Problem 34 - A Graphical Approach to College Algebra

Question

Find each product or quotient. $$\frac{6 n^{2}-5 n-6}{6 n^{2}+5 n-6} \cdot \frac{12 n^{2}-17 n+6}{12 n^{2}-n-6}$$

Step-by-Step Solution

Answer

The simplified product is \(\frac{2n - 3}{2n + 3}\).

Step 1: Factor each Quadratic Expression

Factorize each of the quadratic expressions in the numerators and denominators if possible. \(6n^2 - 5n - 6\) can be factored into \((3n + 2)(2n - 3)\). \(6n^2 + 5n - 6\) factors to \((3n - 2)(2n + 3)\). \(12n^2 - 17n + 6\) factors to \((3n - 2)(4n - 3)\). Lastly, \(12n^2 - n - 6\) can be factored as \((3n + 2)(4n - 3)\).

Step 2: Substitute Factored Forms Back into Expression

Replace the original fractions with their factored counterparts: \[ \frac{(3n + 2)(2n - 3)}{(3n - 2)(2n + 3)} \times \frac{(3n - 2)(4n - 3)}{(3n + 2)(4n - 3)} \]

Step 3: Cancel Common Factors Across the Fractions

Identify common terms on the numerators and denominators of the combined fraction. Cancel out the following pairwise factors: \((3n + 2), (3n - 2),\) and \((4n - 3)\). This simplifies the expression to \(\frac{2n - 3}{2n + 3}\).

Step 4: Final Expression Simplification

After cancelling common factors, the simplified expression is \(\frac{2n - 3}{2n + 3}\).

Key Concepts

Factoring Quadratics

Before diving into complex algebraic expressions, it's essential to understand how to factor quadratic expressions. Quadratics are expressions of the form \(ax^2 + bx + c\). This type of polynomial can be factored into products of binomials, which makes them easier to handle, especially when solving equations or simplifying expressions.
To factor a quadratic expression, you look for two numbers that multiply to \(ac\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(b\) (the coefficient of \(x\)). For example:

• The quadratic \(6n^2 - 5n - 6\) can be factored into \((3n + 2)(2n - 3)\) because \(3 \times -3 = -9\) and \(2 \times 2 = 4\), with a sum of \(-5\).
• Similarly, \(12n^2 - 17n + 6\) factors to \((3n - 2)(4n - 3)\) because \(3 \times -3 = -9\) and \(-2 \times 4 = -8\), giving a total of \(-17\).

Mastering factoring is foundational to simplifying quadratic-related expressions effectively.

Simplifying Rational Expressions

Rational expressions are fractions with polynomials in the numerator and the denominator. Simplifying them involves factoring and then canceling common factors. It's crucial to first factor both the numerator and the denominator.
For example, consider the expression \(\frac{(3n + 2)(2n - 3)}{(3n - 2)(2n + 3)}\). By identifying common terms that appear in both the numerator and denominator, such as \(3n + 2\), you can simplify the expression by canceling them out. Similarly, in the full expression \(\frac{(3n + 2)(2n - 3)}{(3n - 2)(2n + 3)} \times \frac{(3n - 2)(4n - 3)}{(3n + 2)(4n - 3)}\), we see more common terms like \((4n - 3)\) and \((3n - 2)\) to cancel.
Once all possible cancellations are made, you are left with a simplified rational expression, like \(\frac{2n - 3}{2n + 3}\), which is much easier to work with.

Multiplication of Fractions

Multiplying fractions may seem straightforward, but it often involves crucial steps to ensure the result is in its simplest form. The process involves multiplying the numerators together and the denominators together. For example, when multiplying two fractions, such as \(\frac{A}{B} \times \frac{C}{D}\), the resultant fraction is \(\frac{A \times C}{B \times D}\).
In algebra, before multiplying fractions, it's efficient to factor each polynomial in their numerators and denominators first. This allows you to cancel any common factors before performing the multiplication, which significantly simplifies the process. Consider the exercise expression, which after factoring is: \(\frac{(3n + 2)(2n - 3)}{(3n - 2)(2n + 3)} \times \frac{(3n - 2)(4n - 3)}{(3n + 2)(4n - 3)}\).
This expression can be dramatically simplified by canceling common factors, resulting in a straightforward multiplication problem that leads to the final, simplified form: \(\frac{2n - 3}{2n + 3}\). This underscores the importance of simplifying before multiplying to avoid unnecessary complexity.