Problem 7 - College Algebra

Question

Find the decomposition of the partial fraction for the nonrepeating linear factors. \(\frac{3 x-79}{x^{2}-5 x-24}\)

Step-by-Step Solution

Answer

The decomposition is \(\frac{-5}{x - 8} + \frac{8}{x + 3}\).

Step 1: Factor the Denominator

First, we will factor the quadratic expression in the denominator, \(x^2 - 5x - 24\). We look for two numbers that multiply to \(-24\) and add to \(-5\). These numbers are \(-8\) and \(3\). So, we can factor the quadratic as \((x - 8)(x + 3)\).

Step 2: Set Up the Partial Fraction Decomposition

Now that we have factored the denominator into \((x - 8)(x + 3)\), we set up the partial fraction decomposition as follows: \(\frac{3x - 79}{x^2 - 5x - 24} = \frac{A}{x - 8} + \frac{B}{x + 3}\).

Step 3: Multiply Through by the Denominator

Next, we multiply each term in the equation by \((x - 8)(x + 3)\) to clear the denominators. This gives us: \(3x - 79 = A(x + 3) + B(x - 8)\).

Step 4: Expand and Simplify

Expand the right side: \(A(x + 3) = Ax + 3A\) and \(B(x - 8) = Bx - 8B\). Substituting back into the equation: \(3x - 79 = Ax + 3A + Bx - 8B\). Combine like terms: \(3x - 79 = (A + B)x + (3A - 8B)\).

Step 5: Solve for Constants A and B

By comparing coefficients, we have two equations: \(A + B = 3\) and \(3A - 8B = -79\). Solve these equations to find \(A\) and \(B\). Solving the first equation for \(A\), we have \(A = 3 - B\). Substitute this into the second equation: \(3(3 - B) - 8B = -79\), which simplifies to \(9 - 3B - 8B = -79\), or \(9 - 11B = -79\). Solving for \(B\), we get \(B = 8\). Substitute \(B = 8\) back into \(A = 3 - B\) to obtain \(A = -5\).

Step 6: Write the Final Decomposition

Thus, the partial fraction decomposition of the given expression is \(\frac{3x - 79}{x^2 - 5x - 24} = \frac{-5}{x - 8} + \frac{8}{x + 3}\).

Key Concepts

Quadratic Factorization

Quadratic factorization is the process of breaking down a quadratic equation into simpler components. This is essential for problems that involve fraction decomposition. In our problem, the denominator is the quadratic expression \(x^2 - 5x - 24\). We need to factor this expression to proceed.

The goal is to find two numbers that multiply to give the constant term, \(-24\), and add up to the middle term coefficient, \(-5\). If you consider the factors of \(-24\), you'll find that \(-8\) and \(3\) meet these conditions:

• \(-8 \times 3 = -24\)
• \(-8 + 3 = -5\)

So, the quadratic expression can be factored as \((x - 8)(x + 3)\). Now it's expressed in terms of linear factors, making it easier to handle in further steps.

Linear Factors

Linear factors are expressions of the form \((x - a)\), where \(a\) is a constant. In partial fraction decomposition, any quadratic or higher polynomial in the denominator must first be broken down into linear factors.

For our problem, once the denominator \(x^2 - 5x - 24\) is factored into \((x - 8)(x + 3)\), each linear factor can be used in setting up the partial fraction decomposition. Here's how it looks:

• \(\frac{3x - 79}{(x - 8)(x + 3)} = \frac{A}{x - 8} + \frac{B}{x + 3}\)

Every linear factor in the denominator contributes a term to the decomposition.
This method is grounded on the idea that certain fractions can be expressed as a sum or difference of simpler fractions, each associated with a linear factor from the denominator.

Coefficient Comparison

Coefficient comparison is a technique used to solve for unknowns in an equation by comparing the coefficients of the same powers of a variable. After setting up the partial fraction, we equated it to its decomposition and eliminated denominators by multiplying through.

This resulted in:
\(3x - 79 = A(x + 3) + B(x - 8)\).
By expanding the right side, you get:
\(Ax + 3A + Bx - 8B\).
Combining like terms gives:
\((A + B)x + (3A - 8B)\).
From this expression, equate the coefficients of \(x\) and constant terms separately:

• \(A + B = 3\) which matches the coefficient of \(x\).
• \(3A - 8B = -79\) which matches the constant term.

Solving these equations gives us the values of \(A\) and \(B\), which are crucial for the final setup of the partial fraction decomposition.

Solving Systems of Equations

Solving systems of equations involves finding the values of variables that satisfy all equations in the system simultaneously. In our problem, with equations \(A + B = 3\) and \(3A - 8B = -79\), we have a system of linear equations.

To solve:

• Start with the simpler equation, \(A + B = 3\).
• Solve for one variable first. Here, we isolated \(A\): \(A = 3 - B\).

Then, substitute \(A = 3 - B\) into the second equation:
\(3(3 - B) - 8B = -79\).
Simplify and solve for \(B\):

• \(9 - 3B - 8B = -79\)
• Combine \(-3B\) and \(-8B\) to get \(-11B\)
• \(9 - 11B = -79\) leads to \(B = 8\).

Substitute \(B\) back into \(A = 3 - B\) to find \(A = -5\). This method ensures we accurately distribute values that satisfy both equations, finalizing the decomposition process.