Partial Fraction Decomposition of Rational Expressions
Partial fraction decomposition is a fundamental algebraic technique used to rewrite a complex rational expression as a sum of simpler fractions. This method is particularly useful in areas like integral calculus, where integrating complex rational functions is simplified by breaking them down into their basic components. When a rational expression has a denominator that can be factored into distinct linear factors, it can be expressed as a sum of terms, each with one of these factors as its denominator and a constant in its numerator.
The problem asks us to decompose the rational expression given by:
\[\frac{X}{(X+2)(X-3)}\]
Decomposition Process for the Given Expression
To perform the partial fraction decomposition for \(\frac{X}{(X+2)(X-3)}\), we follow a clear, step-by-step process:
- Setting Up the Partial Fraction Form:
The denominator consists of two distinct linear factors: \((X+2)\) and \((X-3)\). Therefore, we can set up the partial fraction decomposition as follows, introducing unknown constants \(A\) and \(B\):
\[\frac{X}{(X+2)(X-3)} = \frac{A}{X+2} + \frac{B}{X-3}\]
Our goal is to find the values of \(A\) and \(B\). - Clearing the Denominators:
To eliminate the denominators, we multiply both sides of the equation by the least common multiple of the denominators, which is \((X+2)(X-3)\):
\[(X+2)(X-3) \times \frac{X}{(X+2)(X-3)} = (X+2)(X-3) \times \left(\frac{A}{X+2} + \frac{B}{X-3}\right)\]
This simplifies to:
\[X = A(X-3) + B(X+2)\]
This equation is an identity, meaning it holds true for all valid values of \(X\). - Solving for Constants \(A\) and \(B\):
We can find the values of \(A\) and \(B\) by strategically choosing values for \(X\) that will simplify the equation.- To find \(A\), set \(X = -2\):
Substitute \(X = -2\) into the equation \(X = A(X-3) + B(X+2)\):
\[-2 = A(-2-3) + B(-2+2)\] \[-2 = A(-5) + B(0)\] \[-2 = -5A\]
Now, solve for \(A\):
\[A = \frac{-2}{-5} = \frac{2}{5}\] - To find \(B\), set \(X = 3\):
Substitute \(X = 3\) into the equation \(X = A(X-3) + B(X+2)\):
\[3 = A(3-3) + B(3+2)\] \[3 = A(0) + B(5)\] \[3 = 5B\]
Now, solve for \(B\):
\[B = \frac{3}{5}\]
- To find \(A\), set \(X = -2\):
- Substituting \(A\) and \(B\) back into the Partial Fraction Form:
With the values of \(A = \frac{2}{5}\) and \(B = \frac{3}{5}\) found, we substitute them back into our initial partial fraction setup:
\[\frac{X}{(X+2)(X-3)} = \frac{\frac{2}{5}}{X+2} + \frac{\frac{3}{5}}{X-3}\]
This can be rewritten by moving the denominators \(5\) down into the main denominator of each term:
\[\frac{X}{(X+2)(X-3)} = \frac{2}{5(X+2)} + \frac{3}{5(X-3)}\]
Summary of Partial Fraction Decomposition Steps
The partial fraction decomposition of the expression \(\frac{X}{(X+2)(X-3)}\) is \(\frac{2}{5(X+2)} + \frac{3}{5(X-3)}\). This result provides a simpler form of the original rational expression.
| Step | Description | Mathematical Expression |
|---|---|---|
| 1 | Set up the decomposition form | \[\frac{X}{(X+2)(X-3)} = \frac{A}{X+2} + \frac{B}{X-3}\] |
| 2 | Clear denominators | \[X = A(X-3) + B(X+2)\] |
| 3 | Solve for \(A\) (using \(X = -2\)) | \[-2 = A(-5) \implies A = \frac{2}{5}\] |
| 4 | Solve for \(B\) (using \(X = 3\)) | \[3 = B(5) \implies B = \frac{3}{5}\] |
| 5 | Substitute \(A\) and \(B\) | \[\frac{X}{(X+2)(X-3)} = \frac{2}{5(X+2)} + \frac{3}{5(X-3)}\] |
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