# Math Help - Someone please help solve this problem

A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.

2. Originally Posted by moomoo762
A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.
$x(x+1)(x+2) = 504$

note the prime factorization of 504 ...

$504 = 7 \cdot 2^3 \cdot 3^2$

3. You should know that the formula for the volume of a rectangular prism is
V = LWH.

So plug in:
$504 = x(x + 1)(x + 2)$

(let's switch sides)
$x(x + 1)(x + 2) = 504$

Multiply out the left side, and afterwards, subtract 504 from both sides. You'll end up with a cubic equation equaling zero. Solve for x. Don't forget to reject any solution that gives you a negative number for any of the lengths.

01

EDIT: beaten to it by skeeter!

4. I'm stuck

I subtract 504 from both sides to make it equal to zero and I get
x^3+5x^2+2x-504=0

What's the best way to solve for x now?

5. Originally Posted by moomoo762
I'm stuck

I subtract 504 from both sides to make it equal to zero and I get
x^3+5x^2+2x-504=0

What's the best way to solve for x now?
rational root theorem

6. Using the rational root test I got 7

I know the other two sides are 8 and 9 but how do I go about finding them?

7. Originally Posted by moomoo762
Using the rational root test I got 7

I know the other two sides are 8 and 9 but how do I go about finding them?
first of all, correct your expansion ...

$x(x+1)(x+2) = 504$

$x^3 + 3x^2 + 2x - 504 = 0$

you know 7 is a root ... use synthetic division

Code:
7]  1.......3........2.......-504
............7.......70........504
-------------------------------
....1......10.......72.........0
note the coefficients of the depressed polynomial ...

$x^2 + 10x + 72 = 0$

since $b^2-4ac < 0$ , the other two roots are imaginary.

$x = 7$ is the only real root

go back to the original equation ...

$x(x+1)(x+2) = 504$

$7(7+1)(7+2) = 504$

8. $x(x+1)(x+2) = 504$

$x(x^2+3x+2)=504$

$x^3+3x^2+2x-504=0$

$(x-7)(x^2+10x+72)=0$

Notice that there are no real solutions for the quadratic factor as $\triangle <0$

So the only real solution is $x=7$

The question stated the sides are $x,\ x+1,\ x+2$ so one side is $7$ and the others are $7+1=8$ and $7+2=9$