Blog: `(4x^2 + 3)/(x - 5)^3` Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

Tuesday, April 14, 2015

$\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}$ Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

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

To decompose this fraction, consider the factor in the denominator. The factor is repeated three times. So the partial fraction of this will have denominator in increasing exponent of the factor.

$\displaystyle\frac{{A}}{{{x}-{5}}}$ , $\displaystyle\frac{{B}}{{{\left({x}-{5}\right)}}^{{2}}}$ , and $\displaystyle\frac{{C}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

$\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}=\frac{{A}}{{{x}-{5}}}+\frac{{B}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{C}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

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

To decompose this fraction, consider the factor in the denominator. The factor is repeated three times. So the partial fraction of this will have denominator in increasing exponent of the factor.

$\displaystyle\frac{{A}}{{{x}-{5}}}$ , $\displaystyle\frac{{B}}{{{\left({x}-{5}\right)}}^{{2}}}$ , and $\displaystyle\frac{{C}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

$\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}=\frac{{A}}{{{x}-{5}}}+\frac{{B}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{C}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

$\displaystyle{{\left({x}-{5}\right)}}^{{3}}\cdot\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}={\left(\frac{{A}}{{{x}-{5}}}+\frac{{B}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{C}}{{{\left({x}-{5}\right)}}^{{3}}}\right)}\cdot{{\left({x}-{5}\right)}}^{{3}}$

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

$\displaystyle{4}{{x}}^{{2}}+{3}={A}{\left({{x}}^{{2}}-{10}{x}+{25}\right)}+{B}{\left({x}-{5}\right)}+{C}$

Then, group together the terms with x^2, the terms with x and the constants.

$\displaystyle{4}{{x}}^{{2}}+{3}={A}{{x}}^{{2}}-{10}{A}{x}+{25}{A}+{B}{x}-{5}{B}+{C}$

$\displaystyle{4}{{x}}^{{2}}+{3}={A}{{x}}^{{2}}+{\left(-{10}{A}+{B}\right)}{x}+{\left({25}{A}-{5}{B}+{C}\right)}$

Set the coefficient of x^2 at the left side equal to the coefficient of x^2 at the right side.

$\displaystyle{4}={A}$ This is the value of A.

Then, set the coefficient of x at the left side equal to the coefficient of x at the right side.

$\displaystyle{0}=-{10}{A}+{B}$ (Let this be EQ1.)

And, set the constant at the left side equal to the constant at the right side.

$\displaystyle{3}={25}{A}-{5}{B}+{C}$ (Let this be EQ2.)

To get the value of B, plug-in A=4 to EQ1.

$\displaystyle{0}=-{10}{A}+{B}$

$\displaystyle{0}=-{10}{\left({4}\right)}+{B}$

$\displaystyle{0}=-{40}+{B}$

$\displaystyle{40}={B}$

And to get the value of C, plug-in A=4 and B=40 to EQ2.

$\displaystyle{3}={25}{A}-{5}{B}+{C}$

$\displaystyle{3}={25}{\left({4}\right)}-{5}{\left({40}\right)}+{C}$

$\displaystyle{3}={100}-{200}+{C}$

$\displaystyle{3}=-{100}+{C}$

$\displaystyle{3}+{100}={C}$

$\displaystyle{103}={C}$

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

$\displaystyle\frac{{4}}{{{x}-{5}}}+\frac{{40}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{103}}{{{\left({x}-{5}\right)}}^{{3}}}$

To check, express the fractions with same denominators.

$\displaystyle\frac{{4}}{{{x}-{5}}}+\frac{{40}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{103}}{{{\left({x}-{5}\right)}}^{{3}}}$

$\displaystyle=\frac{{4}}{{{x}-{5}}}\cdot\frac{{{\left({x}-{5}\right)}}^{{2}}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{40}}{{{\left({x}-{5}\right)}}^{{2}}}\cdot\frac{{{x}-{5}}}{{{x}-{5}}}+\frac{{103}}{{{\left({x}-{5}\right)}}^{{3}}}$

$\displaystyle=\frac{{{4}{\left({{x}}^{{2}}-{10}{x}+{25}\right)}}}{{{\left({x}-{5}\right)}}^{{3}}}+\frac{{{40}{\left({x}-{5}\right)}}}{{{\left({x}-{5}\right)}}^{{3}}}{\)}+\frac{{103}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

$\displaystyle=\frac{{{4}{\left({{x}}^{{2}}-{10}{x}+{25}\right)}+{40}{\left({x}-{5}\right)}+{103}}}{{{\left({x}-{5}\right)}}^{{3}}}$

$\displaystyle=\frac{{{4}{{x}}^{{2}}-{40}{x}+{100}+{40}{x}-{200}+{103}}}{{{\left({x}-{5}\right)}}^{{3}}}$

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

Therefore, $\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}=\frac{{4}}{{{x}-{5}}}+\frac{{40}}{{{\left({x}-{5}\right)}}^{{2}}}+\frac{{103}}{{{\left({x}-{5}\right)}}^{{3}}}$ .

Blog

Tuesday, April 14, 2015

$\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}$ Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

No comments:

Post a Comment

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

Tuesday, April 14, 2015

Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

No comments:

Post a Comment

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

$\displaystyle\frac{{{4}{{x}}^{{2}}+{3}}}{{{\left({x}-{5}\right)}}^{{3}}}$ Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

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