Problem 60 - College Algebra with Modeling and Visualization

Question

Factor the expression completely, if possible. \(z^{2}-169\)

Step-by-Step Solution

Answer

The expression \(z^2 - 169\) factors to \((z - 13)(z + 13)\).

Step 1: Recognize the Form

The given expression is \(z^2 - 169\). We notice that both terms of this expression are perfect squares. It can be written in the form of \(a^2 - b^2\). Here, \(a^2 = z^2\) and \(b^2 = 169\).

Step 2: Identify the Square Roots

Calculate the square roots of each perfect square. \(a = \sqrt{z^2} = z\) and \(b = \sqrt{169} = 13\).

Step 3: Apply the Difference of Squares Formula

We use the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\). Substitute the values we found: \((z - 13)(z + 13)\).

Step 4: Confirm the Factorization

Expand \((z - 13)(z + 13)\) to check if it matches the original expression: \((z - 13)(z + 13) = z^2 + 13z - 13z - 169 = z^2 - 169\). The factorization is correct.

Key Concepts

Difference of Squares

The concept of Difference of Squares is a handy tool for factoring certain types of polynomials. This method applies when a polynomial has a specific form: two perfect squares separated by a subtraction sign, like in the expression \(a^2 - b^2\). Here, both \(a^2\) and \(b^2\) are perfect squares, and the minus sign indicates the difference. This form can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).

• Identify the perfect squares: Check each term in the expression to decide if it's a perfect square. This will involve recognizing numbers like 4, 9, 16, 25, which result from squaring integers.
• Apply the formula: Once confirmed, use the formula by substituting for \(a\) and \(b\) the square roots of the terms.
• Verify by expanding: After factorization, you can expand back to confirm the factorization. This step is important to ensure accuracy.

In the example \(z^2 - 169\), \(z^2\) and 169 are both perfect squares, making the expression a textbook example of the Difference of Squares.

Perfect Squares

To understand how to factor expressions with the difference of squares method, it's crucial to recognize perfect squares. A perfect square is a number or expression that can be the square of an integer. For instance, 25 is a perfect square because it equals \(5^2\), and \(x^2\) is a perfect square as it's \(x\) raised to the power of 2.

Perfect squares are common in factoring:

• When identifying perfect squares in polynomials, look for terms like \(z^2\), 169, or similar that can be expressed as an exponent of 2.
• Calculating the square root: For example, the square root of 49 is 7, and for 81, it’s 9. Recognizing these helps in simplifying the process of factoring polynomials using the difference of squares.

In expressions like \(z^2 - 169\), recognizing \(z^2\) and 169 as perfect squares allows you to use special factoring techniques to simplify expressions quickly.

Square Roots

Square roots are associated closely with perfect squares, as they are the inverse operation. Understanding square roots is fundamental for factoring polynomials, especially when applying the difference of squares method. The square root of a number or expression \(n\) is a value that, when multiplied by itself, gives \(n\).

Key points about square roots:

• Finding square roots: Know basic square roots like \(\sqrt{4} = 2\) or \(\sqrt{25} = 5\). Recognizing these roots quickly is helpful in both computation and recognizing patterns in expressions.
• Utilize in factorizations: When you determine that \(z^2\) and 169 are perfect squares in the expression \(z^2 - 169\), you find their square roots \(z\) and 13 to apply the difference of squares formula efficiently.
• Check results: After completing the factorization, take the square roots of the resulting terms to ensure everything lines up correctly with initial terms through expansion or other methods.

In our example of \(z^2 - 169\), calculating \(\sqrt{z^2}\) and \(\sqrt{169}\) as \(z\) and 13 is essential for proper application of the factorization formula. Knowing how to find square roots makes this step seamless.