Blog: `(2x^3 - 4x^2 - 15x + 5)/(x^2 - 2x - 8)` Write the partial fraction decomposition of the rational expression.

Friday, March 16, 2012

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$ Write the partial fraction decomposition of the rational expression.

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$

Since the above rational expression is an improper rational expression , so the first step is to divide and express the expression as a sum of simpler fractions such that the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator.

Dividing the polynomial using long division method yields,

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}={2}{x}+\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$

Since the polynomials do not completely divide, we have to continue with the partial fractions of the remainder expression,

Now let's factorize the denominator of the remainder expression,

$\displaystyle{{x}}^{{2}}-{2}{x}-{8}={{x}}^{{2}}-{4}{x}+{2}{x}-{8}$

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

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

Let, $\displaystyle\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}=\frac{{A}}{{{x}+{2}}}+\frac{{B}}{{{x}-{4}}}$

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

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

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

$\displaystyle\therefore{\left({x}+{5}\right)}={x}{\left({A}+{B}\right)}-{4}{A}+{2}{B}$

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={1}$ ----- equation 1

$\displaystyle-{4}{A}+{2}{B}={5}$ ------ equation 2

Now solve the above equations to get the solutions of A and B,

From equation 1,

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

substitute the above expression of A in equation 2 ,

$\displaystyle-{4}{\left({1}-{B}\right)}+{2}{B}={5}$

$\displaystyle-{4}+{4}{B}+{2}{B}={5}$

$\displaystyle{6}{B}={5}+{4}$

$\displaystyle{B}=\frac{{9}}{{6}}$

$\displaystyle{B}=\frac{{3}}{{2}}$

Plug the value of B in equation 1,

$\displaystyle{A}+\frac{{3}}{{2}}={1}$

$\displaystyle{A}={1}-\frac{{3}}{{2}}$

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

$\displaystyle\frac{{{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}=-\frac{{1}}{{{2}{\left({x}+{2}\right)}}}+\frac{{3}}{{{2}{\left({x}-{4}\right)}}}$

$\displaystyle\therefore\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}={2}{x}-\frac{{1}}{{{2}{\left({x}+{2}\right)}}}+\frac{{3}}{{{2}{\left({x}-{4}\right)}}}$

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Friday, March 16, 2012

$\displaystyle\frac{{{2}{{x}}^{{3}}-{4}{{x}}^{{2}}-{15}{x}+{5}}}{{{{x}}^{{2}}-{2}{x}-{8}}}$ Write the partial fraction decomposition of the rational expression.

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