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

Sunday, October 18, 2009

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

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

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

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

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

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

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

Equating the coefficients the like terms,

$\displaystyle{A}={0}$

$\displaystyle{B}={2}$

$\displaystyle{4}{A}+{C}={1}$

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

Plug the value of the A and B in the above equations,

$\displaystyle{4}{\left({0}\right)}+{C}={1}$

$\displaystyle{C}={1}$

$\displaystyle{4}{\left({2}\right)}+{D}={8}$

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

$\displaystyle{D}={8}-{8}$

$\displaystyle{D}={0}$

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

Now let's check it algebraically,

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

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

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

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

=LHS

Hence it is verified.

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Sunday, October 18, 2009

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

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