Problem 23 - Algebra and Trigonometry

Question

\(13-44=\) 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 factor the quadratic in the denominator: \(8x^2 - 10x + 3\). We look for two numbers that multiply to \(8 \, \times\, 3 = 24\) and add up to \(-10\). Those numbers are \(-4\) and \(-6\). So, we rewrite the middle term as \(-4x - 6x\) and factor by grouping: \[8x^2 - 4x - 6x + 3 = 4x(2x - 1) - 3(2x - 1) = (4x - 3)(2x - 1)\]

Step 2: Set up Partial Fractions

Write the partial fraction decomposition format for the rational function using the linear factors from the denominator:\[\frac{x}{(4x - 3)(2x - 1)} = \frac{A}{4x - 3} + \frac{B}{2x - 1}\]where \(A\) and \(B\) are constants that we need to determine.

Step 3: Solve for Constants

To find \(A\) and \(B\), multiply through by the common denominator \((4x - 3)(2x - 1)\):\[x = A(2x - 1) + B(4x - 3)\]Expand and combine like terms:\[x = (2A + 4B)x - (A + 3B)\]Equate the coefficients of like terms: 1. \(2A + 4B = 1\)2. \(-(A + 3B) = 0\)From the second equation, \(A + 3B = 0\). Solve these equations to find \(A\) and \(B\).

Step 4: Solve the System of Equations

Solve the system to find the values of \(A\) and \(B\):1. \(A + 3B = 0\)2. \(2A + 4B = 1\)From Equation 1: \(A = -3B\).Substitute \(A = -3B\) in Equation 2:\[2(-3B) + 4B = 1\]\[-6B + 4B = 1\]\[-2B = 1\]\[B = -\frac{1}{2}\]Substitute \(B\) back into \(A = -3B\):\[A = -3\left(-\frac{1}{2}\right) = \frac{3}{2}\]

Step 5: Write the Decomposed Fractions

Now that we've found \(A\) and \(B\), substitute them back into the partial fractions:\[\frac{x}{8x^2 - 10x + 3} = \frac{\frac{3}{2}}{4x - 3} + \frac{-\frac{1}{2}}{2x - 1}\]This is the partial fraction decomposition of the function.

Key Concepts

Factorization

Factorization is a crucial step in the process of partial fraction decomposition. It involves expressing a polynomial as a product of its simpler factors, making it easier to work with.
For the problem at hand, we begin by factoring the quadratic polynomial in the denominator:

• The given polynomial is \(8x^2 - 10x + 3\).
• We need to break it into parts that multiply to give the original polynomial.

To factor, identify two numbers that multiply to \(8 \times 3 = 24\) and add to \(-10\). These numbers are \(-4\) and \(-6\).
The middle term \(-10x\) is rewritten as \(-4x - 6x\) to form logical groupings: \[8x^2 - 4x - 6x + 3\]
By grouping and factoring further, we arrive at: \[4x(2x - 1) - 3(2x - 1) = (4x - 3)(2x - 1)\].
This factorization simplifies the rational function and sets the stage for partial fraction decomposition.

Rational Function

A rational function is an important concept in algebra and calculus. It is a fraction that has a polynomial in both its numerator and denominator.
In our exercise, the rational function is given by: \[\frac{x}{8x^2 - 10x + 3}\]
We aim to decompose it into simpler fractions. Doing this allows us to perform easier calculus operations like integration or to better understand the behavior of the function.

• The key here is recognizing that the denominator factors affect the decomposition structure.

Once the denominator is factored, set up partial fractions based on each factor: \[\frac{A}{4x - 3} + \frac{B}{2x - 1}\]
Constants \(A\) and \(B\) are unknowns we need to solve for, which will complete the decomposition.

System of Equations

In partial fraction decomposition, once the denominators are set, solving for the numerical values involves forming a system of equations. This is where algebraic manipulation is put into practice.
Our system is derived from equating the two expressions:

• Equality is maintained through multiplying and rearranging terms.

From \[x = A(2x - 1) + B(4x - 3)\]
Expanding gives: \[(2A + 4B)x - (A + 3B)\]
This leads to two manageable equations by equating coefficients of powers of \(x\): 1. \(2A + 4B = 1\) 2. \(-(A + 3B) = 0\)
Solving step-by-step:

• From the second equation, derive \(A = -3B\).
• Substitute into the first equation to find \(B = -\frac{1}{2}\).
• Back-solve for \(A\) using \(A = -3B\) to find \(A = \frac{3}{2}\).

These values satisfy the equations, completing the process of solving for the coefficients in the partial fraction decomposition.