Problem 14 - Algebra for College Students

Question

Find the partial fraction decomposition for each rational expression. See answers below. \(\frac{10 x^{2}-73 x+144}{x(x-4)^{2}}\)

Step-by-Step Solution

Answer

\(\frac{9}{x} + \frac{1}{x-4} - \frac{13}{(x-4)^2}\)

Step 1: Identify the Denominator

Analyze the denominator of the rational expression \(x(x-4)^2\). It is composed of the linear factor \(x\) and the repeated linear factor \((x-4)^2\).

Step 2: Set Up the Partial Fraction Decomposition

Set up the decomposition expression for the given rational expression as follows:\[ \frac{A}{x} + \frac{B}{x-4} + \frac{C}{(x-4)^2} \]where \(A\), \(B\), and \(C\) are constants to be determined.

Step 3: Clear the Fraction

Multiply both sides by the original denominator \(x(x-4)^2\) to clear the fractions.\[ 10x^2 - 73x + 144 = A(x-4)^2 + Bx(x-4) + Cx \]

Step 4: Expand and Combine Terms

Expand the right side of the equation:\[ A(x^2 - 8x + 16) + Bx^2 - 4Bx + Cx \]Combine like terms:\[ Ax^2 - 8Ax + 16A + Bx^2 - 4Bx + Cx \]This simplifies to:\[ (A + B)x^2 + (-8A - 4B + C)x + 16A \]

Step 5: Compare Coefficients

Set up equations by comparing coefficients of corresponding powers of \(x\) from both sides:- For \(x^2\): \(A + B = 10\)- For \(x\): \(-8A - 4B + C = -73\)- Constant term: \(16A = 144\)

Step 6: Solve for Constants

Solve for \(A\), \(B\), and \(C\) using the system of equations:1. \(16A = 144\) gives \(A = 9\)2. Substitute \(A = 9\) into \(A + B = 10\) to find \(B = 1\)3. Substitute \(A = 9\) and \(B = 1\) into \(-8A - 4B + C = -73\) to find \(C = -13\).

Step 7: Write the Decomposition

Substitute the values of \(A\), \(B\), and \(C\) back into the partial fractions:\[ \frac{9}{x} + \frac{1}{x-4} - \frac{13}{(x-4)^2} \]

Key Concepts

Rational Expressions

Rational expressions are like fractions, but instead of just having numbers in the numerator and denominator, they include variables as well. Think of them as a way to divide two polynomials. For example, the expression \( \frac{10x^2 - 73x + 144}{x(x-4)^2} \) is a rational expression because it has a polynomial on top and a product of polynomials at the bottom. Why is this important? Rational expressions can be simplified, added, subtracted, multiplied, or divided like regular fractions. However, handling them requires some special techniques, like factoring and reducing, due to the presence of variables. When you're given a rational expression for partial fraction decomposition, the goal is to break it down into simpler fractions. This makes complex integrations or algebraic manipulations more manageable, especially useful in calculus topics.

Linear Factors

In the world of algebra, a linear factor is simply a factor of a polynomial that takes the form of \( (x - a) \), where \( a \) is a constant. Linear factors help in decomposing rational expressions into partial fractions. In our exercise, the denominator is \( x(x-4)^2 \), which consists of the linear factor \( x \) and the repeated linear factor \( (x-4) \). Each of these contributes to setting up our decomposition formula.Let's break it down:

• The factor \( x \) suggests we will have a fraction of the form \( \frac{A}{x} \).
• The factor \( x-4 \), that appears twice, suggests fractions of the form \( \frac{B}{x-4} \) and \( \frac{C}{(x-4)^2} \).

Being able to recognize linear factors is crucial because they guide how we set up the partial fraction decomposition, aiming for each linear factor or power thereof to have its own term in the decomposition.

System of Equations

A system of equations is a collection of two or more equations with common variables. When dealing with partial fraction decomposition, we often end up needing to solve a system of equations to find the constants that make our decomposition valid.In this exercise, once we multiply by the denominator to clear fractions, we expand and combine terms, which allows us to equate coefficients from both sides of the equation. Here's how:

• For the coefficients of \( x^2 \), we set up: \( A + B = 10 \).
• For the coefficients of \( x \), the equation is: \( -8A - 4B + C = -73 \).
• For the constant term (no \( x \)), we have: \( 16A = 144 \).

By solving this system:

• From \( 16A = 144 \), it's straightforward to find \( A = 9 \).
• Substitute \( A \) into \( A + B = 10 \) to find \( B = 1 \).
• Use \( A = 9 \) and \( B = 1 \) in the final equation to solve for \( C = -13 \).

This approach allows us to find precise values for the constants \( A \), \( B \), and \( C \), completing the decomposition process.