Partial fractions with repeated and quadratic factors

When the denominator repeats a factor:

$\frac{f(x)}{(ax+b)(cx+d)^2} \equiv \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$

When the denominator contains a quadratic with no real roots:

$\frac{f(x)}{(ax+b)(cx^2+d)} \equiv \frac{A}{ax+b} + \frac{Bx + C}{cx^2+d}$

Why the repeated factor needs two terms:

• $\dfrac{C}{(cx+d)^2}$ alone cannot produce a piece like $\dfrac{1}{cx+d}$;
• $\dfrac{B}{cx+d}$ alone cannot produce a piece with a squared denominator.

Together they cover both cases.

Why the quadratic needs $Bx + C$. A constant numerator could only ever produce constant multiples of $\dfrac{1}{cx^2+d}$. A numerator like $3x - 1$ would be impossible.

Two tools, used together:

1. Substitute a root of a factor wherever a factor appears. That wipes out the other terms.
2. Compare coefficients of $x^2$, $x$ and the constant. Each comparison is one equation.

With a repeated factor, substituting the two roots gives $A$ and the constant on the squared term, but not $B$. The remaining constants need coefficient comparison, so that step is compulsory here.

The forgotten step: stopping after substitution and assuming you are finished. With three constants you usually need at least one comparison.