The volume in cubic meters of water in an aquarium is given by the polynomial $v(x)=x^3-16x^2+79x-120$. If the depth in meters can be represented by $x-3$, what are the possible dimensions of the rectangular aquarium in terms of $x$? If the aquarium holds $70$ cubic meters, what are the dimensions of the aquarium?
I realize this question has been asked before, but I have done as much work as I can and just need a piece of advice as to where to go from here. I just need help on the second part of the questions (finding the dimensions).
Here is what I have (note that I first wrote the equation in factored form, but it is now in standard):
$0=x^3-16x^2+79x-190$ Where do I go from here?
This cannot be factored to the point where you can find "real, rational, roots", so how do I determine the dimensions? Any ideas would be appreciated. Thanks.
$v(x)=(x-3)(x-5)(x-8)$ means that the dimensions of the aquarium are $(x-3)\times(x-5)\times(x-8)$, as I'm sure you've already determined for part 1. For the second part, as you noted, and observing that $10$ is a root, $x$ satisfies $$x^3-16x^2+79x-190 = (x - 10) (x^2 -6x+19)=0$$ The quadratic in the second statement has a negative discriminant, hence it has no roots and the entire polynomial only has $x=10$ as a real solution. Plugging that in above, the dimensions of the box are $7\times5\times2$.