Source: Charlotte Hartman...a great problem solver!

**Hint:** Consider the general quadratic: x^{2} + px + q = 0, with roots -r and -s.

What do you know: rs = q and r + s = p.

Now, what about the general cubic.....

**Solution Commentary:** Suppose -r, -s, -t are the roots of the general cubic roots of x^{3} + px^{2} +qx + z = 0. Then, we know z = rst, q = rs + st + rt, and p = r + s + t. This can be shown by multiplying out the factors (x + r)(x + s)(x + t).

Do you start to see a pattern here...what would you expect to be true for the general quartic?

Back to the original problem at hand, if r, s, t are the roots of x^{3} + 6x^{2} +9x - 2 = 0, then rs + st + rt = 9. But, the dimensions of the rectangular box then are r, s, t as well...and its total surface area is 2(rs + st + rt) = 2(9) = 18.

Math can be powerful!