Simplify Algebraic Expressions: Domain And Multiplication

When working with algebraic expressions, two crucial steps are determining the domain and performing the multiplication. Understanding these aspects ensures that our operations are valid and that we can simplify the expression effectively. Let's dive into the expression $(3 a-15 b) \cdot \frac{8}{a^2-25 b^2}$ and break down how to find its domain and then tackle the multiplication.

Understanding the Domain of an Algebraic Expression

The domain of an algebraic expression refers to all the possible values for the variables that do not result in an undefined operation. For rational expressions (those with variables in the denominator), the primary concern is avoiding division by zero. In our expression, $(3 a-15 b) \cdot \frac{8}{a^2-25 b^2}$, the denominator is $a^2-25 b^2$. To find the domain, we must identify the values of $a$ and $b$ that would make this denominator equal to zero.

We set the denominator equal to zero and solve for the variables: $a^2 - 25 b^2 = 0$. This equation is a difference of squares, which can be factored as $(a - 5b)(a + 5b) = 0$. For this product to be zero, at least one of the factors must be zero. Therefore, we have two conditions:

1. $a - 5b = 0 a = 5b$
2. $a + 5b = 0 a = -5b$

These conditions tell us that the expression is undefined when $a$ is equal to $5b$ or $-5b$. We can express the domain by stating the restrictions on $a$ and $b$. In terms of $a$, the domain is $a \neq 5b$ and $a \neq -5b$. Alternatively, we can express these restrictions in terms of $b$ by rearranging the equations: $b \neq \frac{a}{5}$ and $b \neq -\frac{a}{5}$. So, the domain is all real numbers for $a$ and $b$ except where $a = 5b$ or $a = -5b$. This ensures that we never divide by zero, maintaining the integrity of our calculations. Properly defining the domain is the first and most critical step before proceeding with any further algebraic manipulation, as it sets the boundaries for valid operations and solutions.

Performing the Multiplication of Algebraic Expressions

Now that we've established the domain, let's focus on performing the multiplication of the given expression: $(3 a-15 b) \cdot \frac{8}{a^2-25 b^2}$. The first step in simplifying this expression is to factor any polynomials that can be factored. Looking at the first part of the expression, $(3 a-15 b)$, we can factor out a common factor of $3$: $3(a - 5b)$.

The denominator of the second fraction, $a^2-25 b^2$, is a difference of squares, which we factored earlier when determining the domain: $(a - 5b)(a + 5b)$.

Now, we can rewrite the expression with the factored terms:

$$3(a - 5b) \cdot \frac{8}{(a - 5b)(a + 5b)}$$

To perform the multiplication, we can think of the first term as a fraction with a denominator of 1:

$$\frac{3(a - 5b)}{1} \cdot \frac{8}{(a - 5b)(a + 5b)}$$

Now, we multiply the numerators together and the denominators together:

$$\frac{3(a - 5b) \cdot 8}{1 \cdot (a - 5b)(a + 5b)}$$

$$\frac{24(a - 5b)}{(a - 5b)(a + 5b)}$$

At this stage, we look for common factors in the numerator and the denominator that can be canceled out. We see that $(a - 5b)$ is a factor in both the numerator and the denominator. Since we've already established the domain where $a \neq 5b$ (which means $a - 5b \neq 0$), we can safely cancel this term:

$$\frac{24 \cancel{(a - 5b)}}{\cancel{(a - 5b)}(a + 5b)}$$

This leaves us with the simplified expression:

$$\frac{24}{a + 5b}$$