Problem 31 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{x^{3}}{\left(x^{2}+9\right)^{2}}$$

Step-by-Step Solution

Answer

The partial-fraction decomposition is \(\frac{x}{x^2+9} - \frac{9x}{(x^2+9)^2}\).

Step 1: Identify the Type of Partial Fractions

The given rational function \(\frac{x^3}{(x^2+9)^2}\) has a non-factorable quadratic denominator raised to a power. This indicates we will need to use the method of partial fraction decomposition with repeated irreducible quadratics.

Step 2: Set Up the Partial Fraction Decomposition

For a repeated irreducible quadratic \((x^2 + 9)^2\), the partial fraction decomposition form will be\[\frac{x^3}{(x^2+9)^2} = \frac{Ax + B}{x^2 + 9} + \frac{Cx + D}{(x^2 + 9)^2}\]where \(A, B, C,\) and \(D\) are constants to be determined.

Step 3: Clear the Fractions

Multiply both sides of the equation by \((x^2 + 9)^2\) to eliminate the denominators, resulting in:\[ x^3 = (Ax + B)(x^2 + 9) + (Cx + D) \]

Step 4: Expand and Collect Like Terms

Expand the right-hand side:\[ Ax^3 + 9Ax + Bx^2 + 9B + Cx + D \]Combine like terms:\[ Ax^3 + Bx^2 + (9A + C)x + (9B + D) \]

Step 5: Solve for Coefficients

Match the coefficients of corresponding terms on both sides:1. \( A = 1 \) (coefficient of \(x^3\))2. \( B = 0 \) (coefficient of \(x^2\))3. \( 9A + C = 0 \) (coefficient of \(x\))4. \( 9B + D = 0 \) (constant term)Given \( A = 1 \) and \( B = 0 \), solve the equations:\( 9 \cdot 1 + C = 0 \Rightarrow C = -9 \)\( 9 \cdot 0 + D = 0 \Rightarrow D = 0 \).

Step 6: Write the Final Decomposition

Substitute the values of \(A, B, C,\) and \(D\) back into the partial fraction decomposition:\[\frac{x^3}{(x^2+9)^2} = \frac{x}{x^2+9} + \frac{-9x}{(x^2+9)^2}\]

Key Concepts

Rational Functions

Rational functions are at the heart of calculus and algebra, and understanding them is essential for tackling a range of mathematical problems. A rational function is simply a fraction where both the numerator and the denominator are polynomials. In simple terms, it looks like this:

• The numerator is a polynomial expression, such as \(x^3\) in our example.
• The denominator is another polynomial, like \( (x^2 + 9)^2 \).

In the function we consider, \( \frac{x^3}{(x^2+9)^2} \), we identify that the denominator is a squared polynomial, making it unique. Partial fraction decomposition helps break down such complex rational expressions into simpler pieces. This technique is particularly useful for integration or when solving differential equations.

Irreducible Quadratic

An irreducible quadratic in the context of rational functions is a quadratic expression that cannot be factored further using real numbers. Let's understand why the expression \(x^2 + 9\) is considered irreducible:

• A typical quadratic \(ax^2 + bx + c\) is irreducible if its discriminant \(b^2 - 4ac\) is negative.
• For \(x^2 + 9\), the discriminant is \(0^2 - 4(1)(9) = -36\), which is indeed negative.

This means it cannot be expressed as a product of two linear factors with real coefficients. In such cases, especially in partial fraction decomposition, we treat these irreducible quadratics carefully, ensuring they are preserved as whole units or raised to various powers, as seen in this problem.

Solving Coefficients

Solving for the coefficients in a partial fraction decomposition is a key step to arriving at the simpler fractions that make up the original rational function. Let's walk through this process:

• We begin by clearing the fractions. Multiply both sides by the denominator to eliminate it. This helps us focus on the numerators.
• Next, expand the numerators and collect like terms. This requires distributing and combining similar terms, keeping an eye on the corresponding degrees of \(x\).
• To solve for the coefficients, compare each side of the equation. This involves setting the coefficients of like terms from both sides as equal. For example, compare terms with \(x^3, x^2, x\), and constant term separately.

In the given exercise, these comparisons yield equations like \(A = 1\), \(9A + C = 0\), and so on, which we solve to find the values of \(A, B, C,\) and \(D\). Once determined, these coefficients allow us to reconstruct the partial fraction decomposition, simplifying the original function.