An algebraic fraction of the form $S=\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials, can be expressed as a sum of simpler algebraic fractions. This is called writing $S$ in its *partial fractions*.

Note that we assume that the degree of $p(x)$ is less than the degree of $q(x)$. If not we divide out the polynomials and work with the remainder.

These simpler algebraic fractions can take the following forms:

Type 1 |
\[\frac{A}{ax+b}\] |
Unrepeated linear factor |

Type 2 |
\[\frac{B}{(ax+b)^m}, \quad m \ge 2\] |
Repeated linear factor |

Type 3 |
\[\frac{Cx+D}{ax^2+bx+c}\] |
Unrepeated quadratic factor |

Type 4 |
\[\frac{Ex+F}{(ax^2+bx+c)^n}, \quad n \ge 2\] |
Repeated quadratic factor |

Which forms exist in the partial fraction decomposition of $S$ depends upon the factorisation of $q(x)$. We know from the The Fundamental Theorem of Algebra that $q(x)$ can be written as a product of factors, each of which is *linear* (of the form $ax+b$) or *quadratic* (of the form $ax^2+bx+c$). Some of these factors may be repeated.

- An
**unrepeated linear**factor gives a partial fraction of**Type 1**. - A
**repeated linear**factor gives partial fractions of**Type 2**. In this case we obtain a sum of partial fractions

\[\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_m}{(ax+b)^m}\]

- An
**unrepeated quadratic**factor gives a partial fraction of**Type 3**. - A
**repeated quadratic**factor gives partial fractions of**Type 4**. In this case we obtain a sum of partial fractions

\[\frac{C_1x+D_1}{ax^2+bx+c}+\frac{C_2x+D_2}{(ax^2+bx+c)^2}+\cdots+\frac{C_nx+D_n}{(ax^2+bx+c)^n}\]

The following worked examples indicate the method of finding the partial fractions.

Express $\dfrac{5x+10}{(x+1)(x+6)}$ as partial fractions.

The partial fractions will take the form:

\[\frac{5x+10}{(x+1)(x+6)} = \frac{A}{x+1} + \frac{B}{x+6}\]

Multiplying by $(x+1)(x+6)$ gives:

\[5x+10 = A(x+6) + B(x+1)\]

Choose $x=-1$, as this will make the $B$ term zero, so we are just left with an equation in $A$ to solve.

\begin{align} 5(-1)+10 &= A(-1+6) + B(0)\\\\ 5&=5A\\\\ A&=1 \end{align}

Choose $x=-6$, as this will make the $A$ term zero, so we are just left with an equation in $B$ to solve.

\begin{align} 5(-6)+10 &= A(0) + B(-6+1)\\\\ -20 &= -5B\\\\ B &= 4 \end{align}

Therefore

\[\dfrac{5x+10}{(x+1)(x+6)} = \frac{1}{x+1} + \frac{4}{x+6}\]

Express $\dfrac{x-2}{(x+1)(x-1)^2}$ as partial fractions.

Firstly, as there is a repeated linear factor $(x-1)^2$ this is a Type 2 partial fraction and combining with the partial fraction from the unrepeated linear factor $x+1$ we have the partial fraction decomposition:

\[\frac{x-2}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\]

Multiply by $(x+1)(x-1)^2$

\[x-2 = A(x-1)^2 + B(x+1)(x-1) + C(x+1)\]

Choose $x=1$

\begin{align} 1-2 &= A(0) + B(0)+ C(1+1)\\\\ -1&=2C\\\\ C&=-\frac{1}{2} \end{align}

Choose $x=-1$

\begin{align} -1-2 &= A(-1-1)^2 + B(0)+ C(0)\\\\ -3 &= 4A\\\\ A &= -\frac{3}{4} \end{align}

Choose $x=0$, and subtitute in the values we have already found for $A$ and $C$.

\begin{align} 0-2 &= -\frac{3}{4}(-1)^2 + B(0+1)(0-1)-\frac{1}{2}(0+1)\\\\ -2 &= -\frac{3}{4} -B -\frac{1}{2}\\\\ -2 &= \frac{5}{4} -B\\\\ B &= \frac{3}{4} \end{align}

Therefore

\[\dfrac{x-2}{(x+1)(x-1)^2}=-\dfrac{3}{4(x+1)} + \dfrac{3}{4(x-1)} - \dfrac{1}{2(x-1)^2}\]

Prof. Robin Johnson shows how to express $\dfrac{2x+1}{(x+1)(x-2)}$ as partial fractions.

Prof. Robin Johnson shows how to express $\dfrac{x+3}{x(x^2-1)}$ as partial fractions.

Prof. Robin Johnson show how to express $\dfrac{1}{(x-1)(x+2)}$ as partial fractions.

Prof. Robin Johnson shows how to express $\dfrac{3-x}{(x^2-1)(2-x)}$ as partial fractions.

Prof. Robin Johnson shows how to express $\dfrac{1}{(x^2-4)(x+2)}$ as partial fractions.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test yourself: Numbas test on partial fractions

- Partial fractions workbook at
**math**centre.