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: 5 in.

B.
height: 21 in.

depth: 5 in.

C.
height: 19 in.

depth: 7 in.

D.
height: 21 in.

depth: 7 in.

Question

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: 5 in.

B.
height: 21 in.

depth: 5 in.

C.
height: 19 in.

depth: 7 in.

D.
height: 21 in.

depth: 7 in.

alekos · Answer

need help?

anonymous · Answer

it is b correct?

alekos · Answer

that's what I got. do you know how to get this?

anonymous · Answer

Given: $$v=x^3+5x^2-57x-189$$ The volume of a box can be written as $$v=height \times depth \times width$$ The width of the box can be written as $$w=x+3=15$$ Which means that x is 12 $$x=15-3=12$$ Plug that into the polynomial to get the volume $$v=12^3+(5\times12^2)-(57\times12)-189=1728+(5\times144)-684-189$$$$=1728+720-873=1575$$ Plug the values into the second volume equation $$1575=h \times d \times 15, h \times d=1575\div15=105$$ Multiply the heights and depths of the choices to find which has a product of 105 $$A: 19\times5=95, B: 21\times5=105$$

alekos · Answer

rebecca. thanks for helping, but it's better to let samantha work it out