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

Saturday, December 22, 2012

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

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

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

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

$\displaystyle\frac{{3}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{x}+{2}\right)}}}=\frac{{A}}{{{x}-{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{x}+{2}}}$

You need to bring the fractions to a common denominator:

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

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

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

Comparing both sides yields:

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

A - B + C = 0 => 2A + C = 0

$\displaystyle{2}{A}-{C}={3}$

Adding the relations yields:

$\displaystyle{4}{A}={3}\Rightarrow{A}=\frac{{3}}{{4}}\Rightarrow{B}=-\frac{{3}}{{4}}\Rightarrow{C}=-\frac{{6}}{{4}}$

Hence, the partial fraction decomposition is $\displaystyle\frac{{3}}{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{x}+{2}\right)}}}=\frac{{3}}{{{4}{x}-{4}}}+\frac{{-{3}{x}-{6}}}{{{4}{{x}}^{{2}}+{4}{x}+{8}}}.$

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Saturday, December 22, 2012

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

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