Case 1 — g(x) Has Distinct (Non-Repeated) Linear Factors

The rule. Suppose $\dfrac{f(x)}{g(x)}$ is a proper fraction and $g(x)$ factors into linear pieces that are all different from one another, say $g(x) = (a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)$ with no factor repeated. Then to every such factor $(ax+b)$ there corresponds exactly one partial fraction of the shape

for some real constant $A$ that we need to find. The full decomposition is the sum of one such term per factor.

The method. Write the assumed decomposition with unknown constants, clear all denominators by multiplying both sides by $g(x)$, and you're left with a polynomial identity — one polynomial equals another, for every value of $x$. There are two equivalent ways to pin down the unknown constants from that identity:

• Convenient substitution (the fast way): since the identity must hold for every $x$, you're free to plug in whatever number you like — including the root of one of the linear factors, which is exactly the value that makes all the other terms on the right vanish and isolates one constant at a time.
• Equating coefficients (the systematic way): expand both sides as polynomials in $x$ and match the coefficient of each power of $x$; this gives a system of linear equations for the unknown constants, useful when the roots aren't "nice" numbers or when you want a cross-check.

Worked example. Resolve $\dfrac{7x-4}{(x-2)(x+3)}$ into partial fractions.

Since $(x-2)$ and $(x+3)$ are distinct linear factors, write

Multiply both sides by $(x-2)(x+3)$:

Put $x=2$ in (1) — this kills the $B$ term: $7(2)-4 = A(2+3) \Rightarrow 10 = 5A \Rightarrow A = 2$.

Put $x=-3$ in (1) — this kills the $A$ term: $7(-3)-4 = B(-3-2) \Rightarrow -25 = -5B \Rightarrow B = 5$.

So

$\frac{7x-4}{(x-2)(x+3)} = \frac{2}{x-2} + \frac{5}{x+3}$ …