3 Proper fractions with repeated linear factors

Sometimes a linear factor appears more than once. For example in

the factor $\left(x+1\right)$ occurs twice. We call it a repeated linear factor . The repeated linear factor ${\left(x+1\right)}^{2}$ produces two partial fractions of the form $\frac{A}{x+1}+\frac{B}{{\left(x+1\right)}^{2}}$ . In general, a repeated linear factor of the form ${\left(ax+b\right)}^{2}$ generates two partial fractions of the form

$\phantom{\rule{2em}{0ex}}\frac{A}{ax+b}+\frac{B}{{\left(ax+b\right)}^{2}}$

This is reasonable since the sum of two such fractions always gives rise to a proper fraction:

$\phantom{\rule{2em}{0ex}}\frac{A}{ax+b}+\frac{B}{{\left(ax+b\right)}^{2}}=\frac{A\left(ax+b\right)}{{\left(ax+b\right)}^{2}}+\frac{B}{{\left(ax+b\right)}^{2}}=\frac{x\left(Aa\right)+Ab+B}{{\left(ax+b\right)}^{2}}$

Key Point 15

A repeated linear factor ${\left(ax+b\right)}^{2}$ in the denominator produces two partial fractions:

$\frac{A}{ax+b}+\frac{B}{{\left(ax+b\right)}^{2}}$

Once again the unknown constants are found by either equating coefficients and/or substituting specific values for $x$ .

Express $\frac{10x+18}{4{x}^{2}+12x+9}$ in partial fractions.

First factorise the denominator:

$\left(2x+3\right)\left(2x+3\right)={\left(2x+3\right)}^{2}$

There is a repeated linear factor $\left(2x+3\right)$ which gives rise to two partial fractions of the form

$\phantom{\rule{2em}{0ex}}\frac{10x+18}{{\left(2x+3\right)}^{2}}=\frac{A}{2x+3}+\frac{B}{{\left(2x+3\right)}^{2}}$

Multiply both sides through by ${\left(2x+3\right)}^{2}$ to obtain the equation to be solved to find $A$ and $B$ :

$10x+18=A\left(2x+3\right)+B$

Now evaluate the constants $A$ and $B$ by equating coefficients:

Equating the $x$ coefficients gives $10=2A$ so $A=5$ . Equating constant terms gives $18=3A+B$ from which $B=3$ .

Finally express the answer in partial fractions:

$\phantom{\rule{2em}{0ex}}\frac{10x+18}{{\left(2x+3\right)}^{2}}=\frac{5}{2x+3}+\frac{3}{{\left(2x+3\right)}^{2}}$

Exercises

Express the following in partial fractions.

 (a)  $\frac{3-x}{{x}^{2}-2x+1}$ , (b)  $-\frac{7x-15}{{\left(x-1\right)}^{2}}$ (c)  $\frac{3x+14}{{x}^{2}+8x+16}$ (d)  $\frac{5x+18}{{\left(x+4\right)}^{2}}$ (e)  $\frac{2{x}^{2}-x+1}{\left(x+1\right){\left(x-1\right)}^{2}}$ (f)  $\frac{5{x}^{2}+23x+24}{\left(2x+3\right){\left(x+2\right)}^{2}}$ (g)  $\frac{6{x}^{2}-30x+25}{{\left(3x-2\right)}^{2}\left(x+7\right)}$ (h)  $\frac{s+2}{{\left(s+1\right)}^{2}}$ (i)  $\frac{2s+3}{{s}^{2}}$ .
 (a)  $-\frac{1}{x-1}+\frac{2}{{\left(x-1\right)}^{2}}$ (b)  $-\frac{7}{x-1}+\frac{8}{{\left(x-1\right)}^{2}}$ (c)  $\frac{3}{x+4}+\frac{2}{{\left(x+4\right)}^{2}}$ (d)  $\frac{5}{x+4}-\frac{2}{{\left(x+4\right)}^{2}}$ (e)  $\frac{1}{x+1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{x-1}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{{\left(x-1\right)}^{2}}$ (f)  $\frac{3}{2x+3}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{1}{x+2}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\frac{2}{{\left(x+2\right)}^{2}}$ (g)  $-\frac{1}{3x-2}+\frac{1}{{\left(3x-2\right)}^{2}}+\frac{1}{x+7}$ (h)  $\frac{1}{s+1}+\frac{1}{{\left(s+1\right)}^{2}}$ (i)  $\frac{2}{s}+\frac{3}{{s}^{2}}$ .