2 Proper fractions with linear factors

Firstly we describe how to calculate partial fractions for proper fractions where the denominator may be written as a product of linear factors. The steps are as follows:

$\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Factorise the denominator.

$\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Each factor will produce a partial fraction. A factor such as $3x+2$ will produce a partial

fraction of the form $\frac{A}{3x+2}$ where $A$ is an unknown constant. In general a linear factor

$ax+b$ will produce a partial fraction $\frac{A}{ax+b}$ . The unknown constants for each partial

fraction may be different and so we will call them $A$ , $B$ , $C$ and so on.

$\phantom{\rule{2em}{0ex}}\bullet \phantom{\rule{1em}{0ex}}$ Evaluate the unknown constants by equating coefficients or using specific values of $x$ .

The sum of the partial fractions is identical to the original algebraic fraction for all values of $x$ .

Key Point 14

A linear factor $ax+b$ in the denominator gives rise to a single partial fraction of the form  $\frac{A}{ax+b}$

The steps involved in expressing a proper fraction as partial fractions are illustrated in the following Example.

Example 41

Express $\frac{7x+10}{2{x}^{2}+5x+3}$ in terms of partial fractions.

Solution

Note that this fraction is proper. The denominator is factorised to give $\left(2x+3\right)\left(x+1\right)$ . Each of the linear factors produces a partial fraction. The factor $2x+3$ produces a partial fraction of the form $\frac{A}{2x+3}$ and the factor $x+1$ produces a partial fraction $\frac{B}{x+1}$ , where $A$ and $B$ are constants which we need to find. We write

$\phantom{\rule{2em}{0ex}}\frac{7x+10}{\left(2x+3\right)\left(x+1\right)}=\frac{A}{2x+3}+\frac{B}{x+1}$

By multiplying both sides by $\left(2x+3\right)\left(x+1\right)$ we obtain

$\phantom{\rule{2em}{0ex}}7x+10=A\left(x+1\right)+B\left(2x+3\right)$ …(*)

We may now let $x$ take any value we choose . By an appropriate choice we can simplify the right-hand side. Let $x=-1$ because this choice eliminates $A$ . We find

$\begin{array}{rcll}7\left(-1\right)+10& =& A\left(0\right)+B\left(-2+3\right)& \text{}\\ 3& =& B\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

so that the constant $B$ must equal 3. The constant $A$ can be found either by substituting some other value for $x$ or alternatively by ‘equating coefficients’.

Observe that, by rearranging the right-hand side, Equation (*) can be written as

$\phantom{\rule{2em}{0ex}}7x+10=\left(A+2B\right)x+\left(A+3B\right)$

Comparing the coefficients of $x$ on both sides we see that $7=A+2B$ . We already know $B=3$ and so

$\begin{array}{rcll}7& =& A+2\left(3\right)& \text{}\\ & =& A+6\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

from which $A=1$ . We can therefore write

$\phantom{\rule{2em}{0ex}}\frac{7x+10}{2{x}^{2}+5x+3}=\frac{1}{2x+3}+\frac{3}{x+1}$

We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right.

Express $\frac{9-4x}{3{x}^{2}-x-2}$ in partial fractions.

First factorise the denominator:

$\left(3x+2\right)\left(x-1\right)$

Because there are two linear factors we write

$\phantom{\rule{2em}{0ex}}\frac{9-4x}{3{x}^{2}-x-2}=\frac{A}{3x+2}+\frac{B}{x-1}$

Multiply both sides by $\left(3x+2\right)\left(x-1\right)$ to obtain the equation from which to find $A$ and $B$ :

$9-4x=A\left(x-1\right)+B\left(3x+2\right)$

Substitute an appropriate value for $x$ to obtain $B$ :

Substitute $x=1$ and get $B=1$

Equating coefficients of $x$ to obtain the value of $A$ :

$-4=A+3B$ , $A=-7$ since $B=1$

Finally, write down the partial fractions:

$\frac{-7}{3x+2}+\frac{1}{x-1}$

Exercises
1. Find the partial fractions of
1. $\frac{5x-1}{\left(x+1\right)\left(x-2\right)}$ ,
2. $\frac{7x+25}{\left(x+4\right)\left(x+3\right)}$ ,
3. $\frac{11x+1}{\left(x-1\right)\left(2x+1\right)}$ .

Check by adding the partial fractions together again.

2. Express each of the following as the sum of partial fractions:
1. $\frac{3}{\left(x+1\right)\left(x+2\right)}$ ,
2. $\frac{5}{{x}^{2}+7x+12}$ ,
3. $\frac{-3}{\left(2x+1\right)\left(x-3\right)}$ ,

1(a)   $\frac{2}{x+1}+\frac{3}{x-2}$ , 1(b)   $\frac{3}{x+4}+\frac{4}{x+3}$ 1(c)   $\frac{4}{x-1}+\frac{3}{2x+1}$ ,

2(a)   $\frac{3}{x+1}-\frac{3}{x+2}$ , 2(b)   $\frac{5}{x+3}-\frac{5}{x+4}$ , 2(c)   $\frac{6}{7\left(2x+1\right)}-\frac{3}{7\left(x-3\right)}$ .