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

Wednesday, March 4, 2015

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

Decompose the denominator:

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

Therefore the fraction decomposition has the form

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{x}-\sqrt{{{2}}}}}+\frac{{C}}{{{x}+\sqrt{{{2}}}}}.$

To find A,B and C multiply both sides by the original denominator:

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

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

Thus $\displaystyle{A}+{B}+{C}={0},$ $\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1}$ and $\displaystyle-{2}{A}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0}.$

$\displaystyle{A}=-{\left({B}+{C}\right)},$

$\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1},$

$\displaystyle{2}{\left({B}+{C}\right)}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0},$ or $\displaystyle{B}{\left({2}-\sqrt{{{2}}}\right)}+{C}{\left({2}+\sqrt{{{2}}}\right)}={0},$ or $\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}+{C}{\left(\sqrt{{{2}}}+{1}\right)}={0}.$

Add and subtract these two equations and obtain

$\displaystyle{2}{B}{\left(\sqrt{{{2}}}-{1}\right)}={1},$ or $\displaystyle{B}=\frac{{1}}{{{2}{\left(\sqrt{{{2}}}-{1}\right)}}}=\frac{{\sqrt{{{2}}}+{1}}}{{2}}$ and

$\displaystyle{2}{C}{\left(\sqrt{{{2}}}+{1}\right)}=-{1},$ or $\displaystyle{C}=-\frac{{1}}{{{2}{\left(\sqrt{{{2}}}+{1}\right)}}}=-\frac{{\sqrt{{{2}}}-{1}}}{{2}}.$

And $\displaystyle{A}=-{\left({B}+{C}\right)}=-{1}.$

Now check this result:

$\displaystyle-\frac{{1}}{{{x}-{1}}}+\frac{{\sqrt{{{2}}}+{1}}}{{2}}\frac{{1}}{{{x}-\sqrt{{{2}}}}}-\frac{{\sqrt{{{2}}}-{1}}}{{2}}\frac{{1}}{{{x}+\sqrt{{{2}}}}}=$

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

$\displaystyle=-\frac{{1}}{{{x}-{1}}}+\frac{{1}}{{2}}\frac{{{2}{x}+{4}}}{{{{x}}^{{2}}-{2}}}=-\frac{{1}}{{{x}-{1}}}+\frac{{{x}+{2}}}{{{{x}}^{{2}}-{2}}}=$

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

which is correct.

Decompose the denominator:

Therefore the fraction decomposition has the form

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{x}-\sqrt{{{2}}}}}+\frac{{C}}{{{x}+\sqrt{{{2}}}}}.$

To find A,B and C multiply both sides by the original denominator:

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

Thus $\displaystyle{A}+{B}+{C}={0},$ $\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1}$ and $\displaystyle-{2}{A}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0}.$

$\displaystyle{A}=-{\left({B}+{C}\right)},$

$\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}-{C}{\left(\sqrt{{{2}}}+{1}\right)}={1},$

$\displaystyle{2}{\left({B}+{C}\right)}-{B}\sqrt{{{2}}}+{C}\sqrt{{{2}}}={0},$ or
$\displaystyle{B}{\left({2}-\sqrt{{{2}}}\right)}+{C}{\left({2}+\sqrt{{{2}}}\right)}={0},$ or
$\displaystyle{B}{\left(\sqrt{{{2}}}-{1}\right)}+{C}{\left(\sqrt{{{2}}}+{1}\right)}={0}.$

Add and subtract these two equations and obtain

$\displaystyle{2}{B}{\left(\sqrt{{{2}}}-{1}\right)}={1},$ or $\displaystyle{B}=\frac{{1}}{{{2}{\left(\sqrt{{{2}}}-{1}\right)}}}=\frac{{\sqrt{{{2}}}+{1}}}{{2}}$ and

$\displaystyle{2}{C}{\left(\sqrt{{{2}}}+{1}\right)}=-{1},$ or $\displaystyle{C}=-\frac{{1}}{{{2}{\left(\sqrt{{{2}}}+{1}\right)}}}=-\frac{{\sqrt{{{2}}}-{1}}}{{2}}.$

And $\displaystyle{A}=-{\left({B}+{C}\right)}=-{1}.$

Now check this result:

$\displaystyle-\frac{{1}}{{{x}-{1}}}+\frac{{\sqrt{{{2}}}+{1}}}{{2}}\frac{{1}}{{{x}-\sqrt{{{2}}}}}-\frac{{\sqrt{{{2}}}-{1}}}{{2}}\frac{{1}}{{{x}+\sqrt{{{2}}}}}=$

$\displaystyle=-\frac{{1}}{{{x}-{1}}}+\frac{{1}}{{2}}\frac{{{2}{x}+{4}}}{{{{x}}^{{2}}-{2}}}=-\frac{{1}}{{{x}-{1}}}+\frac{{{x}+{2}}}{{{{x}}^{{2}}-{2}}}=$

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

which is correct.

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Wednesday, March 4, 2015

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}$ 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"?

Wednesday, March 4, 2015

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;?

$\displaystyle\frac{{x}}{{{{x}}^{{3}}-{{x}}^{{2}}-{2}{x}+{2}}}$ 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"?