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

Friday, April 29, 2016

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

$\displaystyle\frac{{3}}{{{{x}}^{{2}}-{3}{x}}}$

To decompose a fraction, factor the denominator.

$\displaystyle\frac{{3}}{{{x}{\left({x}-{3}\right)}}}$

Then, write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction.

$\displaystyle\frac{{A}}{{x}}$ and $\displaystyle\frac{{B}}{{{x}-{3}}}$

Add these two fractions and set it equal to the given rational expression.

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

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

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

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

Then, plug-in the roots of each factor.

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

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

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

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

$\displaystyle{1}={B}$

For the factor x, its root is x=0.

$\displaystyle{3}={A}{\left({0}-{3}\right)}+{B}{\left({0}\right)}$

$\displaystyle{3}=-{3}{A}$

$\displaystyle\frac{{3}}{{-{3}}}=\frac{{-{3}{A}}}{{-{3}}}$

$\displaystyle-{1}={A}$

So the given rational expression decomposes to:

$\displaystyle-\frac{{1}}{{x}}+\frac{{1}}{{{x}-{3}}}$

This can be re-written as:

$\displaystyle\frac{{1}}{{{x}-{3}}}-\frac{{1}}{{x}}$

To check, express the two fractions with same denominators.

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

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

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