Problem 24 - Algebra for College Students

Question

Give a step-by-step explanation of how to find the partial fraction decomposition of \(\frac{11 x+5}{2 x^{2}+5 x-3}\).

Step-by-Step Solution

Answer

\(\frac{11x+5}{2x^2+5x-3} = \frac{3}{2x-1} + \frac{4}{x+3}\).

Step 1: Factor the Denominator

The first step in finding the partial fraction decomposition is to factor the denominator. The given denominator is \(2x^2 + 5x - 3\). Let's factor this quadratic expression.\[2x^2 + 5x - 3 = (2x - 1)(x + 3)\]Thus, the denominator is factored as \((2x - 1)(x + 3)\).

Step 2: Set Up the Partial Fraction Decomposition

Since the denominator factors into \((2x - 1)(x + 3)\), the partial fraction decomposition will have two terms. We express \(\frac{11x+5}{2x^2+5x-3}\) as:\[\frac{11x+5}{2x^2+5x-3} = \frac{A}{2x-1} + \frac{B}{x+3}\]Here, \(A\) and \(B\) are constants that we need to solve for.

Step 3: Clear the Denominator

To find the values of \(A\) and \(B\), multiply both sides of the equation by the denominator \((2x - 1)(x + 3)\). This gives:\[11x + 5 = A(x + 3) + B(2x - 1)\]Now we have an equation without fractions.

Step 4: Expand and Simplify

Expand the expression on the right-hand side:\[A(x + 3) + B(2x - 1) = Ax + 3A + 2Bx - B\]Combine like terms:\[(A + 2B)x + (3A - B)\]This needs to equal \(11x + 5\).

Step 5: Set Up a System of Equations

Compare coefficients from both sides of the equation for like powers of \(x\):1. For the coefficient of \(x\):\(A + 2B = 11\)2. For the constant term:\(3A - B = 5\)We now have a system of two linear equations.

Step 6: Solve the System of Equations

Solve the equations \(A + 2B = 11\) and \(3A - B = 5\).Multiply the second equation by 2 for easier elimination:\[6A - 2B = 10\]Add to the first equation:\[(A + 2B) + (6A - 2B) = 11 + 10\]\[7A = 21\]Solve for \(A\):\[A = 3\]Substitute \(A = 3\) back into \(A + 2B = 11\):\[3 + 2B = 11\]\[2B = 8\]\[B = 4\]

Step 7: Write the Final Partial Fraction Decomposition

Using the values \(A = 3\) and \(B = 4\), substitute back into the partial fractions:\[\frac{11x+5}{2x^2+5x-3} = \frac{3}{2x-1} + \frac{4}{x+3}\]This expression is the partial fraction decomposition of \(\frac{11x+5}{2x^2+5x-3}\).

Key Concepts

Quadratic Factorization

The first step in partial fraction decomposition often involves factoring the denominator of a rational expression, especially if it is a quadratic. Factoring quadratic expressions is a process where you break them down into simpler linear expressions that can be multiplied together to get the original expression.
In this exercise, the denominator we started with was a quadratic polynomial: \(2x^2 + 5x - 3\).
To factor this polynomial, we looked for two numbers that multiply to \(-6\) (the product of the leading coefficient, \(2\), and the constant term, \(-3\)) and add up to \(5\) (the linear coefficient). The numbers that fit these criteria are \(6\) and \(-1\).

• We rewrite the middle term using these numbers: \(2x^2 + 6x - x - 3\).
• Group terms to find common factors: \(2x(x + 3) - 1(x + 3)\).
• Factor by grouping: \((2x - 1)(x + 3)\).

Thus, the original quadratic is factored as \((2x - 1)(x + 3)\).
Understanding quadratic factoring helps in determining the form of partial fraction decomposition.

Linear Equations

Linear equations are equations of the first degree, meaning they contain variables raised only to the power of one. In the process of partial fraction decomposition, solving linear equations is crucial to finding the constants that make the decomposition valid.
After factoring the quadratic denominator in the exercise into linear factors, we expressed the fraction in terms of these factors with constants \(A\) and \(B\). The decomposition looks like:

• \(\frac{A}{2x-1} + \frac{B}{x+3}\)

Next, we cleared the fraction by multiplying through by the denominator, which set us up with a linear equation in terms of \(x\) and a constant term.
This setup paves the way for constructing a system of linear equations when equating coefficients on both sides of the equation.

Coefficient Comparison

Coefficient comparison is a method used to find values for unknown coefficients in an expression. Once we've cleared the fractions in our partial fraction expression, we end up with an equation that needs to match the original numerator for all values of \(x\). This method involves setting up equations by comparing matching coefficients of powers of \(x\).
In our example:

• The expanded equation from the partial fractions is \((A + 2B)x + (3A - B)\).
• The goal is to match it with \(11x + 5\), the original numerator.

We equate the coefficients of \(x\) and the constant terms:

• For \(x\): \(A + 2B = 11\).
• For constants: \(3A - B = 5\).

This technique of comparing coefficients is key in simplifying the search for appropriate values of \(A\) and \(B\).
It ensures accuracy in breaking down complex expressions into simpler ones.

System of Equations

Once we have expressions for each term in the partial fractions involving unknowns (like \(A\) and \(B\)), solving these expressions involves creating and solving a system of equations.
The setup in partial fractions often results in two or more linear equations which must be solved simultaneously to find the unknown constants.
From this exercise, after equating coefficients, we ended with these equations:

• \(A + 2B = 11\) (from the \(x\) coefficients).
• \(3A - B = 5\) (from the constant terms).

To solve this system, methods such as substitution or elimination can be utilized.
In this example, we multiplied the second equation to make it easier to eliminate \(B\):

• Multiply \(3A - B = 5\) by 2 to get \(6A - 2B = 10\).
• Add the result to \(A + 2B = 11\): \(7A = 21\).
• Solve for \(A\) giving \(A = 3\).
• Substituting \(A = 3\) back into \(A + 2B = 11\) yields \(B = 4\).

Thus, solving the system gives us the needed constants, \(A = 3\) and \(B = 4\).
With these values, our partial fractions become clearly defined.