### Partial Fractions

#### Definition

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.

#### Worked Examples

###### Example 1

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

###### Solution

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

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

###### Example 2

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

###### Solution

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

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

#### Video Examples

###### Example 1

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

###### Example 2

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

###### Example 3

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

###### Example 4

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

###### Example 5

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

#### Workbook

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

#### More Support

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