Problem 42 - Algebra and Trigonometry

Question

Factor the difference of two squares. $$64 x^{2}-81$$

Step-by-Step Solution

Answer

The factored form of \(64x^2−81\) is \( (8x-9)(8x+9) \).

Step 1: Identifying \(a^2\) and \(b^2\)

Given \(64x^{2}-81\), let the first term \(64x^2\) be \(a^2\) and the second term \(81\) be \(b^2\).

Step 2: Identifying \(a\) and \(b\)

To factor \(a^2 - b^2\), we need to identify \(a\) and \(b\). Since \(a^2 = 64x^2\), this implies \(a = \sqrt{64x^2} = 8x\). Furthermore, as \(b^2 = 81\), this implies \(b = \sqrt{81} = 9\).

Step 3: Applying the formula

Substituting \(a = 8x\) and \(b = 9\) into the formula \(a^2−b^2 = (a−b)(a+b)\), we obtain \(64x^2−81=(8x-9)(8x+9)\). Thus, the factored form of \(64x^2−81\) is \( (8x-9)(8x+9) \).