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

Thursday, July 16, 2015

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

First we have to make the fraction a proper one, for this it is necessary to divide the numerator by the denominator with the remainder.

$\displaystyle{{x}}^{{4}}+{2}{{x}}^{{3}}+{4}{{x}}^{{2}}+{8}{x}+{2}={x}{\left({{x}}^{{3}}+{2}{{x}}^{{2}}+{x}\right)}+{3}{{x}}^{{2}}+{8}{x}+{2}.$

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

The proper part has the decomposition form of

$\displaystyle\frac{{{3}{{x}}^{{2}}+{8}{x}+{2}}}{{{x}{{\left({x}+{1}\right)}}^{{2}}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{1}}}+\frac{{C}}{{{\left({x}+{1}\right)}}^{{2}}}.$

Multiply both sides by $\displaystyle{x}{{\left({x}+{1}\right)}}^{{2}}$ and obtain

$\displaystyle{3}{{x}}^{{2}}+{8}{x}+{2}={A}{\left({{x}}^{{2}}+{2}{x}+{1}\right)}+{B}{x}{\left({x}+{1}\right)}+{C}{x}=$

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

So $\displaystyle{A}+{B}={3},$ $\displaystyle{2}{A}+{B}+{C}={8}$ and $\displaystyle{A}={2},$ from this we get $\displaystyle{B}={1}$ and $\displaystyle{C}={3}.$

The answer: $\displaystyle\frac{{{{x}}^{{4}}+{2}{{x}}^{{3}}+{4}{{x}}^{{2}}+{8}{x}+{2}}}{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{x}}}={x}+\frac{{2}}{{x}}+\frac{{1}}{{{x}+{1}}}+\frac{{3}}{{{\left({x}+{1}\right)}}^{{2}}}.$

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Thursday, July 16, 2015

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

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