[DanPatrick10]

I'm having difficulty with this problem:

3r^3 + 6r^2t - 24rt^2

I factored out 3r:

3r[(r^2 + 6rt - 24t^2)]

What do I factor out next?

Book Answer: 3r(r + 4t)(r - 2t)

Using FOIL I can get the original trinomial 3r(r^2 - 2rt + 4rt - 8t^2) = 3r^3 + 6r^2t - 24rt^2

I know I am nuking this:mad:

[rwinterton]

You now have "3r" and a quadratic equation. Quadratic equations can always be factored: Factors (roots) = (-b ± √(b^2-4ac)) / 2a where a, b, and c are the coefficients of the quadratic expression

aX^2 + bX + c

[ebaines]

You have made an error when you factored out the 3r term. You should have:

3r[r^2 + 2rt - 8t^2)

From here it should be fairly easy to factor into the form 3r(r-at)(r+bt). To get the a and b coefficients you need to consider the various combinatons of factors of 8, and find a pair whose difference is 2. Then play with the signs to get it right.