Problem 57 - Beginning and Intermediate Algebra

Question

Factor completely. $$\frac{1}{4} k^{2}-\frac{4}{9}$$

Step-by-Step Solution

Answer

The completely factored expression is: \(\boxed{\left(\frac{3k + 4}{6}\right)\left(\frac{3k - 4}{6}\right)}\).

Step 1: Identify the expression as a difference of squares

Observe that the given expression can be written as: \[\left(\frac{k^2}{4}\right) - \left(\frac{2^2}{3^2}\right),\] where each term is a perfect square. We can now apply the difference of squares formula.

Step 2: Apply the difference of squares formula

Recall that the difference of squares formula states that for any two terms, a and b, the expression \(a^2 - b^2\) can be factored as \((a+b)(a-b)\). Thus, we can apply this formula to our expression: \[\frac{1}{4}k^2 - \frac{4}{9} = \left(\frac{k}{2}\right)^2 - \left(\frac{2}{3}\right)^2\] Now, let \(a = \frac{k}{2}\) and \(b = \frac{2}{3}\). Then, \[\frac{1}{4}k^2 - \frac{4}{9} = (\frac{k}{2} + \frac{2}{3})(\frac{k}{2} - \frac{2}{3})\]

Step 3: Simplify the factored expression

Now, we will simplify each term in the parentheses: \[(\frac{k}{2} + \frac{2}{3})(\frac{k}{2} - \frac{2}{3})\] To add or subtract fractions, we need a common denominator. In this case, the common denominator is 6. So, we have: \[\left(\frac{3k + 4}{6}\right)\left(\frac{3k - 4}{6}\right)\] Since both factors have the same denominator, we can simply write: \[\left(\frac{3k + 4}{6}\right)\left(\frac{3k - 4}{6}\right) = \boxed{\left(\frac{3k + 4}{6}\right)\left(\frac{3k - 4}{6}\right)}\] Our final factored expression is \(\left(\frac{3k + 4}{6}\right)\left(\frac{3k - 4}{6}\right)\).

Key Concepts

Factoring Expressions

Factoring expressions is like unlocking a secret way of writing them. It helps us simplify and understand complex expressions better. One famous method of factoring is called the **difference of squares**. This technique is used when you have an expression in the form of \(a^2 - b^2\), which can be factored into \((a+b)(a-b)\). Here, both \(a\) and \(b\) are perfect squares. In our example, the expression \(\frac{1}{4}k^2 - \frac{4}{9}\) fits this form.

• The first term \(\frac{1}{4}k^2\) is a perfect square, which is \((\frac{k}{2})^2\).
• The second term \(\frac{4}{9}\) is also a perfect square, written as \(\left(\frac{2}{3}\right)^2\).

This allows us to rewrite the expression using the difference of squares formula as \((\frac{k}{2} + \frac{2}{3})(\frac{k}{2} - \frac{2}{3})\). Factoring expressions helps simplify calculations and highlights the expression's inherent structure.

Simplifying Fractions

Simplifying fractions makes expressions easier to interpret and work with. When factoring expressions, it's common to encounter fractions like in the example: \((\frac{k}{2} + \frac{2}{3})(\frac{k}{2} - \frac{2}{3})\). To simplify, we need to perform addition or subtraction within those parentheses, which involves finding a common denominator.For \(\frac{k}{2}\) and \(\frac{2}{3}\), the common denominator is 6.

• Convert \(\frac{k}{2}\) to \(\frac{3k}{6}\) by multiplying the numerator and the denominator by 3.
• Convert \(\frac{2}{3}\) to \(\frac{4}{6}\) by multiplying both by 2.

Using these converted fractions, the expression becomes \(\left(\frac{3k + 4}{6}\right)(\frac{3k - 4}{6})\). This leaves you with a neat product of two expressions, each having a simplified form, making calculations straightforward.

Algebraic Formulas

Understanding algebraic formulas is crucial for solving expressions like the difference of squares. **Algebraic formulas** provide shortcuts and standardized methods for expressing relationships or transforming expressions into different forms. The formula for the difference of squares, \(a^2 - b^2 = (a+b)(a-b)\), is integral in factoring and reducing expressions.In the context of our exercise, recognizing this pattern allowed us to factor a seemingly complex fraction into simpler terms, revealing its underlying structure. By applying this and other algebraic formulas, you can solve problems efficiently.

• They offer clarity and simplify the process of breaking down expressions.
• They also help in predicting outcomes when expressions are expanded or simplified.

Learning and using these formulas gives you powerful tools to tackle a wide range of algebraic problems, turning complex tasks into manageable calculations.