4 Proper fractions with quadratic factors

Sometimes when a denominator is factorised it produces a quadratic term which cannot be factorised into linear factors. One such quadratic factor is ${x}^{2}+x+1$ . This factor produces a partial fraction of the form $\frac{Ax+B}{{x}^{2}+x+1}$ . In general a quadratic factor of the form $a{x}^{2}+bx+c$ produces a single partial fraction of the form $\frac{Ax+B}{a{x}^{2}+bx+c}$ .

Key Point 16

A quadratic factor $a{x}^{2}+bx+c$ in the denominator produces a partial fraction of the form

$\frac{Ax+B}{a{x}^{2}+bx+c}$

Express as partial fractions $\frac{3x+1}{\left({x}^{2}+x+10\right)\left(x-1\right)}$

Note that the quadratic factor cannot be factorised further. We have

$\phantom{\rule{2em}{0ex}}\frac{3x+1}{\left({x}^{2}+x+10\right)\left(x-1\right)}=\frac{Ax+B}{{x}^{2}+x+10}+\frac{C}{x-1}$

First multiply both sides by $\left({x}^{2}+x+10\right)\left(x-1\right)$ :

$\left(Ax+B\right)\left(x-1\right)+C\left({x}^{2}+x+10\right)$

Evaluate $C$ by letting $x=1$ :

$4=12C$ so that $C=\frac{1}{3}$

Equate coefficients of ${x}^{2}$ and hence find $A$ , and then substitute any other value for $x$ (or equate coefficients of $x$ ) to find $B$ :

$-\frac{1}{3}$ , $\frac{7}{3}$ . Finally express in partial fractions:

$\frac{3x+1}{\left({x}^{2}+x+10\right)\left(x-1\right)}=\frac{-\frac{1}{3}x+\frac{7}{3}}{{x}^{2}+x+10}+\frac{\frac{1}{3}}{x-1}=\frac{7-x}{3\left({x}^{2}+x+10\right)}+\frac{1}{3\left(x-1\right)}$