Blog: `(3x + 1)/(2x^3 + 3x^2)` Write the partial fraction decomposition of the rational expression.

Friday, March 10, 2017

$\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

$\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$

To decompose this to partial fractions, factor the denominator.

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

Then, write a fraction for each factor. For the repeated factor x, form a partial fraction for each exponent x from 1 to 2. And assign a variable for each numerators.

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

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

$\displaystyle\frac{{{3}{x}+{1}}}{{{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}=\frac{{A}}{{x}}+\ldots$

$\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$

To decompose this to partial fractions, factor the denominator.

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

Then, write a fraction for each factor. For the repeated factor x, form a partial fraction for each exponent x from 1 to 2. And assign a variable for each numerators.

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

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

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

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

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

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

Then, plug-in the roots of the factors.

For the factor x^2, its root is x=0

$\displaystyle{3}\cdot{0}+{1}={A}\cdot{0}{\left({2}\cdot{0}+{3}\right)}+{B}{\left({2}\cdot{0}+{3}\right)}+{C}\cdot{{0}}^{{2}}$

$\displaystyle{1}={3}{B}$

$\displaystyle\frac{{1}}{{3}}={B}$

For the factor (2x + 3), its root is x=-3/2.

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

$\displaystyle-\frac{{7}}{{2}}={A}{\left(-\frac{{3}}{{2}}\right)}{\left({0}\right)}+{B}{\left({0}\right)}+\frac{{9}}{{4}}{C}$

$\displaystyle-\frac{{7}}{{2}}=\frac{{9}}{{4}}{C}$

$\displaystyle-\frac{{14}}{{9}}={C}$

To get the value of A, assign any value to x, and plug-in the values of B and C to:

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

Let x = 1.

$\displaystyle{3}\cdot{1}+{1}={A}\cdot{1}{\left({2}\cdot{1}+{3}\right)}+\frac{{1}}{{3}}{\left({2}\cdot{1}+{3}\right)}+{\left(-\frac{{14}}{{9}}\right)}\cdot{{1}}^{{2}}$

$\displaystyle{4}={5}{A}+\frac{{5}}{{3}}-\frac{{14}}{{9}}$

$\displaystyle{4}={5}{A}+\frac{{1}}{{9}}$

$\displaystyle{4}-\frac{{1}}{{9}}={5}{A}$

$\displaystyle\frac{{35}}{{9}}={5}{A}$

$\displaystyle\frac{{7}}{{9}}={A}$

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

$\displaystyle\frac{{\frac{{7}}{{9}}}}{{x}}+\frac{{\frac{{1}}{{3}}}}{{{x}}^{{2}}}+\frac{{-\frac{{14}}{{9}}}}{{{2}{x}+{3}}}$

This simplifies to:

$\displaystyle\frac{{7}}{{{9}{x}}}+\frac{{1}}{{{3}{{x}}^{{2}}}}-\frac{{14}}{{{9}{\left({2}{x}+{3}\right)}}}$

To check, express them with same denominators.

$\displaystyle\frac{{7}}{{{9}{x}}}+\frac{{1}}{{{3}{{x}}^{{2}}}}-\frac{{14}}{{{9}{\left({2}{x}+{3}\right)}}}=\frac{{7}}{{{9}{x}}}\cdot\frac{{{x}{\left({2}{x}+{3}\right)}}}{{{x}{\left({2}{x}+{3}\right)}}}+\frac{{1}}{{{3}{{x}}^{{2}}}}\cdot\frac{{{3}{\left({2}{x}+{3}\right)}}}{{{3}{\left({2}{x}+{3}\right)}}}-\frac{{14}}{{{9}{\left({2}{x}+{3}\right)}}}\cdot\frac{{{x}}^{{2}}}{{{x}}^{{2}}}$

$\displaystyle=\frac{{{14}{{x}}^{{2}}+{21}{x}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}+\frac{{{6}{x}+{9}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}-\frac{{{14}{{x}}^{{2}}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}$

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

$\displaystyle=\frac{{{14}{{x}}^{{2}}+{21}{x}+{6}{x}+{9}-{14}{{x}}^{{2}}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}=\frac{{{27}{x}+{9}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}=\frac{{{9}{\left({3}{x}+{1}\right)}}}{{{9}{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}=\frac{{{3}{x}+{1}}}{{{{x}}^{{2}}{\left({2}{x}+{3}\right)}}}$

$\displaystyle=\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$

Therefore, $\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}=\frac{{7}}{{{9}{x}}}+\frac{{1}}{{{3}{{x}}^{{2}}}}-\frac{{14}}{{{9}{\left({2}{x}+{3}\right)}}}$ .

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Friday, March 10, 2017

$\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

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

Friday, March 10, 2017

Write the partial fraction decomposition of the rational expression.

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Post a Comment

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

$\displaystyle\frac{{{3}{x}+{1}}}{{{2}{{x}}^{{3}}+{3}{{x}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

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