The polynomial x 3 + 5x 2 - ­57x -­189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3) in. if the width of the box is 15 in., what are the other two dimensions? ( hint: the height is greater than the depth.) a. height: 19 in. depth: 7 in. b. height: 19 in. depth: 5 in c. height: 21 in. depth: 5 in. d. height: 21 in. depth: 7 in.

V=x^3+5x^2-57x-189
Width: W=(x+3) in = 15 in →x+3=15
Solving for x:
x+3-3=15-3→x=12

With x=12 the Volume would be:
V=(12)^3+5(12)^2-57(12)-189
V=1,728+5(144)-684-189
V=1,728+720-684-189
V=1,575

V=W*D*H
Depth: D
Height: H
with H>D

V=1,575; W=15
Replacing in the equation above:
1,575=15*D*H
Dividing both sides by 15
1,575/15=(12*D*H)/15
105=D*H
3*5*7=D*H
D<H
If D=5→H=3*7→H=21
If D=7→H=3*5→H=15

Option C. height: 21 in. depth: 5 in.

Please, see the attached file for another form to solve the problem

Where is the qeustion

The great thing about this problem is they give you everything you need to solve it easily. so the expression is: 12h+30w12h+30wand so everywhere you have an “h” you place 66. and everywhere you have a “w” you place 33. then the expression will look like this: 12(66)+30(33)12(66)+30(33)from here you multiply and then add. getting the following of: $785,862hope this