Problem 94 - Intermediate Algebra

Question

Look Alikes . . . a. \(\frac{3 d^{2}+6}{4} \div \frac{4 d^{2}+8}{3}\) b. \(\frac{3 d^{2}+6}{4} \cdot \frac{4 d^{2}+8}{3}\)

Step-by-Step Solution

Answer

a. \(\frac{9}{16}\); b. \((d^2 + 2)^2\)

Step 1: Simplify the Numerator

For both parts a and b, begin by simplifying the numerators of the fractions. The numerator for the first fraction in both parts is \(3d^2 + 6\). Factor the common factor of 3 out: \(3(d^2 + 2)\). Similarly, for the second fraction's numerator, \(4d^2 + 8\), factor out the common factor of 4: \(4(d^2 + 2)\).

Step 2: Rewrite the Expressions

Write both expressions with the simplified form obtained in Step 1. For part a, we have: \(\frac{3(d^2 + 2)}{4} \div \frac{4(d^2 + 2)}{3}\). For part b, we have: \(\frac{3(d^2 + 2)}{4} \cdot \frac{4(d^2 + 2)}{3}\).

Step 3: Apply Division Rule to Part a

For part a, recall that dividing by a fraction is the same as multiplying by its reciprocal. Replace the division operation with multiplication by the reciprocal of the second fraction: \(\frac{3(d^2 + 2)}{4} \cdot \frac{3}{4(d^2 + 2)}\).

Step 4: Simplify Part a

Notice that \(d^2 + 2\) appears in both the numerator and denominator of the expression in part a, allowing us to cancel those terms: \(\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}\). Thus, the answer to part a is \(\frac{9}{16}\).

Step 5: Apply Multiplication Rule to Part b

For part b, multiply the numerators together and the denominators together: \(\frac{3(d^2 + 2) \cdot 4(d^2 + 2)}{4 \cdot 3}\).

Step 6: Simplify Part b

Since \(3 \, \text{and} \, 4\) appear in the numerator and denominator, they cancel each other out. Thus, we simplify it to \((d^2 + 2)^2\). Therefore, the answer to part b is \((d^2 + 2)^2\).

Key Concepts

Factoring

Factoring is a crucial tool in algebra that helps to simplify expressions. It involves breaking down a complex expression into simpler components called factors. In the exercise, we had to factor expressions like \(3d^2 + 6\) and \(4d^2 + 8\). Both expressions have a common factor which makes them easy to simplify.

• For \(3d^2 + 6\), the common factor is 3. We can express this as \(3(d^2 + 2)\).
• For \(4d^2 + 8\), the common factor is 4, and it simplifies to \(4(d^2 + 2)\).

Factoring is particularly handy because it can transform a complex problem into a more manageable one. In this exercise, recognizing common factors was the first step in your path to simplifying the algebraic expressions.

Fraction Operations

Fraction operations involve addition, subtraction, multiplication, and division of fractions. In the exercise, focus was primarily on division and multiplication. Division by a fraction is equivalent to multiplying by its reciprocal.

• For part a, the division \(\frac{3(d^2 + 2)}{4} \div \frac{4(d^2 + 2)}{3}\) simplifies to \(\frac{3(d^2 + 2)}{4} \times \frac{3}{4(d^2 + 2)}\).

Similarly, multiplication means directly working on numerators and denominators:

• Part b involves multiplying directly, so \(\frac{3(d^2 + 2)}{4} \times \frac{4(d^2 + 2)}{3}\) becomes \(\frac{3 \times 4 (d^2 + 2)^2}{4 \times 3}\).

Understanding these basic operations is critical for simplifying complex fractional expressions in algebra.

Simplification

Simplification is the process of reducing an expression to its simplest form. This process makes calculations quicker and solutions clearer.

• In part a after rewriting the division as multiplication by the reciprocal, expressions like \(\frac{3(d^2 + 2)}{4} \times \frac{3}{4(d^2 + 2)}\) allow for cancellation of \(d^2 + 2\), leaving \(\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\).


• In part b, after conducting the multiplication, the expression simplifies because both 3 and 4 cancel each other out once in the numerator and denominator, resulting in \((d^2 + 2)^2\).

Effective simplification requires patience and an eye for finding terms that appear both in the numerator and the denominator. It's vital for making algebraic expressions less daunting.

Intermediate Algebra

Intermediate Algebra bridges the gap between basic algebra and more advanced topics. It provides a solid foundation of skills necessary for solving a greater variety of problems. This exercise exemplifies key concepts in intermediate algebra such as:

• Recognizing and factoring common expressions.
• Performing operations with complex fractions.
• Simplifying expressions by canceling common terms.

These skills are crucial as they pave the way for tackling equations and inequalities, working with functions, and understanding algebraic structures. Mastery of intermediate algebra is not only critical for mathematics but also for applying these concepts in real-world problems.