Blog: `(8x - 12)/(x^2(x^2 + 2)^2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Saturday, November 3, 2012

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

You need to decompose the fraction in simple irreducible fractions, such that:

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}=\frac{{A}}{{x}}+\frac{{B}}{{{{x}}^{{2}}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}+{2}}}+\frac{{{E}{x}+{F}}}{{{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$

You need to bring to the same denominator all fractions, such that:

$\displaystyle{8}{x}-{12}={A}{x}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}+{B}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}+{\left({C}{x}+{D}\right)}\cdot{{x}}^{{2}}\cdot{\left({{x}}^{{2}}+{2}\right)}+{\left({E}{x}+{F}\right)}\cdot{{x}}^{{2}}$

$\displaystyle{8}{x}-{12}={A}{x}{\left({{x}}^{{4}}+{4}{{x}}^{{2}}+{4}\right)}+{B}{{x}}^{{4}}+{4}{B}{{x}}^{{2}}+{4}{B}+{\left({C}{{x}}^{{3}}+{D}{{x}}^{{2}}\right)}\cdot{\left({{x}}^{{2}}+{2}\right)}+{E}{{x}}^{{3}}+{F}{{x}}^{{2}}$

$\displaystyle{8}{x}-{12}={A}{{x}}^{{5}}+{4}{A}{{x}}^{{3}}+{4}{A}{x}+{B}{{x}}^{{4}}+{4}{B}{{x}}^{{2}}+{4}{B}+{C}{{x}}^{{5}}+{2}{C}{{x}}^{{3}}+{D}{{x}}^{{4}}+{2}{D}{{x}}^{{2}}+{E}{{x}}^{{3}}+{F}{{x}}^{{2}}$

You need to group the terms having the same power of x:

$\displaystyle{8}{x}-{12}={{x}}^{{5}}{\left({A}+{C}\right)}+{{x}}^{{4}}{\left({B}+{D}\right)}+{{x}}^{{3}}{\left({4}{A}+{2}{C}+{E}\right)}+{{x}}^{{2}}{\left({4}{B}+{2}{D}+{F}\right)}+{x}{\left({4}{A}\right)}++{4}{B}$

Comparing the expressions both sides yields:

$\displaystyle{A}+{C}={0}\Rightarrow{A}=-{C}$

$\displaystyle{B}+{D}={0}\Rightarrow{B}=-{D}$

$\displaystyle{4}{A}+{2}{C}+{E}={0}\Rightarrow{2}{A}+{E}={0}\Rightarrow{E}=-{2}{A}$

$\displaystyle{4}{B}+{2}{D}+{F}={0}\Rightarrow{F}=-{2}{D}$

$\displaystyle{4}{A}={8}\Rightarrow{A}={2}\Rightarrow{C}=-{2}\Rightarrow{E}=-{4}$

$\displaystyle{4}{B}=-{12}\Rightarrow{B}=-{3}\Rightarrow{D}={3}\Rightarrow{F}=-{6}$

Hence, decomposing the fraction, yields $\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}=\frac{{2}}{{x}}-\frac{{3}}{{{{x}}^{{2}}}}+\frac{{-{2}{x}+{3}}}{{{{x}}^{{2}}+{2}}}+\frac{{-{4}{x}-{6}}}{{{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}.$

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Saturday, November 3, 2012

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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