## Elementary Geometry for College Students (6th Edition)

Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then we can factor the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can rearrange the given equation: $w~(w+5)=66$ $w^2+5w=66$ $w^2+5w-66 = 0$ To factor the left side of the equation, we need to find two numbers $r$ and $s$ such that $r+s = 5$ and $r\times s = -66$. We can see that $(11)+(-6) = 5~$ and $(11)\times (-6) = -66$ We can solve the equation as follows: $w~(w+5)=66$ $w^2+5w=66$ $w^2+5w-66 = 0$ $(w+11)~(w-6) = 0$ $w+11 = 0~~$ or $~~w-6 = 0$ $w = -11~~$ or $~~w = 6$ Since the width must be a positive number, $w = 6$ We can find the length: $w+5 = 6+5 = 11$