Blog: `(4x^2 + 2x - 1)/(x^2(x + 1))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Sunday, March 15, 2009

$\displaystyle\frac{{{4}{{x}}^{{2}}+{2}{x}-{1}}}{{{{x}}^{{2}}{\left({x}+{1}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}}^{{2}}}+\frac{{C}}{{{x}+{1}}}$

Multiply through by the LCD $\displaystyle{{x}}^{{2}}{\left({x}+{1}\right)}.$

$\displaystyle{4}{{x}}^{{2}}+{2}{x}-{1}={A}{x}{\left({x}+{1}\right)}+{B}{\left({x}+{1}\right)}+{C}{{x}}^{{2}}$

$\displaystyle{4}{{x}}^{{2}}+{2}{x}-{1}={A}{{x}}^{{2}}+{A}{x}+{B}{x}+{B}+{C}{{x}}^{{2}}$

$\displaystyle{4}{{x}}^{{2}}+{2}{x}-{1}={\left({A}+{C}\right)}{{x}}^{{2}}+{\left({A}+{B}\right)}{x}+{B}$

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

$\displaystyle{B}=-{1}$

$\displaystyle{2}={A}+{B}$

$\displaystyle{2}={A}-{1}$

$\displaystyle{A}={3}$

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

$\displaystyle{4}={3}+{C}$

$\displaystyle{C}={1}$

$\displaystyle{A}={3},{B}=-{1},{C}={1}$

$\displaystyle\frac{{{4}{{x}}^{{2}}+{2}{x}-{1}}}{{{{x}}^{{2}}{\left({x}+{1}\right)}}}=\frac{{3}}{{x}}-\frac{{1}}{{{x}}^{{2}}}+\frac{{1}}{{{x}+{1}}}$

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