Problem 63 - Elementary and Intermediate Algebra

Question

Perform the operations and simplify, if possible. See Example 6 $$\frac{3 n^{2}+5 n-2}{12 n^{2}-13 n+3} \div \frac{n^{2}+3 n+2}{4 n^{2}+5 n-6}$$

Step-by-Step Solution

Answer

Simplified expression: \(\frac{(3n - 1)(4n - 3)}{(4n - 1)(3n - 3)(n + 1)}\)

Step 1: Rewrite the Division as Multiplication

The given expression involves division of two fractions. To simplify, we first rewrite the division as multiplication by the reciprocal. This changes the expression from \[ \frac{3n^2 + 5n - 2}{12n^2 - 13n + 3} \div \frac{n^2 + 3n + 2}{4n^2 + 5n - 6} \]to \[ \frac{3n^2 + 5n - 2}{12n^2 - 13n + 3} \times \frac{4n^2 + 5n - 6}{n^2 + 3n + 2} \]

Step 2: Factor All Polynomials

Next, we factor each polynomial to simplify the expressions. - Factor \(3n^2 + 5n - 2\): This factors to \((3n - 1)(n + 2)\).- Factor \(12n^2 - 13n + 3\): This factors to \((4n - 1)(3n - 3)\).- Factor \(n^2 + 3n + 2\): This factors to \((n + 1)(n + 2)\).- Factor \(4n^2 + 5n - 6\): This factors to \((4n - 3)(n + 2)\).The expression is now: \[ \frac{(3n - 1)(n + 2)}{(4n - 1)(3n - 3)} \times \frac{(4n - 3)(n + 2)}{(n + 1)(n + 2)} \]

Step 3: Cancel Common Factors

Now we cancel out any common factors from the numerator and the denominator across the multiplication:- The factor \((n + 2)\) appears in both the numerator of the first fraction and the denominator of the second fraction, so it can be canceled.After canceling, the expression reduces to: \[ \frac{(3n - 1)(1)}{(4n - 1)(3n - 3)} \times \frac{(4n - 3)(1)}{(n + 1)(1)} \]

Step 4: Multiply the Remaining Factors

With all common factors canceled, multiply the numerators and denominators:The expression becomes: \[ \frac{(3n - 1)(4n - 3)}{(4n - 1)(3n - 3)(n + 1)} \]This is the simplified form of the expression.

Key Concepts

Polynomial Factorization

Polynomial factorization is a crucial skill when working with rational expressions. It involves expressing a polynomial as a product of its factors. This process allows us to simplify fractions by canceling out common factors. To factor a polynomial, look for common terms or use techniques like grouping or the quadratic formula. For instance, take the polynomial \(3n^2 + 5n - 2\). By applying factorization techniques, we can write this as \((3n - 1)(n + 2)\). These factors become crucial when simplifying rational expressions through division.

Being able to factor polynomials accurately is essential when dealing with expressions, especially in algebra. This sets the foundation for algebraic manipulation and simplification. Mastering factorization forms the backbone of making complex algebraic expressions more manageable.

Fraction Operations

Fraction operations such as multiplication and division are vital when working with rational expressions. Dividing fractions is essentially multiplying by the reciprocal. For example, to divide \(\frac{3n^2 + 5n - 2}{12n^2 - 13n + 3}\) by \(\frac{n^2 + 3n + 2}{4n^2 + 5n - 6}\), we first rewrite the division as:

• \(\frac{3n^2 + 5n - 2}{12n^2 - 13n + 3} \times \frac{4n^2 + 5n - 6}{n^2 + 3n + 2}\)

This converts the problem into a multiplication operation, making it easier to factor and simplify the expressions.

Each time you handle rational expressions, remember the reciprocal rule and check for common factors. Simplifying fractions by canceling out these common terms streamlines the problem and allows for easier solutions.

Algebraic Simplification

Algebraic simplification enables you to condense complex expressions into more understandable forms. In rational expressions, look for common factors among numerators and denominators. Once polynomials are factored, cancel out identical terms that appear in both the numerator and denominator. In our example, the factor \((n + 2)\) was present in both parts and was canceled, simplifying the expression.

After canceling, you're left with much simpler components to multiply:

• In the numerator: \((3n - 1)(4n - 3)\)
• In the denominator: \((4n - 1)(3n - 3)(n + 1)\)

Algebraic simplification cuts down on the complexity and number of operations required to solve equations. It's essential for efficiency and accuracy in solving real-world problems.