Problem 18 - 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. \(\frac{8 a b-4 b^{2}}{6 a b}\)

Step-by-Step Solution

Answer

Simplified: \(\frac{2a-b}{3a}\), undefined for \(a = 0\) or \(b = 0\).

Step 1: Factor the Numerator and Denominator

First, factor both the numerator and the denominator. The numerator is \(8ab - 4b^2\), which can be factored as \(4b(2a - b)\). The denominator is \(6ab\), which can be rewritten as \(2 \cdot 3ab\).

Step 2: Simplify the Expression

After factoring, the expression becomes \(\frac{4b(2a-b)}{3 \cdot 2ab}\). Cancel out the common factors of \(4b\) and \(2b\) in the numerator and the denominator. Now, simplify the expression: \(\frac{2a - b}{3a}\).

Step 3: Identify the Undefined Values

The fraction is undefined when the denominator equals zero. Set the denominator, \(6ab\), equal to zero and solve for \(a\) and \(b\). In this case, \(3a\) can't be zero, which implies \(a eq 0\). Also, since the entire denominator contained \(b\), \(b eq 0\) as well.

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial step in simplifying rational expressions. Similar to numbers, polynomials can be broken down into simpler components called factors. To factor a polynomial, look for common factors or special patterns like the difference of squares or a trinomial square.

• In our expression, the numerator was originally \(8ab - 4b^2\). By observing, we notice that both terms have a common factor of \(4b\), allowing us to factor it out and rewrite it as \(4b(2a - b)\).
• The denominator, \(6ab\), can also be broken down by factoring the number 6 into \(2 \cdot 3\), leading to the expression \(2 \cdot 3ab\).

By factoring both parts, we set the stage for easier simplification by identifying components that can be canceled out. This skill is useful in manipulating and simplifying expressions in algebra.

Undefined Fractions

In mathematics, a fraction is considered undefined when its denominator equals zero, as division by zero is impossible. It’s important to identify these cases to correctly simplify rational expressions.

• For example, in our expression \(\frac{8ab - 4b^2}{6ab}\), the denominator \(6ab\) must not be zero.
• If \(6ab = 0\), this would mean either \(a = 0\) or \(b = 0\) would make the expression undefined, as both would nullify the denominator.

By determining these undefined conditions, we can specify the values of variables that are valid for the expression. In this exercise, the fraction is undefined when \(a = 0\) or \(b = 0\), highlighting that both \(a\) and \(b\) must not be zero.

Rational Expressions

A rational expression is simply the ratio of two polynomials. Simplifying these expressions involves similar steps to simplifying numeric fractions: factor and cancel common terms.

• In our example, the rational expression \(\frac{8ab - 4b^2}{6ab}\) represents the ratio of the polynomial on the numerator to the polynomial in the denominator.
• After factoring, common factors in both the numerator and the denominator can be canceled to simplify the expression further. Here it became \(\frac{2a - b}{3a}\).

Working with rational expressions involves careful factoring and attention to the conditions that would lead to undefined values. This ensures that the simplification process adheres to mathematical principles.

Algebraic Fractions

Algebraic fractions are expressions containing variables in the numerator, denominator, or both, forming a ratio of two expressions. Simplifying them mirrors the process of simplifying numeric fractions.

• To simplify our algebraic fraction, \(\frac{8ab - 4b^2}{6ab}\), we first factored it, resulting in \(\frac{4b(2a-b)}{2 \cdot 3ab}\).
• We then canceled the common factor \(4b\) from the numerator and \(2b\) from the denominator, which required careful analysis of each term to ensure nothing essential was lost during simplification.
• Ultimately, this simplification led to \(\frac{2a - b}{3a}\), an expression that is easier to handle and solve.

Mastering this process enhances understanding of algebraic manipulation and supports the solution of more complex problems in algebra.