac{2}{x^2+4}$.

• For $\frac{x^2+3x-5}{(x+1)(x^2+x+1)}$, the setup is $\frac{A}{x+1} + \frac{Bx+C}{x^2+x+1}$. Multiplying gives $x^2+3x-5 = A(x^2+x+1) + (Bx+C)(x+1)$. Let $x=-1$, then $1-3-5 = A(1)$, so $A=-7$. Comparing coefficients: $x^2: 1=A+B → 1=-7+B → B=8$. Constant: $-5=A+C → -5=-7+C → C=2$. The decomposition is $\frac{-7}{x+1} + \frac{8x+2}{x^2+x+1}$.

Property

The partial fraction decomposition of $\frac{P(x)}{Q(x)}$, when $Q(x)$ has a repeated irreducible quadratic factor and the degree of $P(x)$ is less than the degree of $Q(x)$, is

Write the denominators in increasing powers. Use linear expressions ($A_nx+B_n$) for the numerators of each quadratic factor. Multiply both sides by the common denominator, expand the right side, collect like terms, and set coefficients equal to create a system of equations.

Examples

• Decompose $\frac{x^3+3x-1}{(x^2+2)^2}$. We write $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{(x^2+2)^2}$. This gives $x^3+3x-1 = (Ax+B)(x^2+2) + Cx+D = Ax^3+2Ax+Bx^2+2B+Cx+D$. So $x^3+3x-1 = Ax^3+Bx^2+(2A+C)x+(2B+D)$. Comparing coefficients: $A=1$, $B=0$, $2A+C=3 → C=1$, and $2B+D=-1 → D=-1$. The result is $\frac{x}{x^2+2} + \frac{x-1}{(x^2+2)^2}$.

• For $\frac{2x^2}{(x^2+x+1)^2}$, the setup is $\frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{(x^2+x+1)^2}$. This gives $2x^2 = (Ax+B)(x^2+x+1) + Cx+D$. Expanding gives $2x^2 = Ax^3+(A+B)x^2+(A+B+C)x+(B+D)$. So $A=0$, $A+B=2 → B=2$, $A+B+C=0 → C=-2$, and $B+D=0 → D=-2$. The decomposition is $\frac{2}{x^2+x+1} - \frac{2x+2}{(x^2+x+1)^2}$.