Blog: `x/(x^3 - x^2 - 2x + 2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Saturday, September 12, 2009

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Decompose the denominator:

$\displaystyle{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}={{x}}^{{2}}{\left({x}-{1}\right)}-{2}{\left({x}-{1}\right)}={\left({{x}}^{{2}}-{2}\right)}{\left({x}-{1}\right)}={\left({x}-{1}\right)}{\left({x}-\sqrt{{{2}}}\right)}{\left({x}+\sqrt{{{2}}}\right)}.$

Therefore the fraction decomposition has the form

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{x}-\sqrt{{{2}}}}}+\frac{{C}}{{{x}+\sqrt{{{2}}}}}.$

To find A,B and C multiply both sides by the original denominator:

$\displaystyle{x}={A}{\left({{x}}^{{2}}-{2}\right)}+{B}{\left({x}-{1}\right)}{\left({x}+\sqrt{{{2}}}\right)}+{C}{\left({x}-{1}\right)}{\left({x}-\sqrt{{{2}}}\right)},$ or

$\displaystyle{x}={{x}}^{{2}}{\left({A}+{B}+{C}\right)}+{x}{\left({B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}\right)}+{\left(-{2}{A}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}\right)}.$

Thus $\displaystyle{A}+{B}+{C}={0},$ $\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1}$ and $\displaystyle-{2}{A}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0}.$

$\displaystyle{A}=-{\left({B}+{C}\right)},$

$\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1},$

$\displaystyle{2}{\left({B}+{C}\right)}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0},$ or
$\displaystyle{B}{\left({2}-\sqrt{{{2}}}\right)}+{C}{\left({2}+\sqrt{{{2}}}\right)}={0},$ or
$\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}+{C}{\left(\sqrt{{{2}}}+{1}\right)}={0}.$

Add and subtract these two equations and obtain

$\displaystyle{2}{B}{\left(\sqrt{{{2}}}-{1}\right)}={1},$ or $\displaystyle{B}=\frac{{1}}{{{2}{\left(\sqrt{{{2}}}-{1}\right)}}}=\frac{{\sqrt{{{2}}}+{1}}}{{2}}$ and

$\displaystyle{2}{C}{\left(\sqrt{{{2}}}+{1}\right)}=-{1},$ or $\displaystyle{C}=-\frac{{1}}{{{2}{\left(\sqrt{{{2}}}+{1}\right)}}}=-\frac{{\sqrt{{{2}}}-{1}}}{{2}}.$

And $\displaystyle{A}=-{\left({B}+{C}\right)}=-{1}.$

Now check this result:

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

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

$\displaystyle=-\frac{{1}}{{{x}-{1}}}+\frac{{1}}{{2}}\frac{{{2}{x}+{4}}}{{{{x}}^{{2}}-{2}}}=-\frac{{1}}{{{x}-{1}}}+\frac{{{x}+{2}}}{{{{x}}^{{2}}-{2}}}=$

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

which is correct.

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Saturday, September 12, 2009

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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