- Factor the denominator: \(x^2 - 2x - 15 = (x + 3)(x - 5)\).
- Set up the partial fraction decomposition: \(\frac{2x+3}{(x+3)(x-5)} = \frac{A}{x+3} + \frac{B}{x-5}\).
- Solve for the constants \(A = \frac{3}{8}\) and \(B = \frac{13}{8}\) using substitution.
- State the final sum of the two rational expressions: \(\boxed{\frac{3}{8(x+3)} + \frac{13}{8(x-5)}}\).
Explanation
- Factoring the Denominator
We are given the rational expression \(\frac{2x+3}{x^2-2x-15}\) and we need to represent it as the sum of two simpler rational expressions. First, we factor the denominator of the given expression, \(x^2 - 2x - 15\). We look for two numbers that multiply to \(-15\) and add to \(-2\). Those numbers are \(-5\) and \(3\). Thus, \(x^2 - 2x - 15 = (x - 5)(x + 3)\). The expression becomes:
\[\frac{2x+3}{(x-5)(x+3)}\]
- Setting up Partial Fraction Decomposition
To decompose this into simpler rational expressions, we use the method of partial fractions. We set the expression equal to the sum of two terms:
\[\frac{2x+3}{(x+3)(x-5)} = \frac{A}{x+3} + \frac{B}{x-5}\]
where \(A\) and \(B\) are constants to be determined. To find \(A\) and \(B\), we multiply both sides by the denominator \((x+3)(x-5)\):
\[2x+3 = A(x-5) + B(x+3)\]
- Solving for Constants A and B
We use strategic substitution to solve for \(A\) and \(B\).
First, let \(x = 5\):
\[2(5)+3 = A(5-5) + B(5+3)\]
\[13 = 8B \implies B = \frac{13}{8}\]
Next, let \(x = -3\):
\[2(-3)+3 = A(-3-5) + B(-3+3)\]
\[-3 = -8A \implies A = \frac{3}{8}\]
Substituting these values back into our decomposition, we get the two rational expressions:
\[\frac{3/8}{x+3} \quad \text{and} \quad \frac{13/8}{x-5}\]
- Final Conclusion
The original rational expression can be expressed as the sum of two rational expressions:
\[\frac{2x+3}{x^2-2x-15} = \frac{3}{8(x+3)} + \frac{13}{8(x-5)}\]
This provides the decomposition into simplified rational terms as required. The final answer is:
\[\boxed{\frac{3}{8(x+3)} + \frac{13}{8(x-5)}}\]
Examples
Partial fraction decomposition is essential in integral calculus for integrating complex rational functions by breaking them into simpler, easier-to-handle components. It also has practical applications in control systems engineering, where it is used to analyze the transfer functions of linear time-invariant systems by identifying their individual poles and residues. Understanding how to split complex fractions allows engineers and mathematicians to simplify the behavior of a system or function into manageable pieces.