Problem 58 - Beginning and Intermediate Algebra

Question

Factor completely. $$\frac{1}{36} d^{2}-\frac{4}{49}$$

Step-by-Step Solution

Answer

\(\left(\frac{7d+12}{42}\right)\left(\frac{7d-12}{42}\right)\)

Step 1: Identify A and B

In the given expression, \(\frac{1}{36} d^2 - \frac{4}{49}\), identify A and B as: \(A^2 = \frac{1}{36} d^2\) which implies \(A = \frac{d}{6}\) \(B^2 = \frac{4}{49}\) which implies \(B = \frac{2}{7}\)

Step 2: Apply the difference of squares formula

Now that we have identified A and B, apply the difference of squares formula \((A^2-B^2)=(A+B)(A-B)\) to factor the given expression: \[\left(\frac{d}{6} + \frac{2}{7}\right)\left(\frac{d}{6} - \frac{2}{7}\right)\]

Step 3: Simplify the expression

The above expression can be simplified further by finding a common denominator for the fractions within the parentheses: \[\left(\frac{7d+12}{42}\right)\left(\frac{7d-12}{42}\right)\] The completely factored expression is: \[\boxed{\left(\frac{7d+12}{42}\right)\left(\frac{7d-12}{42}\right)}\]

Key Concepts

Difference of Squares

The difference of squares is a unique algebraic pattern that simplifies the task of factoring certain expressions. It takes the form of:

• two perfect squares separated by a subtraction sign, for example, \( A^2 - B^2 \).

To factor an expression like this, we utilize the formula \((A^2-B^2)=(A+B)(A-B)\). The beauty of this formula lies in its ability to break down what seems like a complex polynomial into simpler binomial factors.
In our exercise, we identify the parts as:

• \( A^2 = \frac{1}{36} d^2 \) and \( B^2 = \frac{4}{49} \)
• Thus, \( A = \frac{d}{6} \) and \( B = \frac{2}{7} \)

Applying the formula gives us \((\frac{d}{6} + \frac{2}{7})(\frac{d}{6} - \frac{2}{7})\). This is a perfect application, breaking the polynomial down effectively using the difference of squares pattern.

Factoring Expressions

Factoring expressions is a fundamental skill in algebra that involves rewriting an expression as a product of its factors. In the context of polynomials, this means identifying parts that can be multiplied to arrive back at the original expression.
Our task often begins with recognizing special patterns like the difference of squares, or identifying greatest common factors. In the exercise:

• We recognized \( \frac{1}{36} d^{2} - \frac{4}{49} \) as a difference of squares, which gave us instant insight into how it could be broken down.
• The expression was then separated into two factors: \((A+B)\) and \((A-B)\), with \( A = \frac{d}{6} \) and \( B = \frac{2}{7} \).

This method of factoring not only makes complex expressions manageable but often is crucial in solving equations, simplifying expressions, or in calculus for dropping higher-level terms.

Algebraic Fractions

Algebraic fractions bring a layer of complexity to polynomial manipulation because they involve ratios where variables appear in the numerator, denominator, or both. Simplifying or factoring algebraic fractions often requires balancing or finding common denominators for simplifying or breaking down the expression.
In our problem:

• We dealt with \( \frac{7d+12}{42} \) and \( \frac{7d-12}{42} \), indicating a need to manage fractions carefully.
• Simplifying these fractions involves understanding how denominators and numerators interact, frequently leading to canceled terms or more manageable expressions.

The goal with algebraic fractions is always to express them in their simplest form, just as we aimed to achieve a clean, compact result with \( \frac{7d+12}{42} \cdot \frac{7d-12}{42} \), demonstrating mastery over the algebraic operations.