Question

1. the length of a rectangle is $\frac{w + 3}{w^{2}+w - 12}$ and the width is $\frac{w^{2}+7w + 12}{w^{2}-9}$. what is the area of the rectangle in simplest form?

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. So $A=\frac{w + 3}{w^{2}+w - 12}\times\frac{w^{2}+7w + 12}{w^{2}-9}$.

Step2: Factor the polynomials

Factor $w^{2}+w - 12=(w + 4)(w-3)$, $w^{2}+7w + 12=(w + 3)(w + 4)$ and $w^{2}-9=(w + 3)(w - 3)$. Then $A=\frac{w + 3}{(w + 4)(w-3)}\times\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}$.

Step3: Simplify the product of fractions

Cancel out the common factors. Cancel out $(w + 3)$ and $(w + 4)$ in the numerator and denominator. We get $A=\frac{w + 3}{(w-3)(w - 3)}=\frac{w + 3}{(w - 3)^{2}}$.

Answer:

$\frac{w + 3}{(w - 3)^{2}}$