Problem 23 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x}{8 x^{2}-10 x+3}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{3/2}{4x - 3} - \frac{1/2}{2x - 1}\).

Step 1: Factor the Denominator

First, we need to factor the denominator of the rational function. We have the expression \(8x^2 - 10x + 3\). We'll factor this quadratic expression:\[8x^2 - 10x + 3 = (4x - 3)(2x - 1)\].Thus, the factors of the denominator are \((4x - 3)\) and \((2x - 1)\).

Step 2: Set Up Partial Fraction Decomposition

Since the rational function is \(\frac{x}{8x^2 - 10x + 3}\) and we have factored the denominator into two linear factors, we can express it as the sum of two fractions:\[ \frac{x}{(4x - 3)(2x - 1)} = \frac{A}{4x - 3} + \frac{B}{2x - 1}\]where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Next, multiply both sides by the original denominator \((4x - 3)(2x - 1)\) to clear the fractions:\[ x = A(2x - 1) + B(4x - 3) \].This gives us a polynomial equation that is true for all \(x\).

Step 4: Collect Like Terms

Expand and collect like terms to simplify:\[ x = (2A + 4B)x - A - 3B \].We then compare coefficients on both sides of the equation.

Step 5: Solve for the Coefficients

Set up a system of equations by comparing coefficients:- Coefficient of \(x\): \(2A + 4B = 1\)- Constant term: \(-A - 3B = 0\)Solve this system of equations. From the second equation, we get \(A = -3B\). Substitute this into the first equation:\[2(-3B) + 4B = 1\]\[-6B + 4B = 1\]\[-2B = 1 \Rightarrow B = -\frac{1}{2}\].Then, \(A = -3(-\frac{1}{2}) = \frac{3}{2}\).

Step 6: Write the Decomposed Fractions

With \(A = \frac{3}{2}\) and \(B = -\frac{1}{2}\), write the partial fraction decomposition:\[ \frac{x}{8x^2 - 10x + 3} = \frac{3/2}{4x - 3} - \frac{1/2}{2x - 1} \].

Key Concepts

Rational Function

A rational function is a fraction involving polynomials in the numerator and the denominator. For example, given a rational function like \( \frac{x}{8x^2 - 10x + 3} \), the numerator is \(x\), and the denominator is a quadratic polynomial, \(8x^2 - 10x + 3\). Rational functions are fundamental in algebra and calculus because they form the backbone for understanding various mathematical behaviors and applications.

Key Characteristics of Rational Functions:

• They can depict vertical asymptotes, dictated by the roots of the denominator, when it equals zero.
• The degree of the numerator and denominator can tell us about the function's horizontal asymptotes.
• These functions can be simplified or decomposed using various algebraic techniques, like partial fraction decomposition.

Understanding rational functions is essential for applications in areas such as optimization and calculus, where they frequently appear.

Factoring Quadratics

Factoring quadratic expressions is crucial when dealing with rational functions. The process involves rewriting the quadratic in a product of simpler expressions. Here, considering the example \(8x^2 - 10x + 3\), it can be factored as \((4x - 3)(2x - 1)\).

How to Factor Quadratics:

• Identify and apply the quadratic formula to find potential roots, if simple factoring is not obvious.
• Use techniques like completing the square, or others such as the factoring by grouping method, which involve splitting the middle term to assist in making a factorable expression.

Factoring is essential because it transforms the quadratic into a product of two linear factors. This makes it manageable for tasks such as finding roots or simplifying expressions, paving the way for partial fraction decomposition.

Linear Factors

Once a polynomial is simplified into its linear components, these are known as linear factors. For the quadratic \(8x^2 - 10x + 3\) mentioned earlier, the linear factors are \((4x - 3)\) and \((2x - 1)\). Linear factors are integral to the process of decomposition, where each factor represents a potential term in the decomposed form of the rational function.

Understanding Linear Factors:

• Each factor represents a root of the polynomial when set equal to zero.
• They make working with polynomial equations much more straightforward, helping transform higher-degree equations into simpler, linear ones.
• They are utilized in partial fraction decomposition to set up the initial decomposition expression, allowing algebraic techniques to deduce unknown coefficients.

Turning polynomials into linear factors is a critical simplification step that underpins much of higher algebra and calculus.

System of Equations

To complete the partial fraction decomposition, solving a system of equations is necessary to find constants that make the decomposition valid. In the example given, the expression \(x = A(2x - 1) + B(4x - 3)\) is expanded and yields a system of linear equations when compared to \(x\). These equations determine the coefficients \(A\) and \(B\).

Steps to Solve a System of Equations:

• Expand the equation and collect like terms, looking at both sides of the equality to set up equations based on coefficients of corresponding terms.
• Use substitution or elimination methods to find the unknown constants. Here, substitution is used to solve for \(B\) and then \(A\).
• Ultimately, rewrite the rational function in its decomposed form using these values.

Solving the system of equations correctly ensures that the decomposed form accurately represents the original rational function over its defined domain.