Question

find the area of a rectangle with dimensions: (x + 2y) and (x - 2y)
x² - 8xy + 4y²
x² + 8xy + 4y²
x² + 4y²
x² - 4y²

Explanation:

Step1: Recall area formula

The area $A$ of a rectangle is $A = l\times w$, where $l$ is the length and $w$ is the width. Here, $l=(x + 2y)$ and $w=(x - 2y)$.

Step2: Apply difference - of - squares formula

We know that $(a + b)(a - b)=a^{2}-b^{2}$. In our case, $a = x$ and $b = 2y$. So, $(x + 2y)(x - 2y)=x^{2}-(2y)^{2}$.

Step3: Simplify the expression

$(2y)^{2}=4y^{2}$, so $x^{2}-(2y)^{2}=x^{2}-4y^{2}$.

Answer:

$x^{2}-4y^{2}$