Question

problem 1: start the area model
use the next three slides to factor $4x^{2}-9$
$4x^{2}-9=$
start by labeling the area model for rectangles i and iv. next,
multiply the main diagonal and enter the product into the table.

Explanation:

Step1: Recognize the difference of squares formula

The formula for the difference of squares is \(a^{2}-b^{2}=(a + b)(a - b)\). For the expression \(4x^{2}-9\), we can rewrite \(4x^{2}\) as \((2x)^{2}\) and \(9\) as \(3^{2}\). So \(a = 2x\) and \(b=3\).

Step2: Apply the difference of squares formula

Using the formula \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a = 2x\) and \(b = 3\), we get \((2x+3)(2x - 3)\).

For the area - model:

• Rectangle I: If we consider the factored form \((2x + 3)(2x-3)\) as a rectangle with length \((2x + 3)\) and width \((2x-3)\), and we split it into four sub - rectangles. The area of rectangle I (if we assume the split is based on the terms of the binomials) is \((2x)\times(2x)=4x^{2}\)
• Rectangle IV: The area of rectangle IV is \((3)\times(- 3)=-9\)
• Product of the main diagonal: The product of the areas of rectangle I and rectangle IV is \(4x^{2}\times(-9)=-36x^{2}\)

Answer:

• Factored form: \((2x + 3)(2x-3)\)
• Rectangle I: \(4x^{2}\)
• Rectangle IV: \(-9\)
• Product of Main Diagonal: \(-36x^{2}\)