Blog: `(8x - 12)/(x^2(x^2 + 2)^2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Friday, October 7, 2016

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}=\frac{{A}}{{x}}+\frac{{B}}{{{{x}}^{{2}}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}+{2}}}+\frac{{{E}{x}+{F}}}{{{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$

You need to bring to the same denominator all fractions, such that:

$\displaystyle{8}{x}-{12}={A}{x}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}+{B}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}+{\left({C}{x}+{D}\right)}\cdot{{x}}^{{2}}\cdot{\left({{x}}^{{2}}+{2}\right)}+{\left({E}{x}+{F}\right)}\cdot{{x}}^{{2}}$

$\displaystyle{8}{x}-{12}={A}{x}{\left({{x}}^{{4}}+{4}{{x}}^{{2}}+{4}\right)}+{B}{{x}}^{{4}}+{4}{B}{{x}}^{{2}}+{4}{B}+{\left({C}{{x}}^{{3}}+{D}{{x}}^{{2}}\right)}\cdot{\left({{x}}^{{2}}+{2}\right)}+{E}{{x}}^{{3}}+{F}{{x}}^{{2}}$

$\displaystyle{8}{x}-{12}={A}{{x}}^{{5}}+{4}{A}{{x}}^{{3}}\ldots$

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

You need to bring to the same denominator all fractions, such that:

$\displaystyle{8}{x}-{12}={A}{{x}}^{{5}}+{4}{A}{{x}}^{{3}}+{4}{A}{x}+{B}{{x}}^{{4}}+{4}{B}{{x}}^{{2}}+{4}{B}+{C}{{x}}^{{5}}+{2}{C}{{x}}^{{3}}+{D}{{x}}^{{4}}+{2}{D}{{x}}^{{2}}+{E}{{x}}^{{3}}+{F}{{x}}^{{2}}$

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

$\displaystyle{8}{x}-{12}={{x}}^{{5}}{\left({A}+{C}\right)}+{{x}}^{{4}}{\left({B}+{D}\right)}+{{x}}^{{3}}{\left({4}{A}+{2}{C}+{E}\right)}+{{x}}^{{2}}{\left({4}{B}+{2}{D}+{F}\right)}+{x}{\left({4}{A}\right)}++{4}{B}$

Comparing the expressions both sides yields:

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

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

$\displaystyle{4}{A}+{2}{C}+{E}={0}\Rightarrow{2}{A}+{E}={0}\Rightarrow{E}=-{2}{A}$

$\displaystyle{4}{B}+{2}{D}+{F}={0}\Rightarrow{F}=-{2}{D}$

$\displaystyle{4}{A}={8}\Rightarrow{A}={2}\Rightarrow{C}=-{2}\Rightarrow{E}=-{4}$

$\displaystyle{4}{B}=-{12}\Rightarrow{B}=-{3}\Rightarrow{D}={3}\Rightarrow{F}=-{6}$

Hence, decomposing the fraction, yields $\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}=\frac{{2}}{{x}}-\frac{{3}}{{{{x}}^{{2}}}}+\frac{{-{2}{x}+{3}}}{{{{x}}^{{2}}+{2}}}+\frac{{-{4}{x}-{6}}}{{{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}.$

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Friday, October 7, 2016

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Friday, October 7, 2016

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\frac{{{8}{x}-{12}}}{{{{x}}^{{2}}{{\left({{x}}^{{2}}+{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?