Blog: `(x^2 + 2x + 8)/(x^2 + 4)^2` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Sunday, December 7, 2008

$\displaystyle\frac{{{{x}}^{{2}}+{2}{x}+{8}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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

$\displaystyle\frac{{{{x}}^{{2}}+{2}{x}+{8}}}{{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}}=\frac{{{A}{x}+{B}}}{{{{x}}^{{2}}+{4}}}+\frac{{{C}{x}+{D}}}{{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}}$

Multiply through by the LCD $\displaystyle{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}.$

$\displaystyle{{x}}^{{2}}+{2}{x}+{8}={\left({A}{x}+{B}\right)}{\left({{x}}^{{2}}+{4}\right)}+{\left({C}{x}+{D}\right)}$

$\displaystyle{{x}}^{{2}}+{2}{x}+{8}={A}{{x}}^{{3}}+{B}{{x}}^{{2}}+{4}{A}{x}+{4}{B}+{C}{x}+{D}$

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

Equate coefficients of like terms. Then solve for A, B, C, D.

$\displaystyle{0}={A}$

$\displaystyle{1}={B}$

$\displaystyle{2}={4}{A}+{C}$

$\displaystyle{2}={4}{\left({0}\right)}+{C}$

$\displaystyle{2}={C}$

$\displaystyle{8}={4}{B}+{D}$

$\displaystyle{8}={4}{\left({1}\right)}+{D}$

$\displaystyle{4}={D}$

$\displaystyle{A}={0},{B}={1},{C}={2},{D}={4}$

$\displaystyle\frac{{{{x}}^{{2}}+{2}{x}+{8}}}{{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}}=\frac{{{0}{x}+{1}}}{{{{x}}^{{2}}+{4}}}+\frac{{{2}{x}+{4}}}{{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}}=\frac{{1}}{{{{x}}^{{2}}+{4}}}+\frac{{{2}{x}+{4}}}{{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}}$

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Sunday, December 7, 2008

$\displaystyle\frac{{{{x}}^{{2}}+{2}{x}+{8}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{2}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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