Problem 11 - Algebra 2 and Trigonometry

Question

In \(3-12,\) multiply and express each product in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{2 a+4}{6 a} \cdot \frac{3 a^{2}}{a^{2}+2 a} $$

Step-by-Step Solution

Answer

The simplified product is \(a\), and the fractions are undefined for \(a = 0\) and \(a = -2\).

Step 1: Factor each expression

First, we need to factor the numerators and denominators of each fraction to simplify the multiplication. Start by factoring the first expression \( \frac{2a + 4}{6a} \):- The numerator \(2a + 4\) can be factored as \(2(a + 2)\).- The denominator is \(6a\).For the second expression \( \frac{3a^2}{a^2 + 2a} \):- The numerator is \(3a^2\).- The denominator \(a^2 + 2a\) can be factored as \(a(a + 2)\).

Step 2: Write the product of the fractions

The multiplication of these expressions becomes:\[\frac{2(a + 2)}{6a} \cdot \frac{3a^2}{a(a + 2)}\]

Step 3: Simplify by canceling common factors

Now, we can simplify by canceling common factors in the numerator and denominator across the entire expression:- \(a+2\) is a common factor and can be canceled out.- One \(a\) from \(3a^2\) cancels one \(a\) in the denominator \(6a\).After canceling, the expression simplifies to:\[\frac{2 \cdot 3a}{6}\]

Step 4: Simplify the expression further

Now simplify \(\frac{2 \cdot 3a}{6}\):- Combine the numerators: \(6a\).- Divide the entire expression by 6:\[\frac{6a}{6} = a\]

Step 5: Determine excluded values for variables

Identify the values where the original fractions are undefined.- The original denomiators are \(6a\) and \(a(a + 2)\).- \(6a = 0\) when \(a = 0\).- \(a(a + 2) = 0\) implies \(a = 0\) or \(a = -2\).- Therefore, the variable values that make the original fractions undefined are \(a = 0, -2\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial step when working with rational expressions. It allows us to break down complex expressions into simpler components that are easier to manage. In our exercise, both numerators and denominators needed factoring to simplify the multiplication process.
For instance, the expression \(2a + 4\) in the numerator can be beautifully simplified by pulling out the greatest common factor (GCF), which is 2. Here, \(2a + 4\) becomes \(2(a + 2)\). Similarly, in denominators like \(a^2 + 2a\), a single \(a\) can be factored out, resulting in \(a(a + 2)\).
Understanding how to factor polynomials efficiently is key, as it not only simplifies calculations but also aids in identifying terms that can be canceled later on.

Simplifying Expressions

Once polynomials are factored, simplifying expressions becomes straightforward. This involves canceling out common factors, making the expression shorter and easier to evaluate. In our exercise, the goal was to simplify:

• Identify common factors across the board: both \(a+2\) from the first numerator and second denominator were canceled.
• The \(a\) factor in \(3a^2\) was reduced by the \(a\) in the denominator \(6a\).

After canceling these, the expression simplified to \(\frac{6a}{6}\). Further simplifying by dividing both numerator and denominator by 6, we arrived at the simplified result \(a\). Simplifying not only clarifies computations but enhances accuracy in handling expressions.

Excluded Values in Fractions

Understanding excluded values in fractions is essential as it defines where an expression becomes undefined. In rational expressions, a fraction is undefined when the denominator is zero, prompting us to investigate these scenarios:

• The initial denominator \(6a\) implies that any zero value for \(a\) would make the fraction not valid since \(6a = 0\).
• Similarly, for the denominator \(a(a + 2)\), solving \(a^2 + 2a = 0\) reveals that \(a = 0\) or \(a = -2\) also makes the expression undefined.

Thus, the excluded values in this exercise are \(a = 0\) and \(a = -2\). It's crucial to highlight these values when finalizing any simplified expression to ensure completeness in the solution.