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 righthand 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 righthand 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.
Task!
Express $\frac{94x}{3{x}^{2}x2}$ in partial fractions.
First factorise the denominator:
$\left(3x+2\right)\left(x1\right)$
Because there are two linear factors we write
$\phantom{\rule{2em}{0ex}}\frac{94x}{3{x}^{2}x2}=\frac{A}{3x+2}+\frac{B}{x1}$
Multiply both sides by $\left(3x+2\right)\left(x1\right)$ to obtain the equation from which to find $A$ and $B$ :
$94x=A\left(x1\right)+B\left(3x+2\right)$
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:
$\frac{7}{3x+2}+\frac{1}{x1}$
Exercises

Find the partial fractions of
 $\frac{5x1}{\left(x+1\right)\left(x2\right)}$ ,
 $\frac{7x+25}{\left(x+4\right)\left(x+3\right)}$ ,

$\frac{11x+1}{\left(x1\right)\left(2x+1\right)}$
.
Check by adding the partial fractions together again.

Express each of the following as the sum of partial fractions:
 $\frac{3}{\left(x+1\right)\left(x+2\right)}$ ,
 $\frac{5}{{x}^{2}+7x+12}$ ,
 $\frac{3}{\left(2x+1\right)\left(x3\right)}$ ,
1(a) $\frac{2}{x+1}+\frac{3}{x2}$ , 1(b) $\frac{3}{x+4}+\frac{4}{x+3}$ 1(c) $\frac{4}{x1}+\frac{3}{2x+1}$ ,
2(a) $\frac{3}{x+1}\frac{3}{x+2}$ , 2(b) $\frac{5}{x+3}\frac{5}{x+4}$ , 2(c) $\frac{6}{7\left(2x+1\right)}\frac{3}{7\left(x3\right)}$ .