SOLUTION: A poster has a length 4 inches greather than its width w. Demetrius will place a frame 2 inches wide around the poster. The area of the poster and frame together is 672 square inch

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: A poster has a length 4 inches greather than its width w. Demetrius will place a frame 2 inches wide around the poster. The area of the poster and frame together is 672 square inch      Log On


   

Question 66106: A poster has a length 4 inches greather than its width w. Demetrius will place a frame 2 inches wide around the poster. The area of the poster and frame together is 672 square inches. Which factored equation could be used to find the width of the poster?
Answer Choices:
A (w+30)(w-20)=0
B (w+4)(w+8)=0
C (w+2)(w+6)=672
D (w-32)(w+20)=0
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let's find the equation for the area of the poster + frame.
The area of the poster alone is:
A+=+L%2AW but the total size with the frame is 4 inches more on each side, so the area then becomes:
A+=+%28L%2B4%29%28W%2B4%29 and since the length, L = W+4, you can substiute this to get:
A+=+%28%28W%2B4%29%2B4%29%28W%2B4%29 and the total area is given as 672 sq.in, so...
%28%28W%2B4%29%2B4%29%28W%2B4%29+=+672 Simplifying this, you get:
%28W%2B8%29%28W%2B4%29+=+672
W%5E2+%2B+12W+%2B+32+=+672 Now subtract 672 from both sides.
W%5E2+%2B+12W+-+640+=+0 Now that we have the quadratic equation, we can factor it, if possible:
W%5E2+%2B+12W+-+640+=+%28W+-+20%29%28W+%2B+32%29 so:
%28W-20%29%28W%2B32%29+=+0
This does not match exactly any of the given possible answers, does it?