Question

Find the partial fraction decomposition of each rational expression. $$ \frac{1}{(2 x+3)(4 x-1)} $$

Short answer

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

Step-by-step solution

- Write the Partial Fraction Form

Express the given fraction as a sum of partial fractions. For the given rational expression \[ \frac{1}{(2x+3)(4x-1)} \], assume it can be written as the sum of two fractions: \[ \frac{A}{2x+3} + \frac{B}{4x-1} \]

- Form the Equation

Combine the fractions on the right-hand side to form a single fraction with the common denominator: \[ \frac{A(4x-1) + B(2x+3)}{(2x+3)(4x-1)} = \frac{1}{(2x+3)(4x-1)} \]. Since the denominators are the same, equate the numerators: \[ A(4x-1) + B(2x+3) = 1 \]

- Solve for A and B

Expand the equation: \[ 4Ax - A + 2Bx + 3B = 1 \]. Combine like terms: \[ (4A + 2B)x + (-A + 3B) = 1 \]. For this to hold true for all x, the coefficients of x and the constant term must be equal to the corresponding coefficients on the right-hand side. Therefore, equate: \[ 4A + 2B = 0 \] \[ -A + 3B = 1 \]

- Solve the System of Equations

Solve the system of linear equations: \[ 4A + 2B = 0 \] and \[ -A + 3B = 1 \]. First, solve for A from the first equation: \[ 4A = -2B \] implying \[ A = -\frac{1}{2}B \]. Plug this value into the second equation: \[ -\frac{1}{2}B + 3B = 1 \] simplifies to \[ \frac{5}{2}B = 1 \] thus, \[ B = \frac{2}{5} \]. Substituting back to find A: \[ A = -\frac{1}{2} \times \frac{2}{5} = -\frac{1}{5} \]

- Write the Partial Fraction Decomposition

Now that A and B are known, substitute them back into the partial fractions: \[ \frac{1}{(2x+3)(4x-1)} = \frac{A}{2x+3} + \frac{B}{4x-1} \], \[-\frac{1}{5(2x+3)} + \frac{2}{5(4x-1)} \]

Key concepts

Rational Expressions

A rational expression is a fraction where both the numerator and the denominator are polynomials. It's like a combination of algebraic expressions and fractions. For example, in our given problem, \(\frac{1}{(2 x+3)(4 x-1)}\) is a rational expression. The key characteristics of rational expressions include:

- The numerator and denominator must be polynomials.
- The denominator cannot be zero, as division by zero is undefined.

Rational expressions are often simplified or decomposed into partial fractions to make them easier to work with, especially in calculus and algebra.

System of Equations

When solving for \(A\) and \(B\) in partial fraction decomposition, we often end up with a system of linear equations. This happens because we equate the coefficients of like terms to solve for unknowns. For instance, in the exercise:
\[A(4x-1) + B(2x+3) = 1\]
Expanding and collecting terms gives us two equations:
\[4A + 2B = 0\] and \[-A + 3B = 1\]

- One way to solve this system is using substitution.
- First, solve one equation for one variable, then substitute it into the other equation.

This method ensures we find the correct values for our coefficients.

Linear Algebra

Partial fraction decomposition and systems of equations draw heavily upon concepts from linear algebra. Linear algebra deals with vectors, matrices, and linear transformations. In our context, the system of linear equations can be viewed as a matrix equation. The system:
\( \begin{align*} 4A + 2B &= 0 \ -A + 3B &= 1 \end{align*} \)
can be represented in matrix form as:
\( \begin{pmatrix} 4 & 2 \ -1 & 3 \end{pmatrix} \begin{pmatrix}A \ B \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \)
Solving matrix equations involves finding the inverse or using various methods such as Gaussian elimination. Understanding linear algebra helps in visualizing and solving these problems efficiently.