MATH1231 - Partial Fractions

## Case 1: Denominator splits into distinct linear factors.

• We factorise it from
(1)
\begin{align} \frac {x+c}{x^2 + ax + bx + ab} \rightarrow \frac {x+c}{(x + a)(x+b)} \end{align}
• From there we write it as equal to
(2)
\begin{align} \frac {A}{x+a} + \frac {B}{x+b} \end{align}
• And multiply these into one form:
(3)
\begin{align} \frac{A(x+b) + B(x+a)}{(x+a)(x+b)} \end{align}
• And hence
(4)
\begin{equation} A(x+b) + B(x+a) = x+c. \end{equation}
• Solve for A and B

## Case 2: Denominator has a repeated linear factor

• E.g.
(5)
\begin{align} \frac {x+c}{(x+4)^3} \end{align}
• This goes to
(6)
\begin{align} \frac {A}{x+4} + \frac {B}{(x+4)^2} + \frac{C}{(x+4)^3} \end{align}
• From here we factorise through to get
(7)
\begin{align} \frac{(x+4)^2 + B(x+4) + C}{(x+4)^3} \end{align}
• And solve for A, B, C

## Case 3: Denominator has an irreducible quadratic factor

• E.g.
(8)
\begin{align} \frac {x^2 + x}{(x-1)(x^2 + 9)} \rightarrow \frac {A}{x-1} + \frac {Bx + C}{x^2 + 9} \end{align}
• Our numerator constants always have one less x than the denominator, as seen above
• And there is always enough constants to have one without an x

## Case 4: Denominator has a repeated irreducible quadratic factor

• We shouldn’t have to deal with this, but you basically apply cases 2 and 3 – so you have an Ax+B on top (and a Cx + D and etc) each over an increasing power of your irreducible quadratic factor.
page revision: 7, last edited: 12 Aug 2011 00:20