Partial Fractions

Tags: math

A way to simplify repeated terms in a algebraic fraction, based on shared factorised terms.

For example:

\begin{align*} \frac{3x^2 - 6x + 3}{12x^2 - 6x -6} &= \frac{3(x^{2-2x+1)}{6(2x}2-x-1)} \\

                                  &= \frac{3(x-1)^2}{6(2x+1)(x-1)} \\\\
                                  &= \frac{x-1}{2(2x+1)}

\end{align*}

Generally we solve partial fractions when the numerator is heavier than or equal to the denominator, that is the largest power of x on the numerator is larger than the largest power of x in the denominator. We can fix this by removing the fractional term from the highest powers of \( x \) and including a non-top heavy fractional component.

For example consider the algebraic fraction:

\begin{align} \frac{4x^3-5x+8}{(x-5)(x+1)} = \frac{5x^3-5x+8}{x2-4x-5} \label{eq:alg-frac-example} \end{align}

This equation has a cubic on the numerator and a quadratic on the denominator so it's top heavy. To solve this we can construct a table containing the denominator across the first column and the terms of the non-fractional simplified components across the first row.

Table 1: Decomposition of the algebraic fraction example in eq:alg-frac-example. Each cell is equivalent to the row value times the column value. tbl:alg-frac-example

| || \( 4x \) | \( 16 \) | Remainder | |-------------|:-------------:|:-------------:|:-----------:| | \( x^2 \) || \( x^3 \) | \( 16x^2 \) | \( 79x \) | | \( -4x \) || \( -16^2 \) | \( -64x \) | \( 88 \) | | \( -5 \) || \( -20x \) | \( -80 \) | |

In tbl:alg-frac-example you can see how we simplify eq:alg-frac-example. The first column produces the required \( x^3 \) term we need for the numerator, the second column produces the necessary \( x^2 \) term needed to cancel out the \( x^2 \) term from the first column and sum to get the \( x^2 \) term we need for the numerator. Observe the diagonals of the body of the table all contain the same power of \( x \), therefore we can just incrementally build up the terms in the columns such that the diagonals sum together to give the numerator of the original algebraic fraction.

The solution to eq:alg-frac-example as shown by tbl:alg-frac-example is \[ 4x + 16 + \frac{79x+88}{x^2-4x-5} \]

Links to this note

Cubic Equation