Problem 25 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{4 x^{2}-32 x+72}{(x+1)(x-5)^{2}}$$

Step-by-Step Solution

Answer

The partial-fraction decomposition is \( \frac{3}{x+1} + \frac{1}{x-5} + \frac{6}{(x-5)^{2}} \).

Step 1: Set Up the Partial Fractions

The given rational function \( \frac{4x^{2} - 32x + 72}{(x+1)(x-5)^{2}} \) can be decomposed into the sum of fractions where the denominators are the factors of the original denominator. Here, \( \frac{4x^{2} - 32x + 72}{(x+1)(x-5)^{2}} = \frac{A}{x+1} + \frac{B}{x-5} + \frac{C}{(x-5)^{2}} \). This setup accounts for the linear factor \((x+1)\) with a single term and the repeated quadratic factor \((x-5)\) with one term for each power (\(x-5\) and \((x-5)^{2}\)).

Step 2: Multiply Through by the Denominator

Clear the fractions by multiplying both sides of the equation by the common denominator, \((x+1)(x-5)^{2}\), \[ 4x^{2} - 32x + 72 = A(x-5)^{2} + B(x+1)(x-5) + C(x+1) \].This removes the denominators, allowing us to solve for the constants \(A\), \(B\), and \(C\).

Step 3: Expand and Simplify the Right Side

Expand the terms on the right side of the equation:- \(A(x-5)^{2} = A(x^2 - 10x + 25)\)- \(B(x+1)(x-5) = B(x^2 - 4x - 5)\)- \(C(x+1) = Cx + C\).Combine these expansions to get a single polynomial equation:\[ 4x^{2} - 32x + 72 = Ax^2 - 10Ax + 25A + Bx^2 - 4Bx - 5B + Cx + C \].Combine like terms to simplify.

Step 4: Equate Coefficients

Equate the coefficients from both sides of the equation:- For \(x^2\): \(A + B = 4\)- For \(x\): \(-10A - 4B + C = -32\)- Constant term: \(25A - 5B + C = 72\).These provide a system of equations to solve for \(A\), \(B\), and \(C\).

Step 5: Solve the System of Equations

Solve the system of equations:1. \(A + B = 4\)2. \(-10A - 4B + C = -32\)3. \(25A - 5B + C = 72\).Start by isolating two variables in terms of one, and substitute to solve:- From (1), \(B = 4 - A\).Substitute \(B = 4 - A\) into equations (2) and (3) to solve for \(A\) and \(C\). Then find \(B\):- Solve simultaneously for the values: \(A = 3\), \(B = 1\), \(C = 6\).

Step 6: Write the Partial Fraction Decomposition

Place the solved values of \(A\), \(B\), and \(C\) back into the partial fraction form:\[ \frac{3}{x+1} + \frac{1}{x-5} + \frac{6}{(x-5)^{2}} \].Thus, the partial-fraction decomposition is complete.

Key Concepts

Rational Functions

A rational function is one of the fundamental building blocks in algebra. It is defined as a function that can be expressed as the quotient of two polynomials. More concretely, it takes the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial expressions and \( Q(x) \) is not equal to zero. This ensures the function is properly defined everywhere except where \( Q(x) = 0 \).
In the original exercise, the given rational function is \( \frac{4x^2 - 32x + 72}{(x+1)(x-5)^2} \). Here, the numerator, \( 4x^2 - 32x + 72 \), is a polynomial of degree 2, and the denominator, \( (x+1)(x-5)^2 \), is a product of linear factors raised to their respective powers.

Understanding rational functions involves identifying their components and behavior, such as:

• Zeros: These occur where the numerator is zero and indicate where the function crosses the x-axis.
• Vertical asymptotes: Found where the denominator equals zero, indicating points where the function goes to infinity.
• Degree and simplification: Often, the highest degree of the polynomial in the numerator relative to the denominator suggests the end behavior or limits of the function as \( x \to \pm\infty \).

This knowledge facilitates the decomposition process as seen in problems like partial fraction decomposition.

System of Equations

A system of equations is a set of two or more equations with a common set of unknowns. Solving such systems is crucial in problems like partial fraction decomposition where unknown coefficients need to be determined. In solving systems of equations, the most common methods are substitution, elimination, or using matrices.

In the given solution, the partial fraction decomposition involved setting up equations based on the equality of polynomial coefficients. For the equations:

• \( A + B = 4 \)
• \(-10A - 4B + C = -32 \)
• \( 25A - 5B + C = 72 \)

The solution process involves:

• Substituting one equation into the others to eliminate one variable.
• Using the resulting equations to further isolate and solve for the remaining variables.
• Simplifying to find the values of \( A, B, \) and \( C \).

This systematic approach is valuable because it breaks down complex problems into simpler, solvable steps. Each step builds upon the previous one, leading to a solution that satisfies all conditions.

Polynomial Expansion

Polynomial expansion is the process of expanding expressions that involve polynomials, such as multiplying out binomials or applying formulas like the distributive property. This technique is key in simplifying expressions and preparing them for comparison, as was necessary in the original exercise.

The main focus areas when expanding include:

• Distributive Property: Apply this property to expand an expression, \((a+b)(c+d) = ac + ad + bc + bd\).
• Combining Like Terms: After expansion, combine terms that have the same degree to simplify the expression.
• Being Careful with Signs: Properly accounting for positive and negative signs during distribution and combining.

In the exercise's solution, expansion helped rewrite the terms like \( A(x-5)^2 \) and \( B(x+1)(x-5) \) in a form that allowed for easy coefficient comparison:

• \( A(x-5)^2 = A(x^2 - 10x + 25) \)
• \( B(x+1)(x-5) = B(x^2 - 4x - 5) \)

Through careful expansion and simplification, the original rational expression could be matched on both sides of the equation, making solving for \( A, B, \) and \( C \) possible. Understanding polynomial expansion is crucial for topics where rewriting and comparison of algebraic expressions is necessary.