Problem 42 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{2 x^{2}-x+8}{\left(x^{2}+4\right)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{2}{x^2 + 4} - \frac{x}{(x^2 + 4)^2}.\)

Step 1: Identify the Denominator Type

The denominator of the given function is \((x^2 + 4)^2\), which is a repeated irreducible quadratic factor.

Step 2: Set Up the Partial Fraction Decomposition

For a repeated irreducible quadratic factor \((x^2 + 4)^2\), the partial fraction decomposition takes the form:\[\frac{2x^2 - x + 8}{(x^2 + 4)^2} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2}.\] We need to find the constants \(A\), \(B\), \(C\), and \(D\).

Step 3: Clear the Fractions

Multiply both sides of the equation by \((x^2 + 4)^2\) to eliminate the denominators:\[2x^2 - x + 8 = (Ax + B)(x^2 + 4) + (Cx + D).\]

Step 4: Expand and Collect Like Terms

Expand the right side of the equation: \((Ax + B)(x^2 + 4) = Ax^3 + 4Ax + Bx^2 + 4B.\) Adding \((Cx + D)\) gives the complete expression: \(Ax^3 + Bx^2 + (4A + C)x + (4B + D).\)

Step 5: Equate Coefficients

By equating coefficients from both sides of the equation, we get:- Coefficient of \(x^3\): \(A = 0\).- Coefficient of \(x^2\): \(B = 2\).- Coefficient of \(x\): \(4A + C = -1\; \Rightarrow 0 + C = -1\; \Rightarrow C = -1\).- Constant term: \(4B + D = 8\; \Rightarrow 8 + D = 8\; \Rightarrow D = 0\).

Step 6: Write the Final Decomposition

Substitute the values found for \(A, B, C,\) and \(D\) into the decomposition:\[\frac{2x^2 - x + 8}{(x^2 + 4)^2} = \frac{2}{x^2 + 4} - \frac{x}{(x^2 + 4)^2}.\]

Key Concepts

Rational Functions

A rational function is simply a quotient of two polynomials. In other words, you'll encounter one polynomial divided by another. In many cases, you can think of them as fractions but involving variables. Rational functions come in handy whenever you need to represent situations with division of quantities. Key features include:

• A numerator and a denominator which are polynomials.
• Possible restrictions on the variable, derived from the denominator, as division by zero is undefined.

In our exercise, the rational function is \(\frac{2x^2 - x + 8}{(x^2 + 4)^2}.\)Notice that the numerator is `2x^2 - x + 8` and the denominator is `(x^2 + 4)^2`.The partial fraction decomposition process involves expressing such complex rational functions into simpler fractions, making analysis and integration easier.

Irreducible Quadratic Factor

An irreducible quadratic factor refers to a polynomial of degree 2 that cannot be factored into real or rational components. It remains in its quadratic form, especially when considering complex numbers still cannot break it down further in real terms.In the exercise, \(x^2 + 4\) is an irreducible quadratic factor. This means it does not break down into linear factors over the real numbers. Instead, we maintain it as it is when setting up partial fractions.When dealing with repeated irreducible quadratic factors, as seen in \((x^2 + 4)^2\), we generate terms by creating separate fractions for each power of the quadratic factor, just like we did to \(\frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2}\). Every term contributes to forming the full partial fraction decomposition.

Equating Coefficients

Equating coefficients is a method for finding unknowns in algebraic expressions. Simply put, it's comparing corresponding parts (or coefficients) of polynomials.After multiplying and expanding expressions as per the partial fractions set up, we equate the coefficients of like terms on both sides of the equation. During this step, terms with the same power of \(x\) are set equal, allowing us to solve for unknown variables, which are typically constants in the decomposition.For instance, in our task, after equating terms from both sides, we determined:

• \(A = 0\)
• \(B = 2\)
• \(C = -1\)
• \(D = 0\)

This solving technique brings structure, as analyzing each coefficient helps in simplifying and classifying the required outputs.

Polynomial Long Division

Polynomial long division is a systematic method to simplify rational expressions or divide one polynomial by another. Although not directly used in the exercise given, it provides valuable insights beneficial before performing decompositions. The procedure for polynomial long division follows steps similar to numerical long division:

• Divide the leading term of the dividend by the leading term of the divisor
• Multiply the entire divisor by this result
• Subtract from the original dividend
• Bring down the next term and repeat

Understanding this process allows students to recognize when partial fraction is necessary as it determines when further breakdown or decomposition of complex rational functions is required. Mastering division aids in managing complex fraction structures effectively, mitigating confusion down the line.