Problem 24 - Algebra 2 and Trigonometry

Question

Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. \(\frac{x^{2}-7 x+12}{x^{2}+2 x-15}\)

Step-by-Step Solution

Answer

Simplified form: \(\frac{x-4}{x+5}\); undefined at \(x=3\) and \(x=-5\).

Step 1: Factor the numerator

The numerator is \(x^2 - 7x + 12\). We need to factor this quadratic expression. We look for two numbers that multiply to \(12\) and add to \(-7\). These numbers are \(-3\) and \(-4\). Thus, we factor it as \((x-3)(x-4)\).

Step 2: Factor the denominator

The denominator is \(x^2 + 2x - 15\). To factor it, we look for two numbers that multiply to \(-15\) and add to \(2\). These numbers are \(-3\) and \(5\). So, we factor it as \((x-3)(x+5)\).

Step 3: Simplify the expression

The rational expression now looks like \(\frac{(x-3)(x-4)}{(x-3)(x+5)}\). We can simplify by canceling the common factor \((x-3)\) from both the numerator and the denominator. This leaves us with \(\frac{x-4}{x+5}\).

Step 4: Determine undefined values

The expression is undefined anywhere the denominator equals zero. Originally, the denominator was \((x-3)(x+5)\), so we solve \((x-3)(x+5)=0\). This occurs at \(x=3\) and \(x=-5\). Thus, the fraction is undefined at \(x=3\) and \(x=-5\).

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a fundamental skill in algebra that allows us to simplify and solve equations more easily. A quadratic expression usually takes the form \(ax^2 + bx + c\). To factor these, we need to find two numbers that not only multiply to give the constant term \(c\) but also add up to the linear coefficient \(b\). This method is often referred to as "factoring by trial and error".

In the exercise we discussed, the expression \(x^2 - 7x + 12\) was factored by identifying the numbers \(-3\) and \(-4\) because their product is \(12\), and they sum up to \(-7\). Similarly, for the expression \(x^2 + 2x - 15\), the correct pair of numbers is \(-3\) and \(5\), as these multiply to \(-15\) and add up to \(2\).

• First Step: Identify the coefficient of the square term, the linear term, and the constant.
• Second Step: Find two numbers that multiply to the constant term and add up to the coefficient of the linear term.
• Third Step: Rewrite the middle term using the two numbers found and factor by grouping if necessary.

By mastering this technique, you'll be able to handle more complex algebraic expressions with greater ease.

Undefined Values in Rational Expressions

Understanding when a rational expression is undefined is crucial when simplifying these types of expressions. A rational expression is undefined wherever its denominator equals zero since division by zero is not possible. This step is vital not only to simplifying the expression but also to understanding its domain.

In our exercise, we found the original denominator was \((x-3)(x+5)\). To find where the expression is undefined, set each factor in the denominator equal to zero and solve for \(x\).

• The factor \((x-3)\) gives the solution \(x=3\).
• The factor \((x+5)\) gives the solution \(x=-5\).

Therefore, the expression is undefined at \(x=3\) and \(x=-5\). These values are sometimes referred to as "excluded values" because they are not part of the solution set. Identifying undefined values helps define the domain of a rational expression precisely.

Canceling Common Factors

Canceling common factors is a simplification process specific to rational expressions, similar to reducing fractions in arithmetic. After factoring both the numerator and the denominator, we often find common factors; these can be "canceled out" to simplify the expression.

Consider the rational expression \(\frac{(x-3)(x-4)}{(x-3)(x+5)}\). Both the numerator and denominator contain the factor \((x-3)\). This allows us to cancel \((x-3)\) from both, resulting in the simpler desque
ul>How to Cancel Common Factors:

First, completely factor both the numerator and the denominator.

Next, identify any factors that appear in both lists.

Cancel these common factors, rewriting the expression without them.

After canceling, the example expression simplifies to \(\frac{x-4}{x+5}\). Important: Note that while canceling, we must have previously identified any undefined values. Factors canceled were valid in the context of the original expression but need to be excluded from the domain.