Problem 15 - College Algebra

Question

Find the decomposition of the partial fraction for the nonrepeating linear factors. \(\frac{6 x}{x^{2}-4}\)

Step-by-Step Solution

Answer

The partial fraction decomposition of \(\frac{6x}{x^2 - 4}\) is \(\frac{3}{x-2} + \frac{3}{x+2}\).

Step 1: Factor the Denominator

The first step is to factor the denominator of the given fraction. The denominator is \(x^2 - 4\), which is a difference of squares. This can be factored into \((x - 2)(x + 2)\). This gives us the fraction \(\frac{6x}{(x - 2)(x + 2)}\).

Step 2: Set Up the Partial Fraction Decomposition

We express \(\frac{6x}{(x - 2)(x + 2)}\) as the sum of two fractions: \(\frac{A}{x-2} + \frac{B}{x+2}\). Here, \(A\) and \(B\) are constants to be determined.

Step 3: Create a Common Denominator

To combine the two fractions \(\frac{A}{x-2} + \frac{B}{x+2}\) into a single expression, we find a common denominator: \((x - 2)(x + 2)\). This results in the equation \(\frac{A(x+2) + B(x-2)}{(x-2)(x+2)}\).

Step 4: Equate Numerators

Since the denominators are already equal, equate the numerators: \(A(x+2) + B(x-2) = 6x\). This expands to \(Ax + 2A + Bx - 2B = 6x\).

Step 5: Simplify and Collect Like Terms

Combine the terms: \((A + B)x + (2A - 2B) = 6x + 0\). By comparing coefficients, solve for \(A\) and \(B\).

Step 6: Solve for Constants A and B

By comparing coefficients of \(x\), we have: \(A + B = 6\). From the constant terms, \(2A - 2B = 0\), which simplifies to \(A = B\). Substitute \(A = B\) into \(A + B = 6\), yielding \(2A = 6\), so \(A = 3\) and \(B = 3\).

Step 7: Write the Final Decomposition

The partial fraction decomposition is then \(\frac{6x}{(x-2)(x+2)} = \frac{3}{x-2} + \frac{3}{x+2}\).

Key Concepts

Linear Factors

In partial fraction decomposition, understanding linear factors is crucial. Linear factors are expressions of the form \(ax + b\) where \(a\) and \(b\) are constants. These factors are essential for breaking down complex fractions into simpler parts. For example, when examining the fraction

• \(\frac{6x}{x^2 - 4}\),

it's helpful to recognize that \(x^2 - 4\) can be factored into

• \((x-2)(x+2)\).

This step converts the expression into one with nonrepeating linear factors. Each of these factors can have a fraction associated with it, and our aim is to express the original fraction as a sum of fractions with these factors in the denominator.

Difference of Squares

Understanding the difference of squares is a handy tool in algebra, especially for factoring quadratic expressions. A difference of squares is when you have two perfect squares subtracted from each other, such as

• \(x^2 - 4\).

This particular expression is a difference of squares because it can be rewritten as

• \((x)^2 - (2)^2\).

The formula for factoring a difference of squares is

• \((a^2 - b^2) = (a - b)(a + b)\).

Applying this to our example,

• \((x^2 - 4)\)

becomes

• \((x-2)(x+2)\).

Recognizing this allows us to set up the partial fraction decomposition by splitting the original fraction into parts that are easier to manipulate.

Equating Numerators

Once the partial fraction decomposition setup is in place, equating numerators becomes a vital step. We take the common denominator of the individual fractions and write their numerators equal to each other. In our case, we expressed

• \(\frac{6x}{(x-2)(x+2)}\)

as

• \(\frac{A}{x-2} + \frac{B}{x+2}\).

Multiplying each side by the common denominator \((x-2)(x+2)\), we get the equation for the numerators:

• \(A(x+2) + B(x-2) = 6x\).

This step doesn't involve the denominators anymore and focuses purely on the numerators. By expanding and simplifying, we equate corresponding coefficients from both sides of the equation.

Coefficient Comparison

In the final stages of solving partial faction decomposition, coefficient comparison allows us to find the values of unknown constants. From the equation

• \(A(x+2) + B(x-2) = 6x\),

we expand to get

• \(Ax + 2A + Bx - 2B\).

This combines to

• \((A + B)x + (2A - 2B) = 6x + 0\).

By comparing coefficients of the like terms on each side:

• For the \(x\) term: \(A + B = 6\)
• For the constant terms: \(2A - 2B = 0\)

Solving these equations, we find that \(A = B\). Substituting makes it simple to solve for the exact values, \(A = 3\) and \(B = 3\). This method of comparing each term by their coefficients is a finite and structured way to solve for the unknowns in partial fraction decomposition.