Problem 8 - 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. $$ \frac{a^{2}-100}{3 a} \cdot \frac{a^{2}}{2 a-20} $$

Step-by-Step Solution

Answer

The product simplifies to \(\frac{(a+10)a}{6}\), undefined when \(a=0\) or \(a=10\).

Step 1: Factor All Parts of the Expression

First, recognize special patterns and factor each part of the expression.For \(a^2 - 100\), use the difference of squares: \(a^2 - 100 = (a + 10)(a - 10)\).For \(2a - 20\), factor out the greatest common factor: \(2a - 20 = 2(a - 10)\).Now, rewrite the expression using these factors:\[ \frac{(a+10)(a-10)}{3a} \cdot \frac{a^2}{2(a-10)} \]

Step 2: Multiply the Expressions

Multiply the numerators together and the denominators together.Numerator: \((a + 10)(a - 10) \cdot a^2\).Denominator: \(3a \cdot 2(a - 10)\).This gives us:\[ \frac{(a+10)(a-10)a^2}{6a(a-10)} \]

Step 3: Simplify the Expression by Canceling Common Factors

Identify common factors in the numerator and the denominator. We see:- \((a-10)\) is a common factor.- \(a\) can be canceled once from the numerator and denominator.Cancel out these factors:\[ \frac{(a+10)a}{6} \]

Step 4: Determine Undefined Values

The original expression is undefined when the denominator is zero. For the original denominators, find when:- \(3a = 0\) results in \(a = 0\).- \(2(a-10) = 0\) results in \(a = 10\).Thus, the expression is undefined for \(a = 0\) and \(a = 10\).

Step 5: State the Simplified Expression

The expression in its simplest form with restrictions is:\[ \frac{(a+10)a}{6} \]with \(a eq 0\) and \(a eq 10\).

Key Concepts

Factoring Expressions

Factoring expressions is like finding the ingredients that make up a particular math problem. Imagine you have a cake, but you want to know what went into making it. When you factor a polynomial expression, it's similar to finding those ingredients—the numbers and variables that multiply together to give you the original expression.

For example, in the given problem, we had the expression \(a^2 - 100\). Recognizing it as a 'difference of squares', we factor it to get \((a + 10)(a - 10)\). This means \(a^2 - 100\) is the product of \(a + 10\) and \(a - 10\). By breaking it down, or factoring, we simplify the problem significantly.

Often, you'll find common patterns like the difference of squares, which makes factoring easier:

• Look for pairs like \(x^2 - y^2\) that factor to \((x - y)(x + y)\).
• Recognize when you can factor by grouping or find a common factor to simplify expressions.

Being comfortable with factoring can make solving equations easier since you work with simpler, more manageable pieces.

Difference of Squares

The difference of squares is a crucial concept in algebra that simplifies problems greatly. It's called the 'difference' of squares because you're subtracting one squared term from another. The general formula is:\[ x^2 - y^2 = (x - y)(x + y) \]
For instance, in the expression \(a^2 - 100\), you identify it as a difference of squares because \(100\) is a perfect square (\(10^2\)) and \(a^2\) is obviously a squared variable. So, it factors neatly into \((a - 10)(a + 10)\).

Why is this important?

• It simplifies multiplication and division of expressions, making the math easier.
• Identifying factorable patterns helps in solving equations and simplifying fractions.

Always watch out for terms that can be rewritten as squares; applying this concept can save you time and effort.

Undefined Values

Undefined values in algebra often arise when you try to divide by zero. Since division by zero is undefined in mathematics, it's crucial to identify where this might happen in any given expression. In the original exercise, you'd want to determine if there are any values of \(a\) that make the denominator zero.

In the expression \( \frac{(a + 10)(a - 10)a^2}{6a(a - 10)} \), the denominator consists of terms \(6a(a - 10)\). If \(a\) equals zero, the whole denominator goes to zero because of the \(a\) term. Similarly, if \(a = 10\), the term \((a - 10)\) goes to zero, which again makes the denominator zero.

To find undefined values:

• Set the denominator equal to zero and solve the equation.
• Any solutions you find are the undefined values for your variable.
• In the problem above, \(a = 0\) and \(a = 10\) cause the expression to be undefined.

Knowing where an expression is undefined helps in understanding its domain and avoiding mathematical errors.