Problem 27 - Algebra and Trigonometry

Question

Find the partial fraction decomposition of the rational function. \(\frac{2 x}{4 x^{2}+12 x+9}\)

Step-by-Step Solution

Answer

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

Step 1: Factor the Denominator

The denominator of the fraction \(4x^2 + 12x + 9\) can be factored. Notice that it fits the perfect square trinomial form \((ax + b)^2\). First, recognize that \((2x + 3)^2 = 4x^2 + 12x + 9\). Therefore, factor the denominator as \((2x + 3)^2\).

Step 2: Set up the Partial Fraction Decomposition

The decomposition is based on the form of the factor in the denominator. Since \((2x + 3)^2\) is a repeated linear factor, the decomposition takes the form: \[ \frac{2x}{(2x+3)^2} = \frac{A}{2x+3} + \frac{B}{(2x+3)^2} \] where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply every term by \((2x+3)^2\) to eliminate the denominators: \[ 2x = A(2x+3) + B \]

Step 4: Solve for Constants

Expand and simplify the expression: \[ 2x = 2Ax + 3A + B \] Group terms by powers of \(x\): \[ 2x = (2A)x + (3A + B) \]. For the equation to hold for all \(x\), the coefficients must match, giving us the system: \[ 2A = 2 \] and \[ 3A + B = 0 \].

Step 5: Solve the System of Equations

From the equation \(2A = 2\), we find \(A = 1\). Substitute \(A = 1\) into the second equation: \[ 3(1) + B = 0 \Rightarrow 3 + B = 0 \Rightarrow B = -3 \].

Step 6: Write the Partial Fraction Decomposition

Substitute \(A\) and \(B\) into the decomposition equation: \[ \frac{2x}{(2x+3)^2} = \frac{1}{2x+3} - \frac{3}{(2x+3)^2} \]. This is the partial fraction decomposition of the rational function.

Key Concepts

Rational Functions

Rational functions are expressions that involve a ratio of two polynomials. In mathematical terms, a rational function can be represented as \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \).
For example, in the problem above, \( \frac{2x}{4x^2 + 12x + 9} \) is a rational function. The polynomial in the numerator is \( 2x \), while the polynomial in the denominator is \( 4x^2 + 12x + 9 \). Rational functions are significant in calculus and algebra due to their properties and applications.

• They can display a variety of behaviors such as horizontal asymptotes, vertical asymptotes, and holes, which are key in graphing.
• Partial fraction decomposition is an essential technique used to integrate rational functions in calculus.

Understanding how to work with rational functions is crucial for simplifying complex expressions and solving higher-level math problems.

Factoring Trinomials

Factoring trinomials is an essential skill in algebra that often involves breaking down a polynomial into the product of simpler expressions. For trinomials like \( ax^2 + bx + c \), factoring involves finding two binomials \((mx + n)(px + q)\) that multiply to the original expression.
In the context of the given exercise, the denominator \(4x^2 + 12x + 9\) is factored by recognizing it as a perfect square trinomial. This means it can be expressed as \((2x + 3)^2\).
Recognizing special forms such as the perfect square can make factoring more efficient:

• Check if the first and last terms are perfect squares.
• Check if the middle term is twice the product of the square roots of the first and last terms.

Once a trinomial is factored, solving the problem becomes simpler because you can break it down into more manageable parts.

Linear Factors

Linear factors are polynomials of the first degree, typically in the form \(ax + b\). They play a crucial role in simplifying expressions and solving equations.
In the partial fraction decomposition process, we break rational functions into simpler terms involving linear factors. For instance, in the exercise, the factorized denominator \((2x + 3)^2\) contains the repeated linear factor \((2x + 3)\), allowing us to decompose the fraction into:

• \(\frac{A}{2x+3}\)
• \(\frac{B}{(2x+3)^2}\)

These forms are vital in finding the constants \(A\) and \(B\) that make the partial fraction equivalent to the original rational function. Recognizing and working with linear factors simplifies the process of solving equations and performing further algebraic operations.

System of Equations

A system of equations is a set of two or more equations that you solve simultaneously. In mathematics, solving a system of equations allows you to find values that satisfy all equations at the same time.
During partial fraction decomposition, determining the constants in the decomposed terms involves setting up and solving a system of equations. In this exercise, once the rational function is expressed in partial fractions, we equate the coefficients of similar powers of \(x\) to get:
\[ 2A = 2 \]
\[ 3A + B = 0 \]
We then solve these equations step by step:

• From \(2A = 2\), solve for \(A\).
• Substitute \(A\) in \(3A + B = 0\) to find \(B\).

Solving the system gives \(A = 1\) and \(B = -3\), allowing us to complete the partial fraction decomposition.