Problem 25 - Precalculus Mathematics for Calculus

Question

Perform the multiplication or division and simplify. $$\frac{x^{2}-2 x-15}{x^{2}-9} \cdot \frac{x+3}{x-5}$$

Step-by-Step Solution

Answer

\(\frac{1}{x - 3}\)

Step 1: Factor the Numerator and Denominator

First, we will factor the quadratic expressions in both numerators and denominators. For the numerator \(x^2 - 2x - 15\), we need to find two numbers that multiply to \(-15\) and add to \(-2\). These numbers are \(-5\) and \(3\). Thus, \(x^2 - 2x - 15 = (x - 5)(x + 3)\).For the denominator \(x^2 - 9\), which is a difference of squares, we have: \(x^2 - 9 = (x - 3)(x + 3)\).

Step 2: Write the Expression with Factored Forms

Substitute the factored forms into the expression: \[ \frac{(x - 5)(x + 3)}{(x - 3)(x + 3)} \cdot \frac{x + 3}{x - 5} \]

Step 3: Simplify by Canceling Common Factors

Next, we cancel common factors in the numerator and the denominator. We can cancel \((x + 3)\) from \(\frac{(x - 5)(x + 3)}{(x - 3)(x + 3)}\) and \(\frac{x + 3}{x - 5}\) leaving us with:\[ \frac{x - 5}{x - 3} \cdot \frac{1}{x - 5} \]

Step 4: Perform Further Simplification

Now, we simplify the resulting expression by canceling the \((x - 5)\) factor: \[ \frac{1}{x - 3} \]

Step 5: Verification and Conclusion

Verify the simplification by ensuring all expressions were properly canceled and confirm the assumptions, such as \(x eq 3, -3, 5\), so no divisions by zero occur. The simplified result is correct.

Key Concepts

Factoring Quadratics

Understanding how to factor quadratics is vital for simplifying rational expressions and solving quadratic equations. A quadratic expression is typically in the form of \( ax^2 + bx + c \). In this exercise, the quadratic part was \( x^2 - 2x - 15 \).
To factor it, we look for two numbers that multiply to the constant term \(-15\), and meanwhile, add up to the middle coefficient \(-2\). Here, the numbers \(-5\) and \(3\) are the perfect candidates, fitting both conditions. Factoring yields:

• \( (x - 5)(x + 3) \)

Always check your factors by expanding them; multiply the binomials back out to ensure you return to the original quadratic. In our case, multiplying \((x-5)(x+3)\) gives \(x^2 - 2x - 15 \) again. This confirms our factorization is correct. Getting comfortable with this process will help you tackle other polynomials effectively.
Remember, practice is key in mastering factoring!

Simplifying Expressions

When simplifying expressions, especially ones involving fractions with polynomials, you need vigilance and patience. First, factor both numerators and denominators as much as possible. In our given exercise, after factoring, the expression becomes:
\[\frac{(x - 5)(x + 3)}{(x - 3)(x + 3)} \cdot \frac{x + 3}{x - 5}\]Next step is to cancel out any common factors in the numerator and denominator. This process of canceling involves:

• Checking for any terms that appear in both the numerator and denominator.
• Eliminating them wherever possible to simplify the expression.

This often transforms a seemingly complex expression into something much simpler. We canceled \((x + 3)\) and then \((x - 5)\), simplifying down to \(\frac{1}{x - 3}\). Simplification makes math not just easier but also more elegant.

Difference of Squares

The difference of squares is a specific kind of quadratic expression characterized by a subtraction operation between two perfect squares. This pattern can be defined as \( a^2 - b^2 = (a - b)(a + b) \). Recognizing and applying this identity simplifies factorizations significantly.
In the exercise’s denominator, we had the expression \( x^2 - 9 \), a classic example of a difference of squares which can be factored as:

• \( (x - 3)(x + 3) \)

Understanding the difference of squares helps in quickly breaking down and simplifying what might first appear to be a complex polynomial. Practice this concept with different numbers and variables to see how powerful the difference of squares can be in algebra.
Identifying these patterns reduces the complexity of algebraic tasks and makes future calculations easier.