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 inC. 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

Answer: Option C. height: 21 in. depth: 5 in.

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