Problem 33 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{t^2 + 9} \).

Step 1: Identify the Expression

We are given the expression \( \frac{t-3}{t^{2}+9} \cdot \frac{t+3}{t^{2}-9} \). This involves multiplying two fractions. Our task is to perform this multiplication and simplify the result.

Step 2: Factor Denominators

Notice that the denominators can be factored: - \( t^2 - 9 \) is a difference of squares and can be factored as \( (t-3)(t+3) \). The first denominator, \( t^2 + 9 \), is a sum of squares, and it is not factorable over the reals.

Step 3: Perform the Multiplication

Multiply across the numerators and the denominators: - Numerator: \((t-3)(t+3)\) - Denominator: \((t^2 + 9)((t-3)(t+3))\)The multiplication leads to the expression: \[ \frac{(t-3)(t+3)}{(t^2+9)((t-3)(t+3))} \]

Step 4: Simplify the Expression

Cancel the common factor \((t-3)(t+3)\) in both the numerator and the denominator: \[ \frac{1}{t^2 + 9} \]After canceling, the simplification is straightforward.

Key Concepts

Multiplication of Fractions

Multiplying fractions is a straightforward process. To multiply two fractions, you simply multiply the numerators together and the denominators together. For example, if you have fractions \( \frac{a}{b} \) and \( \frac{c}{d} \), their product is \( \frac{ac}{bd} \).
Here's how it plays out in our exercise:

• First, look at the numerators: \((t-3)\) and \((t+3)\). Multiply them together to get \( (t-3)(t+3) \).
• Next, consider the denominators: \(t^2+9\) and \(t^2-9\). Multiply these to get \((t^2+9)(t^2-9)\).

After setting up your multiplication, you'll end up with: \[ \frac{(t-3)(t+3)}{(t^2+9)((t-3)(t+3))} \]Focusing on multiplication first makes it clearer to manage the next steps, particularly when prepping for simplification.

Simplifying Fractions

Once you've multiplied the fractions, the next step is simplifying. Simplifying fractions means reducing the fraction to its lowest terms.
In our exercise, we had:\[\frac{(t-3)(t+3)}{(t^2+9)((t-3)(t+3))}\]Simplification is done by canceling out the common factors in the numerator and the denominator. In this case, the expression \((t-3)(t+3)\) appears both in the numerator and in the denominator.

• This can be removed, as dividing the same expression by itself results in 1, given it's not zero.
• Once removed, the fraction simplifies to \( \frac{1}{t^2 + 9} \).

Simplification is key in algebra as it helps produce the most straightforward form of the solution, making further analysis or computation easier.

Factoring Polynomials

Factoring polynomials is often a critical step in simplifying algebraic fractions. It involves breaking down a polynomial into a product of simpler polynomials.
In the original exercise, we dealt with the expressions \(t^2 + 9\) and \(t^2 - 9\). Let's explore these:

• The expression \(t^2 - 9\) is a difference of squares, which can be factored into \((t-3)(t+3)\).
• This is a perfect example of using the difference of squares method, where \(a^2 - b^2 = (a - b)(a + b)\).
• On the other hand, \(t^2 + 9\) is a sum of squares, which generally cannot be factored into real number polynomials.

Factoring polynomials not only helps in multiplication and simplification but also in recognizing patterns and making complex algebraic manipulations manageable. It is a skill that strengthens overall problem-solving capabilities in algebra.