Blog: `(x + 1)/(x^2 - x - 6)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Tuesday, February 23, 2010

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}$

Let's factorize the denominator,

$\displaystyle{{x}}^{{2}}-{x}-{6}={{x}}^{{2}}-{3}{x}+{2}{x}-{6}$

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

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

$\displaystyle\therefore\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}=\frac{{{x}+{1}}}{{{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}$

Now let $\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}=\frac{{A}}{{{x}-{3}}}+\frac{{B}}{{{x}+{2}}}$

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}=\frac{{{A}{\left({x}+{2}\right)}+{B}{\left({x}-{3}\right)}}}{{{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}=\frac{{{A}{x}+{2}{A}+{B}{x}-{3}{B}}}{{{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}$

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

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

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={1}$

$\displaystyle{2}{A}-{3}{B}={1}$

Now let's solve the above two equations to get the values of A and B,

express B in terms of A from the first equation,

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

substitute the expression of B in the second equation,

$\displaystyle{2}{A}-{3}{\left({1}-{A}\right)}={1}$

$\displaystyle{2}{A}-{3}+{3}{A}={1}$

$\displaystyle{5}{A}-{3}={1}$

$\displaystyle{5}{A}={1}+{3}$

$\displaystyle{5}{A}={4}$

$\displaystyle{A}=\frac{{4}}{{5}}$

Plug the value of A in the first equation,

$\displaystyle\frac{{4}}{{5}}+{B}={1}$

$\displaystyle{B}={1}-\frac{{4}}{{5}}$

$\displaystyle{B}=\frac{{1}}{{5}}$

$\displaystyle\therefore\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}=\frac{{4}}{{{5}{\left({x}-{3}\right)}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}$

Now let's check the above result,

RHS= $\displaystyle\frac{{4}}{{{5}{\left({x}-{3}\right)}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{4}{\left({x}+{2}\right)}+{1}{\left({x}-{3}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{4}{x}+{8}+{x}-{3}}}{{{5}{\left({{x}}^{{2}}+{2}{x}-{3}{x}-{6}\right)}}}$

$\displaystyle=\frac{{{5}{x}+{5}}}{{{5}{\left({{x}}^{{2}}-{x}-{6}\right)}}}$

$\displaystyle=\frac{{{5}{\left({x}+{1}\right)}}}{{{5}{\left({{x}}^{{2}}-{x}-{6}\right)}}}$

$\displaystyle=\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}$

= LHS

Hence it is verified.

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Tuesday, February 23, 2010

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{2}}-{x}-{6}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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