Problem 9 - College Algebra

Question

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

Step-by-Step Solution

Answer

The partial fraction decomposition of \( x/(x-3)(x-2) \) is \( 3/(x - 3) + 2/(x - 2) \)

Step 1 - Set up the partial fraction decomposition

Start by expressing the given fraction as a sum of two simpler fractions. Given \( \frac{x}{(x-3)(x-2)} \), express it as: \( \frac{x}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2} \) where A and B are constants we need to solve.

Step 2 - Clear the denominators

Multiply every term by the least common multiple of the denominators to clear the denominators. The equation becomes: \( x = A(x-2) + B(x-3) \)

Step 3 - Set up the system of equations

Now, choose suitable values for x to make coefficients disappear and solve for A and B. By setting \( x = 2 \) and \( x = 3 \), we get two equations: \( 2 = A(0) + B(1) => B = 2 \) and \( 3 = A(1) + B(0) => A = 3 \)

Step 4 - Substitute A and B back into the expression

Lastly, substitute A and B back into the expression \( \frac{A}{x-3} + \frac{B}{x-2} \). Thus, the partial fraction decomposition of \( x/(x-3)(x-2) \) is \( 3/(x - 3) + 2/(x - 2) \)

Key Concepts

Rational Expression

A rational expression is basically a fraction where both the numerator and the denominator are polynomials. The importance of understanding rational expressions lies in the vast applications they have in algebra. For example, in the given problem, \(\frac{x}{(x-3)(x-2)}\), is a rational expression. It involves a polynomial in the numerator, \(x\), and one in the denominator, \((x-3)(x-2)\).
Rational expressions often need to be simplified to solve equations easily or for integration purposes in calculus. Simplifying a rational expression might involve factorization or breaking down the expression into a sum of simpler fractions—this process is known as partial fraction decomposition. One of the major goals of partial fraction decomposition is to express a complex expression like the one given into simpler, more manageable pieces.

System of Equations

A system of equations consists of multiple equations that share some common variables, which we aim to solve simultaneously. In the process of partial fraction decomposition, a system of equations is set up to determine the constants that simplify the fraction. For the expression \(\frac{x}{(x-3)(x-2)}\), after assuming its breakdown into simpler fractions, we have:\[\frac{x}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}\]This equation involves two unknown constants, \(A\) and \(B\).
By clearing the denominators, we transform the problem into a system of linear equations:

• \(x = A(x-2) + B(x-3)\)

Selecting suitable values for \(x\), such as \(x = 2\) and \(x = 3\), helps us isolate terms to solve for \(A\) and \(B\). Thus, we end up with:

• \(2 = B\)
• \(3 = A\)

Solving this system of equations allows us to find the values of the constants, \(A\) and \(B\), which are key to simplifying the original rational expression.

Constants

Constants are values that remain unchanged throughout the scope of the problem. They are crucial when dealing with partial fraction decomposition because they define the simplified version of the original rational expression. In our context, constants \(A\) and \(B\) are employed to represent parts of the simplified fractions.
Their determination is necessary to complete the decomposition. First, by creating a new equation by multiplying through by the common denominator, we are able to eliminate the fractions:\[x = A(x-2) + B(x-3)\]By choosing particular values for \(x\), we simplify this equation to isolate \(A\) and \(B\).

• For \(x=2\): the term related to \(A\) vanishes, allowing us to directly find \(B = 2\).
• For \(x=3\): the term with \(B\) becomes zero, letting us solve for \(A = 3\).

Eventually, these constants are substituted back into the partial fraction expression, resulting in the final decomposition:\(\frac{3}{x-3} + \frac{2}{x-2}\). This clearly structures the problem into manageable components, highlighting the utility of constants in mathematical decomposition.