The volume of the box is represented by (x^2+5x+6)(x+5). The ends are (x+2). What is the bottome of the box?

Answer 1

Hi,

Volume equals length x width x height. Your equation of (x^2+5x+6)(x+5) factors into (x + 3)(x + 2)(x +5), so one of those is the length, one is the width, and one is the height. Since you say that the ends are (x+2), that means it is the height. So,
(x + 3)(x +5) must be the length and the width in either order. To find the bottome, multiply these together and you get x^2 + 8x + 15.

This is the area of the bottom of your box, x^2 + 8x + 15.

I hope this helps!

Answer 2

:)

x^2 +5x+6 = (x+2)(x+3)

if the ends are (x+2) then one other dimension is (x+3) and (x+5)

the bottom could be either (x+2) by (x+5) or (x+2) by (x+3). or (x+3) by (x+5). You need more info to know how the box is sitting to know which side is the bottom!

I was contemplating while doing the problem that the question had good intentions by an instructor, however, the answer is not so clear as to factoring the quadratic and ASSUMING the dimensions are those factors.

think of this: suppose I say I have a box of dimensions 2 by 6 by 8 and that the volume of the box is 2*12*4? do you see that just because a volume has some representation that the dimensions of the volume do not have to have a direct consequence, but would still have to be interrelated.

let's look at the problem above specifically. let x=1

you have volume of 72 according to the representation.

does that mean that the sides must necessarily be 3, 4 and 6?
of course not! you are only assured that a side has length (x+2) as that's what was stated in the problem. the other two sides are just realted in such a way that their product is 24. So the other sides could be 4 and 6, or 12 and 2, or an infinite set of numbers provided their product is 24.

Don't be fooled in how the problem was presented!

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Answer 3

If the ends are (x + 2) wide AND (x + 3) high, then the bottom is (x + 2) wide and (x + 5) long.
With the information given, there are 3 "easy" combinations :
(x + 2) by (x + 3)
(x + 2) by (x + 5)
(x + 3) by (x + 5)

There being no constraint on the dimensions save that one is (x + 2)m the other two could be any dimensions whose product can be represented by (x + 3)(x + 5). ((x + 3)(x + 5))^(1/2) is one possibility which comes to mind.

Answer 4

in case the box is a cuboid , volume = l* b* h
= (x^2+5x+6)(x+5)
(x^2+5x+6) = x^2 + 3x + 2x +6
= (x+2 ) (x +3)

therefore bottom = x+3

Answer 5

(x^2+5x+6)/(x+2)=(x+3)
So the bottoms are x+3 and x+5

Answer 6

I agree

(x+2), (x+3) and (x+5) are the sides

Ana