Problem 46 - College Algebra with Modeling and Visualization

Question

Simplify the expression. $$ \frac{x^{2}-16}{x+3} \div \frac{x+4}{x^{2}-9} $$

Step-by-Step Solution

Answer

The simplified expression is \((x-4)(x-3)\).

Step 1: Understand Division of Fractions

Before simplifying, recognize that dividing by a fraction is equivalent to multiplying by its reciprocal. For the given expression \(\frac{x^{2}-16}{x+3} \div \frac{x+4}{x^{2}-9}\), you can rewrite it as \(\frac{x^{2}-16}{x+3} \times \frac{x^{2}-9}{x+4}\).

Step 2: Factor the Numerators and Denominators

Next, factor each part of the expression. The numerator \(x^2 - 16\) is a difference of squares and factors into \((x-4)(x+4)\). The denominator \(x^2 - 9\) is also a difference of squares and factors into \((x-3)(x+3)\). The expression becomes \(\frac{(x-4)(x+4)}{x+3} \times \frac{(x-3)(x+3)}{x+4}\).

Step 3: Multiply the Expressions

Now that you have factored the expressions, multiply the numerators together and the denominators together. This results in \(\frac{(x-4)(x+4)(x-3)(x+3)}{(x+3)(x+4)}\).

Step 4: Simplify by Cancelling Common Factors

Identify the common factors present in the numerator and the denominator. Both \(x+3\) and \(x+4\) are present in both the numerator and denominator, allowing them to be cancelled. After cancelling, the simplified expression is \((x-4)(x-3)\).

Step 5: Write Final Answer

After cancelling the common factors, you're left with the simplified expression \((x-4)(x-3)\). Since there are no more common factors, this is the final simplified form.

Key Concepts

Difference of Squares

The difference of squares is an important algebraic concept. It refers to expressions in the form of \(a^2 - b^2\), which can be factored as \((a - b)(a + b)\). In our exercise, both \(x^2-16\) and \(x^2-9\) are examples of difference of squares. For \(x^2 - 16\), it can be factored into \((x-4)(x+4)\) because \(16\) is \(4^2\). Similarly, \(x^2 - 9\) factors into \((x-3)(x+3)\) since \(9\) is \(3^2\). Understanding how to factor these quickly can simplify solving many algebraic fractions.
Recognizing patterns like the difference of squares reduces complex expressions to simple products. This method is a powerful tool for simplifying expressions in algebra.

Factoring

Factoring involves breaking down an expression into products of simpler expressions. It simplifies algebraic expressions and is crucial in solving quadratic equations, simplifying fractions, and integrating polynomials.
In our exercise, we factor the expression \(x^2 - 16\) into \((x-4)(x+4)\) and \(x^2 - 9\) into \((x-3)(x+3)\).

• Factoring allows us to simplify expressions by identifying and cancelling common factors.
• It's like reversing the process of expanding an expression; instead of multiplying terms together, we find what was originally multiplied together.
• Knowing common factor forms, like the difference of squares, speeds up the process.

Effective factoring is a key skill in simplifying and solving algebraic problems.

Fraction Division

Fraction division might seem challenging initially, but it hinges on a simple rule: dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).
In the given problem, we start with the division \(\frac{x^{2}-16}{x+3} \div \frac{x+4}{x^{2}-9}\). Remember:

• Change division into multiplication by flipping the second fraction.
• The expression becomes \(\frac{x^{2}-16}{x+3} \times \frac{x^{2}-9}{x+4}\), simplifying the task from division to multiplication.

This transformation lets us leverage our skills in multiplying fractions through numerators and denominators. Thus, transforming problems that seem complicated at first glance into more manageable ones.

Simplifying Algebraic Expressions

Simplifying algebraic expressions means reducing expressions to their simplest form, often through division of common factors. This involves cancelling out any terms in the numerator and the denominator that are the same.
For our problem, after multiplying the factored elements, we have an expression \(\frac{(x-4)(x+4)(x-3)(x+3)}{(x+3)(x+4)}\) to simplify:

• Identify common factors in both the numerator and the denominator: \((x+3)\) and \((x+4)\).
• Cancel these common terms to simplify to \((x-4)(x-3)\).
• Ensuring no common factors remain results in the most reduced expression possible.

Effective simplification is crucial for solving equations efficiently and accurately in algebra.