Problem 49 - Prealgebra and Introductory Algebra

Question

Multiply. $$\frac{x^{2}+2 x-35}{x^{2}+4 x-21} \cdot \frac{x^{2}+3 x-18}{x^{2}+9 x+18}$$

Step-by-Step Solution

Answer

The solution to the problem is \(\frac{x-5}{x+3}\).

Step 1 - Simplify the fractions

Begin by factoring the numerator and the denominator of each fraction. The factored form of first fraction is \[\frac{(x-5)(x+7)}{(x-3)(x+7)}\] and the factored form of second fraction is \[\frac{(x-3)(x+6)}{(x+3)(x+6)}\].

Step 2 - Further Simplification

We can now cancel out the terms that appear both in the numerator and the denominator. We observe that the terms (x+7) and (x+6) are the same, one in the numerator and the other in the denominator. They thus cancel each other out. What remains for the first fraction is \[\frac{x-5}{x-3}\] and the second fraction is \[\frac{x-3}{x+3}\].

Step 3 - Multiplication

Multiply the remaining terms to get the final answer. This is done by multiplying the numerators with each other and the denominators with each other. The final result is \[\frac{(x-5)(x-3)}{(x-3)(x+3)}\]

Key Concepts

Factoring Polynomials

Before diving into multiplying polynomial expressions, we need to understand how to factor polynomials. Factoring polynomials is essentially breaking down a larger polynomial into simpler ones that, when multiplied, give back the original polynomial. It’s a bit like reversing multiplication.

In the given exercise, we are dealing with quadratic polynomials, such as \(x^2 + 2x - 35\). To factor such polynomials, look for two numbers that multiply to the constant term (in this case, \(-35\)) and add to the middle coefficient (in this case, \(2\)). Here, \(5\) and \(-7\) satisfy these conditions, so \(x^2 + 2x - 35\) factors into \((x-5)(x+7)\).

Key steps in factoring polynomials:

• Identify the polynomial you want to factor.
• Find two numbers that complete the conditions (multiply to the constant term and add to the middle coefficient).
• Write the polynomial in its factored form, using these numbers.

Factoring makes complex expressions easier to handle, setting the stage for simplifying rational expressions.

Rational Expressions

Rational expressions, much like regular fractions, involve a numerator and a denominator. However, in the case of rational expressions, both the numerator and the denominator are polynomials.

In our exercise, the rational expression is represented as the product of two expressions:\[\frac{x^{2}+2x-35}{x^{2}+4x-21} \cdot \frac{x^{2}+3x-18}{x^{2}+9x+18}\]

The goal with rational expressions is often to simplify them by factoring and canceling common factors (which we will discuss in the next section). This process is crucial in making the expressions manageable and easier to multiply or divide.

Working with rational expressions:

• Ensure each part of the expression is fully factored.
• Identify and cancel any common factors in the numerators and denominators.
• Simplify the expression to its lowest terms, if possible.

These steps bring clarity and simplicity to operations with rational expressions.

Canceling Common Factors

After factoring both the numerators and the denominators, the next step is to cancel common factors in rational expressions. This is similar to simplifying numeric fractions, like canceling a common factor of 2 in \(\frac{4}{6}\) to get \(\frac{2}{3}\).

In the exercise, after factoring, the expressions look like:\[\frac{(x-5)(x+7)}{(x-3)(x+7)} \cdot \frac{(x-3)(x+6)}{(x+3)(x+6)}\]

We can see both \((x+7)\) and \((x+6)\) appear in both numerators and denominators, meaning they can be canceled out. This reduces the complex expression to simpler terms:
\[\frac{x-5}{x-3} \cdot \frac{x-3}{x+3}\]
The process of canceling common factors:

• Factor the expressions completely.
• Identify terms that are identical both in the numerator and denominator.
• Cancel these terms, simplifying the expression.

Canceling is integral in simplifying expressions, ensuring that the final expression is as straightforward as possible.