(male narrator)

In Part 1 of this video, we began looking

at picture frames and finding the width of a frame

by drawing a picture. Here, we have

another problem where an 8 by 12 inch drawing

is put in a frame– drawing a picture

of an 8 by 12 inch drawing put in a frame

of uniform width. We remember that the frame is

on the left and right sides, and so,

we have an x or an unknown distance

on both sides. Now the new width

is 8 plus 2x. Similarly, we have frame

on top and bottom, giving us 12 plus 2x. We are told

that the area of the frame is equal to the area

of the picture. We’ll have to do a little work

to find out what this will be. The picture itself is

an 8 by 12 rectangle: 8 times 12

is equal to 96. The picture has

an area of 96… which means that the frame,

being the same, must also have

an area of 96. All of it together

in the large rectangle, then, would be the picture, 96,

plus the frame, 96, giving us 192. The total area, when we multiply

the width by the length, should equal that area of 192: 8 plus 2x, times 12,

plus 2x, equals the 192. We can start solving this

by FOILing out what’s left to get 96, plus 16x, plus 24x,

plus 4x squared, equals 192. Combining like terms

and putting things in order gives us 4x squared, plus 40x,

plus 96, equals 192. In order to start solving,

we want the equation to equal 0, so we will subtract 192

from both sides. This gives us 4x squared,

plus 40x, minus 96, equals 0. We are ready to start solving

by factoring the equation. Always start with the

greatest common factor of 4, leaving x squared, plus 10x,

minus 24, equals 0. We can continue factoring

that trinomial to be x, plus 12,

times x, minus 2, equals 0. To solve for x, we take each factor with an x

and set it equal to 0. These equations solve

quite nicely by subtracting 12 to tell us that x

is equal to -12; or adding 2 to tell us

that x is equal to 2. Recall that x is

the width of the frame, and so, we would not have

a negative width on our frame. Throwing

the negative answer out, the only possible width

for our frame is a 2-inch frame. As we set up our equations

for our frame problems, we must remember that the frame

is on top and bottom, so we have two x’s; and left and right, giving us

two x’s on that side as well.