Question

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

Short answer

The partial fraction decomposition is \[ \frac{2}{7(3x-2)} + \frac{1}{7(2x+1)} \].

Step-by-step solution

Set up the Partial Fraction Decomposition

For the rational expression \(\frac{x}{(3x-2)(2x+1)}\), assume it can be decomposed into partial fractions of the form \[ \frac{x}{(3x-2)(2x+1)} = \frac{A}{3x-2} + \frac{B}{2x+1} \] where \(A\) and \(B\) are constants to be determined.

Multiply Both Sides by the Denominator

Multiply both sides of the equation by \((3x-2)(2x+1)\) to clear the fractions: \[ x = A(2x+1) + B(3x-2) \]

Expand and Simplify

Expand the right-hand side: \[ x = 2Ax + A + 3Bx - 2B \] Combine like terms: \[ x = (2A + 3B)x + (A - 2B) \]

Equate Coefficients

For the identity to hold for all \(x\), the coefficients of like terms on both sides of the equation must be equal. Therefore, we get two equations: \[ \begin{cases} 2A + 3B = 1 \ A - 2B = 0 \end{cases} \]

Solve the System of Equations

Solve the system of equations. First solve for \(A\) in terms of \(B\) from the second equation: \[ A = 2B \] Substitute this into the first equation: \[ 2(2B) + 3B = 1 \] Simplify and solve for \(B\): \[ 4B + 3B = 1 \ 7B = 1 \ B = \frac{1}{7} \] Now substitute back to find \(A\): \[ A = 2 \times \frac{1}{7} = \frac{2}{7} \]

Write the Partial Fraction Decomposition

Substitute \(A\) and \(B\) back into the partial fractions: \[ \frac{x}{(3x-2)(2x+1)} = \frac{\frac{2}{7}}{3x-2} + \frac{\frac{1}{7}}{2x+1} \] which simplifies to: \[ \frac{x}{(3x-2)(2x+1)} = \frac{2}{7(3x-2)} + \frac{1}{7(2x+1)} \]

Key concepts

Rational Expression

A rational expression is a fraction where both the numerator and the denominator are polynomials. For example, in the expression \(\frac{x}{(3x-2)(2x+1)}\), the numerator is \(x\) and the denominator is the product of two binomials \((3x-2) \text{ and } (2x+1)\).
Rational expressions can often be simplified or decomposed using methods like partial fraction decomposition. This technique makes it easier to perform calculus operations, such as integration, on the function.
By breaking down a complex fraction into simpler fractions, we can focus on less complicated pieces, making calculations more manageable.

System of Equations

When you decompose a rational expression into partial fractions, you often need to solve for unknown constants. This usually involves setting up and solving a system of equations.
In the given problem, we assume the decomposition form \(\frac{x}{(3x-2)(2x+1)} = \frac{A}{3x-2} + \frac{B}{2x+1}\).
Multiplying both sides by the denominator clears the fractions and gives an equation involving both \(A\) and \(B\):

• \(x = A(2x+1) + B(3x-2)\)

Expanding and combining like terms yield another equation in \(A\) and \(B\):

• \(x = (2A + 3B)x + (A - 2B)\)

For this equation to be true for all \(x\), the coefficients of corresponding terms on both sides must be equal. This gives us a system of linear equations:

• \(2A + 3B = 1\)
• \(A - 2B = 0\)

Solving this system involves techniques such as substitution or elimination, helping us find the values of \(A\) and \(B\).

Coefficients

Coefficients are the numerical factors multiplying the variables in a polynomial term. Understanding how to work with coefficients is crucial when solving partial fraction decompositions.
In the decomposition problem, we end up with an equation involving terms like \(A(2x+1)\) and \(B(3x-2)\). Expanding these terms results in coefficients attached to the variable \(x\) and to the constant terms. For instance,

• Expanding \(A(2x+1)\) gives \(2Ax + A\)
• Expanding \(B(3x-2)\) gives \(3Bx - 2B\)

When we combine like terms, we get:

• \(x = (2A + 3B)x + (A - 2B)\)

For this to hold true for all \(x\), the coefficients on both sides for the same powers of \(x\) must be equal. This yields:

• \((2A + 3B)x = x \rightarrow 2A + 3B = 1\)
• \((A - 2B) = 0\)

These coefficients help us set up a system of linear equations to solve for \(A\) and \(B\). Understanding how to manipulate and equate these coefficients is vital to finding the correct values.