Blog: `(x + 6)/(x^3 - 3x^2 - 4x + 12)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Monday, November 28, 2016

$\displaystyle\frac{{{x}+{6}}}{{{{x}}^{{3}}-{3}{{x}}^{{2}}-{4}{x}+{12}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{x}+{6}}}{{{{x}}^{{3}}-{3}{{x}}^{{2}}-{4}{x}+{12}}}$

To decompose this to partial fractions, factor the denominator.

$\displaystyle{{x}}^{{3}}-{3}{{x}}^{{2}}-{4}{x}+{12}$

$\displaystyle={\left({{x}}^{{3}}-{3}{{x}}^{{2}}\right)}+{\left(-{4}{x}+{12}\right)}$

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

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

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

Then, write a fraction for each factor. Since the numerators are still unknown, assign a variable for each numerator.

$\displaystyle\frac{{A}}{{{x}-{3}}}$ , $\displaystyle\frac{{B}}{{{x}-{2}}}$ and $\displaystyle\frac{{C}}{{{x}+{2}}}$

Add these three fractions and set it equal to the given fraction.

$\displaystyle\frac{{{x}+{6}}}{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}=\frac{{A}}{{{x}-{3}}}+\frac{{B}}{{{x}-{2}}}+\frac{{C}}{{{x}+{2}}}$

To solve for the values of A, B and C, eliminate the fractions in the equation. So, multiply both sides by the LCD.

$\displaystyle{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}\cdot\frac{{{x}+{6}}}{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}={\left(\frac{{A}}{{{x}-{3}}}+\frac{{B}}{{{x}-{2}}}+\frac{{C}}{{{x}+{2}}}\right)}\cdot{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}$

$\displaystyle{x}+{6}={A}{\left({x}-{2}\right)}{\left({x}+{2}\right)}+{B}{\left({x}-{3}\right)}{\left({x}+{2}\right)}+{C}{\left({x}-{3}\right)}{\left({x}-{2}\right)}$

Then, plug-in the roots of each factor.

For the factor (x-2), its root is x=2.

$\displaystyle{2}+{6}={A}{\left({2}-{2}\right)}{\left({2}+{2}\right)}+{B}{\left({2}-{3}\right)}{\left({2}+{2}\right)}+{C}{\left({2}-{3}\right)}{\left({2}-{2}\right)}$

$\displaystyle{8}={A}{\left({0}\right)}{\left({4}\right)}+{B}{\left(-{1}\right)}{\left({4}\right)}+{C}{\left(-{1}\right)}{\left({0}\right)}$

$\displaystyle{8}=-{4}{B}$

$\displaystyle\frac{{8}}{{-{4}}}=\frac{{-{4}{B}}}{{-{4}}}$

$\displaystyle-{2}={B}$

For the factor (x + 2), its root is x=-2.

$\displaystyle-{2}+{6}={A}{\left(-{2}-{2}\right)}{\left(-{2}+{2}\right)}+{B}{\left(-{2}-{3}\right)}{\left(-{2}+{2}\right)}+{C}{\left(-{2}-{3}\right)}{\left(-{2}-{2}\right)}$

$\displaystyle{4}={A}{\left(-{4}\right)}{\left({0}\right)}+{B}{\left(-{5}\right)}{\left({0}\right)}+{C}{\left(-{5}\right)}{\left(-{4}\right)}$

$\displaystyle{4}={20}{C}$

$\displaystyle\frac{{4}}{{20}}=\frac{{{20}{C}}}{{20}}$

$\displaystyle\frac{{1}}{{5}}={C}$

And for the factor (x-3), its root is x=3.

$\displaystyle{3}+{6}={A}{\left({3}-{2}\right)}{\left({3}+{2}\right)}+{B}{\left({3}-{3}\right)}{\left({3}+{2}\right)}+{C}{\left({3}-{3}\right)}{\left({3}-{2}\right)}$

$\displaystyle{9}={A}{\left({1}\right)}{\left({5}\right)}+{B}{\left({0}\right)}{\left({5}\right)}+{C}{\left({0}\right)}{\left({1}\right)}$

$\displaystyle{9}={5}{A}$

$\displaystyle\frac{{9}}{{5}}=\frac{{{5}{A}}}{{5}}$

$\displaystyle\frac{{9}}{{5}}={A}$

So the partial fraction decomposition of the given rational expression is:

$\displaystyle\frac{{\frac{{9}}{{5}}}}{{{x}-{3}}}+\frac{{-{2}}}{{{x}-{2}}}+\frac{{\frac{{1}}{{5}}}}{{{x}+{2}}}$

And this simplifies to:

$\displaystyle=\frac{{9}}{{{5}{\left({x}-{3}\right)}}}-\frac{{2}}{{{x}-{2}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}$

To check, express the three fractions with same denominators.

$\displaystyle\frac{{9}}{{{5}{\left({x}-{3}\right)}}}-\frac{{2}}{{{x}-{2}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{9}}{{{5}{\left({x}-{3}\right)}}}\cdot\frac{{{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}{{{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}-\frac{{2}}{{{x}-{2}}}\cdot\frac{{{5}{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}\cdot\frac{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}}}{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}}}$

$\displaystyle=\frac{{{9}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}-\frac{{{10}{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}+\frac{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{9}{\left({{x}}^{{2}}-{4}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}-\frac{{{10}{\left({{x}}^{{2}}-{x}-{6}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}+\frac{{{{x}}^{{2}}-{5}{x}+{6}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

Now that they have same denominators, proceed to add/subtract them.

$\displaystyle=\frac{{{9}{\left({{x}}^{{2}}-{4}\right)}-{10}{\left({{x}}^{{2}}-{x}-{6}\right)}+{{x}}^{{2}}-{5}{x}+{6}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{9}{{x}}^{{2}}-{36}-{10}{{x}}^{{2}}+{10}{x}+{60}+{{x}}^{{2}}-{5}{x}+{6}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{5}{x}+{30}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{5}{\left({x}+{6}\right)}}}{{{5}{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

$\displaystyle=\frac{{{x}+{6}}}{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}$

Therefore, $\displaystyle\frac{{{x}+{6}}}{{{\left({x}-{3}\right)}{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}=\frac{{9}}{{{5}{\left({x}-{3}\right)}}}-\frac{{2}}{{{x}-{2}}}+\frac{{1}}{{{5}{\left({x}+{2}\right)}}}$ .

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Monday, November 28, 2016

$\displaystyle\frac{{{x}+{6}}}{{{{x}}^{{3}}-{3}{{x}}^{{2}}-{4}{x}+{12}}}$ 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"?

Monday, November 28, 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;?

$\displaystyle\frac{{{x}+{6}}}{{{{x}}^{{3}}-{3}{{x}}^{{2}}-{4}{x}+{12}}}$ 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"?