Problem 18 - College Algebra

Question

For the following exercises, multiply the rational expressions and express the product in simplest form. $$ \frac{6 b^{2}+13 b+6}{4 b^{2}-9} \cdot \frac{6 b^{2}+31 b-30}{18 b^{2}-3 b-10} $$

Step-by-Step Solution

Answer

The product in simplest form is \( \frac{(3b+2)(6b-5)(b+6)}{(2b-3)(6b+5)(3b-2)} \).

Step 1: Factor the First Rational Expression Numerator

The numerator of the first rational expression is \( 6b^2 + 13b + 6 \). We need to factor it by finding two numbers that multiply to \( 6 \times 6 = 36 \) and add to \( 13 \). These numbers are \( 9 \) and \( 4 \). So, we rewrite the expression as \( 6b^2 + 9b + 4b + 6 \). Factoring by grouping gives us \((3b + 2)(2b + 3)\).

Step 2: Factor the First Rational Expression Denominator

The denominator of the first rational expression is \( 4b^2 - 9 \), which is a difference of squares. It can be factored as \((2b - 3)(2b + 3)\).

Step 3: Factor the Second Rational Expression Numerator

The numerator of the second rational expression is \( 6b^2 + 31b - 30 \). We look for two numbers that multiply to \(-180\) (\(6 \times -30\)) and add to \(31\). These are \(36\) and \(-5\). Rewriting the expression gives us \(6b^2 + 36b - 5b - 30\). Factoring by grouping, we get \((6b - 5)(b + 6)\).

Step 4: Factor the Second Rational Expression Denominator

The denominator of the second rational expression is \( 18b^2 - 3b - 10 \). We need to find two numbers that multiply to \(-180\) (\(18 \times -10\)) and add to \(-3\). These numbers are \(-15\) and \(12\). So, the expression becomes \(18b^2 + 12b - 15b - 10\). Factoring by grouping, we obtain \((6b + 5)(3b - 2)\).

Step 5: Multiply and Simplify the Expressions

After factoring, the expression is \( \frac{(3b+2)(2b+3)}{(2b-3)(2b+3)} \cdot \frac{(6b-5)(b+6)}{(6b+5)(3b-2)} \). Before multiplying, cancel the common terms: \(2b+3\) cancels with \(2b+3\), leaving \( \frac{(3b+2)(6b-5)(b+6)}{(2b-3)(6b+5)(3b-2)} \). Thus, the simplified product is \( \frac{(3b+2)(6b-5)(b+6)}{(2b-3)(6b+5)(3b-2)} \).

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill needed to manipulate and simplify expressions. The goal is to express a polynomial as a product of its factors. These factors are either prime polynomials or products of other polynomials.

To factor a quadratic polynomial like \(ax^2 + bx + c\), you should:

• Identify two numbers that multiply to \(a \cdot c\) and add to \(b\).
• Rewrite the middle term using these two numbers.
• Group terms and factor out the common factors from each group.

By breaking down the original polynomial into simpler components, you can make solving problems much more straightforward. For example, in the exercise, we factored polynomials like \(6b^2 + 13b + 6\) step-by-step into \((3b + 2)(2b + 3)\) by applying these principles.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form. This process requires combining like terms, factoring, and cancelling out common factors where possible. When dealing with rational expressions, the main goal is to express the fraction in its simplest form without altering its value.

Here's how you can simplify rational expressions:

• Factor the numerators and denominators completely.
• Cancel out any common factors between the numerator and the denominator.

In the provided exercise, after factoring the polynomials, we cancelled the term \((2b+3)\) that appeared in both the numerator and the denominator. This step is critical, as it significantly reduces the complexity of the expression, making further computation easier and more straightforward.

Difference of Squares

The difference of squares is a special type of binomial that can be factored using a straightforward formula. This pattern appears when you have two squared terms separated by a subtraction sign and looks like \(a^2 - b^2\).
The difference of squares can be expressed as:

• \(a^2 - b^2 = (a - b)(a + b)\)

This concept is essential because it allows for quick factorization of specific types of polynomials.

In our exercise, the term \(4b^2 - 9\) is a difference of squares, which we factored as \((2b - 3)(2b + 3)\). Recognizing and applying the difference of squares formula helps tremendously in simplifying and solving many algebraic problems. It is a handy tool in your algebra toolkit.