9.4 Partial Fractions

Partial Fraction Decomposition

Partial fraction decomposition reverses the process of adding fractions. Instead of combining simple fractions into one complex fraction, you're breaking a complex rational expression back apart into a sum of simpler ones. This skill shows up constantly in calculus (especially integration), so mastering it now gives you a serious head start.

Decomposition of Rational Expressions

The core idea: if you have a rational expression $\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is less than the degree of $Q(x)$, and $Q(x)$ factors into distinct linear factors, you can split it into simpler fractions.

For example, if $Q(x) = (x-a)(x-b)(x-c)$, the decomposition looks like:

$\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$

Before you start: Make sure the fraction is proper (degree of numerator < degree of denominator). If it isn't, use polynomial long division first, then decompose the remainder.

Steps to find A, B, and C:

1. Factor the denominator completely.
2. Set up the partial fraction form with unknown constants ($A$, $B$, $C$, etc.).
3. Multiply both sides by the common denominator to clear all fractions.
4. Solve for the constants using one of two methods:
   • Substitution method: Plug in the values that make each factor zero. For instance, substituting $x = a$ eliminates every term except the one with $A$, so you can solve for $A$ directly.
   • Equating coefficients: Expand the right side, then match the coefficients of each power of $x$ on both sides. This creates a system of equations you can solve.

Example: Decompose $\frac{5x + 1}{(x-1)(x+2)}$

1. Set up: $\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$
2. Multiply both sides by $(x-1)(x+2)$: $5x + 1 = A(x+2) + B(x-1)$
3. Substitute $x = 1$: $6 = 3A$, so $A = 2$
4. Substitute $x = -2$: $-9 = -3B$, so $B = 3$
5. Result: $\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2}$

Partial Fractions with Repeated Factors

When a linear factor appears more than once in the denominator, you need a separate term for each power of that factor, from 1 up to the highest power.

For a repeated factor $(x-a)^n$, include:

$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$

A common mistake is writing only one fraction for a repeated factor. If you see $(x-3)^3$ in the denominator, you need three separate terms with denominators $(x-3)$, $(x-3)^2$, and $(x-3)^3$.

To solve for the coefficients:

1. Multiply both sides by the full common denominator.
2. Substitute $x = a$ to find $A_n$ (the constant for the highest power) directly.
3. Use either substitution of other convenient $x$-values or equating coefficients to find the remaining constants.

Example: For $\frac{3x^2 + 1}{(x-1)^2(x+2)}$, the setup is:

$\frac{3x^2+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$

Notice there are two terms for the $(x-1)$ factor because it's squared.

Quadratic Factors in Partial Fractions

When the denominator contains an irreducible quadratic factor (one that can't be factored over the reals, meaning its discriminant $b^2 - 4ac < 0$), the numerator of that partial fraction must be linear, not constant.

For an irreducible quadratic factor $ax^2 + bx + c$, the corresponding term is:

$\frac{Ax + B}{ax^2 + bx + c}$

The numerator is $Ax + B$ (degree 1) because it must be exactly one degree less than the denominator factor (degree 2). This is different from linear factors, where the numerator is just a constant.

Example setup: For $\frac{2x^2 + 3}{(x-1)(x^2+4)}$, the decomposition is:

$\frac{2x^2+3}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+4}$

You'd then multiply through by $(x-1)(x^2+4)$ and use substitution ($x = 1$ gives $A$ quickly) combined with equating coefficients to find $B$ and $C$.

Repeated Quadratics in Decomposition

Just as repeated linear factors need multiple terms, repeated irreducible quadratic factors do too. For $(ax^2 + bx + c)^n$, include:

$\frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}$

Each term has a linear numerator, and you write one term for each power from 1 through $n$.

These problems generate larger systems of equations, so equating coefficients is typically the most reliable approach. Be organized: expand everything carefully, group by powers of $x$, and solve the resulting system step by step.

Summary of Decomposition Forms

Applications in Calculus and Differential Equations

You won't need this for your pre-calc exam, but here's why this topic exists in the curriculum: in calculus, integrating a fraction like $\frac{5x+1}{(x-1)(x+2)}$ directly is difficult. Once you decompose it into $\frac{2}{x-1} + \frac{3}{x+2}$, each piece integrates easily using basic logarithm rules. Partial fractions also appear when solving certain differential equations using Laplace transforms. The decomposition skills you build now will pay off later.