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Magazine Chapter 6: Problem 94
Look Alikes . . . a. \(\frac{3 d^{2}+6}{4} \div \frac{4 d^{2}+8}{3}\) b. \(\frac{3 d^{2}+6}{4} \cdot \frac{4 d^{2}+8}{3}\)
Short Answer
Expert verified
a. \(\frac{9}{16}\); b. \((d^2 + 2)^2\)
Step by step solution
01
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: \(4(d^2 + 2)\).
02
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: \(\frac{3(d^2 + 2)}{4} \cdot \frac{4(d^2 + 2)}{3}\).
03
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: \(\frac{3(d^2 + 2)}{4} \cdot \frac{3}{4(d^2 + 2)}\).
04
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 \(\frac{9}{16}\).
05
Apply Multiplication Rule to Part b
For part b, multiply the numerators together and the denominators together: \(\frac{3(d^2 + 2) \cdot 4(d^2 + 2)}{4 \cdot 3}\).
06
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 \((d^2 + 2)^2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(4(d^2 + 2)\).
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 \(\frac{3(d^2 + 2)}{4} \times \frac{3}{4(d^2 + 2)}\).
- Part b involves multiplying directly, so \(\frac{3(d^2 + 2)}{4} \times \frac{4(d^2 + 2)}{3}\) becomes \(\frac{3 \times 4 (d^2 + 2)^2}{4 \times 3}\).
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 \((d^2 + 2)^2\).
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.
