Blog: `(x + 2)/(x(x^2 - 9))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Monday, August 4, 2008

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{x}+{2}}}{{{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

Now let $\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{3}}}+\frac{{C}}{{{x}-{3}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{A}{\left({x}+{3}\right)}{\left({x}-{3}\right)}+{B}{\left({x}\right)}{\left({x}-{3}\right)}+{C}{\left({x}\right)}{\left({x}+{3}\right)}}}{{{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{A}{\left({{x}}^{{2}}-{9}\right)}+{B}{\left({{x}}^{{2}}-{3}{x}\right)}+{C}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle\therefore{\left({x}+{2}\right)}={A}{\left({{x}}^{{2}}-{9}\right)}+{B}{\left({{x}}^{{2}}-{3}{x}\right)}+{C}{\left({{x}}^{{2}}+{3}{x}\right)}$

$\displaystyle{x}+{2}={A}{{x}}^{{2}}-{9}{A}+{B}{{x}}^{{2}}-{3}{B}{x}+{C}{{x}}^{{2}}+{3}{C}{x}$

$\displaystyle{x}+{2}={\left({A}+{B}+{C}\right)}{{x}}^{{2}}+{\left(-{3}{B}+{3}{C}\right)}{x}-{9}{A}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}+{C}={0}$

$\displaystyle-{3}{B}+{3}{C}={1}$

$\displaystyle-{9}{A}={2}$

Solve the above three equations to get the values of A,B and C,

$\displaystyle{A}=-\frac{{2}}{{9}}$

Bach substitute the value of A in the first equation,

$\displaystyle-\frac{{2}}{{9}}+{B}+{C}={0}$

$\displaystyle{B}+{C}=\frac{{2}}{{9}}$

From the above equation ,express C in terms of B

$\displaystyle{C}=\frac{{2}}{{9}}-{B}$

Substitute the expression of C in the second equation,

$\displaystyle-{3}{B}+{3}{\left(\frac{{2}}{{9}}-{B}\right)}={1}$

$\displaystyle-{3}{B}+\frac{{2}}{{3}}-{3}{B}={1}$

$\displaystyle-{6}{B}={1}-\frac{{2}}{{3}}$

$\displaystyle-{6}{B}=\frac{{1}}{{3}}$

$\displaystyle{B}=-\frac{{1}}{{18}}$

Now plug the value of B in the expression of C,

$\displaystyle{C}=\frac{{2}}{{9}}-{\left(-\frac{{1}}{{18}}\right)}$

$\displaystyle{C}=\frac{{2}}{{9}}+\frac{{1}}{{18}}$

$\displaystyle{C}=\frac{{{2}\cdot{2}+{1}}}{{18}}$

$\displaystyle{C}=\frac{{5}}{{18}}$

$\displaystyle\therefore\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=-\frac{{2}}{{{9}{x}}}-\frac{{1}}{{{18}{\left({x}+{3}\right)}}}+\frac{{5}}{{{18}{\left({x}-{3}\right)}}}$

Now let's check the result,

RHS= $\displaystyle-\frac{{2}}{{{9}{x}}}-\frac{{1}}{{{18}{\left({x}+{3}\right)}}}+\frac{{5}}{{{18}{\left({x}-{3}\right)}}}$

$\displaystyle=\frac{{-{2}\cdot{2}{\left({x}+{3}\right)}{\left({x}-{3}\right)}-{x}{\left({x}-{3}\right)}+{5}{x}{\left({x}+{3}\right)}}}{{{18}{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

$\displaystyle=\frac{{-{4}{\left({{x}}^{{2}}-{9}\right)}-{\left({{x}}^{{2}}-{3}{x}\right)}+{5}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{-{4}{{x}}^{{2}}+{36}-{{x}}^{{2}}+{3}{x}+{5}{{x}}^{{2}}+{15}{x}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{18}{x}+{36}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{18}{\left({x}+{2}\right)}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

=LHS

Hence it is verified.

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Monday, August 4, 2008

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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