Blog: `(2x - 3)/(x - 1)^2` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Sunday, August 15, 2010

$\displaystyle\frac{{{2}{x}-{3}}}{{{\left({x}-{1}\right)}}^{{2}}}$

Let $\displaystyle\frac{{{2}{x}-{3}}}{{{\left({x}-{1}\right)}}^{{2}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle\frac{{{2}{x}-{3}}}{{{\left({x}-{1}\right)}}^{{2}}}=\frac{{{A}{\left({x}-{1}\right)}+{B}}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle\frac{{{2}{x}-{3}}}{{{\left({x}-{1}\right)}}^{{2}}}=\frac{{{A}{x}-{A}+{B}}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle\therefore{\left({2}{x}-{3}\right)}={A}{x}-{A}+{B}$

comparing the coefficients of the like terms,

$\displaystyle{A}={2}$

$\displaystyle-{A}+{B}=-{3}$

Plug in the value of A in the above equation,

$\displaystyle-{2}+{B}=-{3}$

$\displaystyle{B}=-{3}+{2}$

$\displaystyle{B}=-{1}$

$\displaystyle\therefore\frac{{{2}{x}-{3}}}{{{\left({x}-{1}\right)}}^{{2}}}=\frac{{2}}{{{x}-{1}}}-\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}$

Now let's check the above result,

RHS= $\displaystyle\frac{{2}}{{{x}-{1}}}-\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle=\frac{{{2}{\left({x}-{1}\right)}-{1}}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle=\frac{{{2}{x}-{2}-{1}}}{{{\left({x}-{1}\right)}}^{{2}}}$

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

$\displaystyle={L}{H}{S}$

Hence it is verified.

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