What Are Partial Fractions?
Partial fractions are the simpler fractions you get when you split a single rational expression — a fraction whose numerator and denominator are both polynomials — into a sum of fractions with smaller denominators. The process of doing this is called partial fraction decomposition.
Formally, a rational expression $\dfrac{P(x)}{Q(x)}$ is rewritten as a sum of fractions, each having one factor of $Q(x)$ as its denominator. The method depends entirely on how $Q(x)$ factors, which is why factorization is the first step every time.
One condition matters before you start: the expression must be proper, meaning the degree of the numerator is less than the degree of the denominator. If it is improper (numerator degree $\geq$ denominator degree), do polynomial long division first to split off the whole-number part, then decompose what remains.
What Are the Cases for Partial Fraction Decomposition?
The form of the decomposition is set by the type of factor in the denominator. There are three you need at this level, plus the repeated-quadratic extension.
Denominator factor | Decomposition form |
|---|---|
Distinct linear, $(ax + b)$ | $\dfrac{A}{ax + b}$ |
Repeated linear, $(ax + b)^n$ | $\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax + b)^2} + \cdots + \dfrac{A_n}{(ax + b)^n}$ |
Irreducible quadratic, $(ax^2 + bx + c)$ | $\dfrac{Ax + B}{ax^2 + bx + c}$ |
Two rules drive the whole table. A linear factor gets a constant numerator $A$. A quadratic factor (one that does not factor further over the reals) gets a linear numerator $Ax + B$. And a repeated factor raised to the power $n$ needs one term for every power from $1$ up to $n$, not just the highest.
How Do You Decompose Into Partial Fractions?
The procedure is the same every time, once you know which case you are in:
Factor the denominator completely and confirm the expression is proper (do long division first if it is not).
Write the decomposition with unknown constants ($A$, $B$, …) using the case table above.
Clear the denominators by multiplying both sides by the original denominator.
Solve for the constants — either by substituting smart values of $x$ (each chosen to zero out a factor) or by comparing coefficients of like powers.
Write the final sum with the constants filled in.
Step $4$ has two routes, and choosing well saves time. Substituting strategic $x$-values (the cover-up idea) is fastest when the denominator has distinct linear factors. Comparing coefficients is the reliable fallback for repeated or quadratic factors. You can mix them.
Examples of Partial Fractions
Example 1
Decompose $\dfrac{4x + 12}{x^2 + 4x}$ into partial fractions.
Factor the denominator: $x^2 + 4x = x(x + 4)$, two distinct linear factors. Write the form:
$$\frac{4x + 12}{x(x + 4)} = \frac{A}{x} + \frac{B}{x + 4}$$
Clear denominators by multiplying through by $x(x + 4)$:
$$4x + 12 = A(x + 4) + Bx$$
Substitute $x = 0$: $;12 = A(4)$, so $A = 3$. Substitute $x = -4$: $;4(-4) + 12 = B(-4)$, so $-4 = -4B$, giving $B = 1$.
$$\frac{4x + 12}{x^2 + 4x} = \frac{3}{x} + \frac{1}{x + 4}$$
Final answer: $\dfrac{3}{x} + \dfrac{1}{x + 4}$.
Example 2
Decompose $\dfrac{5x - 4}{(x - 1)(x + 2)}$.
A tempting shortcut is to write each piece with the whole denominator under it, $\frac{5x-4}{x-1} + \frac{5x-4}{x+2}$, as if "splitting" just copies the numerator over each factor. Add those back and you get $\frac{(5x-4)(x+2) + (5x-4)(x-1)}{(x-1)(x+2)}$, whose numerator is degree $2$, not the degree $1$ we started with. That cannot be right.
The decomposition uses unknown constants, not copies of the numerator:
$$\frac{5x - 4}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$$ $$5x - 4 = A(x + 2) + B(x - 1)$$
Substitute $x = 1$: $;1 = 3A$, so $A = \dfrac{1}{3}$. Substitute $x = -2$: $;-14 = -3B$, so $B = \dfrac{14}{3}$.
$$\frac{5x - 4}{(x - 1)(x + 2)} = \frac{1}{3(x - 1)} + \frac{14}{3(x + 2)}$$
Final answer: $\dfrac{1}{3(x - 1)} + \dfrac{14}{3(x + 2)}$.
Example 3
Decompose $\dfrac{3x + 1}{(x - 2)^2}$, a repeated linear factor.
A repeated factor of power $2$ needs a term for each power, $1$ and $2$:
$$\frac{3x + 1}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2}$$ $$3x + 1 = A(x - 2) + B$$
Substitute $x = 2$: $;7 = B$. Compare $x$-coefficients: $;3 = A$.
$$\frac{3x + 1}{(x - 2)^2} = \frac{3}{x - 2} + \frac{7}{(x - 2)^2}$$
Final answer: $\dfrac{3}{x - 2} + \dfrac{7}{(x - 2)^2}$.
Example 4
Decompose $\dfrac{2x^2 + 3}{x(x^2 + 1)}$, with an irreducible quadratic factor.
The factor $x^2 + 1$ does not factor over the reals, so it gets a linear numerator $Bx + C$:
$$\frac{2x^2 + 3}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$$ $$2x^2 + 3 = A(x^2 + 1) + (Bx + C)x$$
Substitute $x = 0$: $;3 = A$. Compare $x^2$-coefficients: $;2 = A + B$, so $B = 2 - 3 = -1$. Compare $x$-coefficients: $;0 = C$.
$$\frac{2x^2 + 3}{x(x^2 + 1)} = \frac{3}{x} + \frac{-x}{x^2 + 1}$$
Final answer: $\dfrac{3}{x} - \dfrac{x}{x^2 + 1}$.
Example 5
Decompose the improper fraction $\dfrac{x^2 + 1}{x^2 - 1}$.
The numerator and denominator have the same degree, so the expression is improper — divide first. Since $\dfrac{x^2 + 1}{x^2 - 1} = 1 + \dfrac{2}{x^2 - 1}$, decompose the proper remainder $\dfrac{2}{x^2 - 1} = \dfrac{2}{(x-1)(x+1)}$:
$$\frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$ $$2 = A(x + 1) + B(x - 1)$$
Substitute $x = 1$: $;2 = 2A$, so $A = 1$. Substitute $x = -1$: $;2 = -2B$, so $B = -1$.
$$\frac{x^2 + 1}{x^2 - 1} = 1 + \frac{1}{x - 1} - \frac{1}{x + 1}$$
Final answer: $1 + \dfrac{1}{x - 1} - \dfrac{1}{x + 1}$.
Example 6
Decompose $\dfrac{x + 7}{x^2 - x - 6}$.
Factor the denominator: $x^2 - x - 6 = (x - 3)(x + 2)$.
$$\frac{x + 7}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2}$$ $$x + 7 = A(x + 2) + B(x - 3)$$
Substitute $x = 3$: $;10 = 5A$, so $A = 2$. Substitute $x = -2$: $;5 = -5B$, so $B = -1$.
$$\frac{x + 7}{x^2 - x - 6} = \frac{2}{x - 3} - \frac{1}{x + 2}$$
Final answer: $\dfrac{2}{x - 3} - \dfrac{1}{x + 2}$.
Why Partial Fractions Matter
Partial fractions exist for one practical reason: a sum of simple fractions is enormously easier to work with than one complicated fraction. Many operations that are impossible on $\dfrac{4x+12}{x^2+4x}$ become routine on $\dfrac{3}{x} + \dfrac{1}{x+4}$, because each simple piece has a known, off-the-shelf answer.
Where the method pays off:
Integration in calculus. You cannot integrate $\dfrac{4x+12}{x^2+4x}$ directly, but each decomposed piece integrates to a logarithm instantly. This is the single most common reason students meet partial fractions.
Inverse Laplace and Z-transforms. Engineers decompose a system's transfer function into partial fractions because each simple piece corresponds to a known signal — the bridge from frequency domain back to time.
Series expansions. Splitting a fraction makes its power-series expansion tractable.
The destination here is calculus: partial fractions are taught in algebra precisely so that, a course or two later, integrating any rational function becomes a mechanical procedure rather than a dead end. The algebra you do now is the setup for an integral you will meet soon.
Where Decomposition Goes Off the Rails
Mistake 1: Wrong numerator form over a quadratic factor
Where it slips in: When the denominator has an irreducible quadratic factor.
Don't do this: Writing $\dfrac{A}{x^2 + 1}$, a constant numerator over a quadratic.
The correct way: A quadratic factor needs a linear numerator: $\dfrac{Ax + B}{x^2 + 1}$. The rule is that the numerator is always one degree lower than its denominator factor — constant over linear, linear over quadratic. The student who carries the constant-numerator habit from the linear case into the quadratic case loses a degree of freedom and gets stuck.
Mistake 2: Missing terms for a repeated factor
Where it slips in: When a factor is raised to a power greater than one.
Don't do this: Writing $\dfrac{A}{(x - 2)^2}$ alone for $\dfrac{3x+1}{(x-2)^2}$.
The correct way: A factor of power $n$ needs every power from $1$ to $n$: $\dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}$. You cannot tell from the combined fraction which lower powers were present in the original sum, so you must allow for all of them. Dropping the $\dfrac{A}{x-2}$ term is the quiet reason a "correct method" still gives a wrong decomposition.
Mistake 3: Forgetting to divide first when the fraction is improper
Where it slips in: When the numerator's degree is $\geq$ the denominator's.
Don't do this: Trying to decompose $\dfrac{x^2 + 1}{x^2 - 1}$ straight into $\dfrac{A}{x-1} + \dfrac{B}{x+1}$ without dividing.
The correct way: Decomposition only works on proper fractions. Do long division first to peel off the polynomial part, then decompose the proper remainder. The reader who skips the proper-fraction check applies the right machinery to the wrong object and gets an inconsistent system of equations.
Conclusion
Partial fractions split one rational expression into a sum of simpler fractions — the reverse of adding fractions over a common denominator.
The decomposition form depends on the denominator's factors: constant over a linear factor, linear over a quadratic, and one term per power for a repeated factor.
The fraction must be proper first; if it is improper, do polynomial long division before decomposing.
Solve for the constants by substituting strategic $x$-values or by comparing coefficients.
The most common mistakes are the wrong numerator form over a quadratic, missing terms for a repeated factor, and skipping the long-division step.
Practice These to Solidify Your Understanding
Decompose these in order, then add your pieces back to check:
$\dfrac{7}{(x - 1)(x + 6)}$
$\dfrac{2x + 1}{(x + 3)^2}$
$\dfrac{x + 4}{x(x^2 + 4)}$
$\dfrac{x^2}{x^2 - 9}$ (improper — divide first)
If question $4$ gives you an inconsistent system, return to the "divide first" mistake above. To work through partial fractions with a live trainer, explore Bhanzu's algebra tutor or help with algebra, or browse math classes online. Want a trainer to walk through more partial-fraction problems? Book a free demo class.
Read More
Polynomials — the numerators and denominators of rational expressions.
Simplifying Rational Expressions — the broader family of polynomial fractions.
Factorization of Quadratic Polynomials — factoring the denominators you decompose.
Factored Form — recognising a fully factored denominator before decomposing.
Dividing Polynomials — the division skill behind handling improper fractions.
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