$d(t)=-3t^3+45t^2+600t$ where $0\leq t \leq 10$
What is the value of $t$ if $d(t) = 3750$?
Practice question from my textbook I'm having a little trouble with for some reason. Would like to have the steps clarified In case this kind of question shows up in a future test. Thanks.
You have to solve the cubic equation $-3t^3+45t^2+600t = 3750$ which is equivalent to $$t^3 -15t^2 -200t +1250 = 0. \tag{1}$$
In my opinion the best strategy is to verify first whether it has a rational root. Here all coefficients are integers and it is well-known that a rational root (if it exists) must be an integer which divides $1250$. We have $1250 = 2 \cdot 5^4$ and a simple trial shows that $t = 5$ is a root. For the other roots you need to solve $$t^2 - 10t -250 = 0$$ which gives $t = 5 \pm 5\sqrt{11}$. Among these three roots $t = 5$ is the only one between $0$ and $10$.
Cubic equations can also be solved via the Cardano formula. However, we know that $(1)$ has three real roots, thus we are in the so-called casus irreducibilis. This means that the roots of $(1)$ are represented as sums of two complex numbers - and this is not very nice.
You're solving a cubic here. While there is a general solution, it's no picnic. For most "elementary" (school-style) problems, you won't be expected to apply the full cubic solution. Instead you'll be expected to apply some combination of Rational Root Theorem, Factor Theorem and Polynomial Long Division (or a shortcut like Synthetic Division) to reduce the degree to a quadratic you can solve with methods you already know. But this requires the cubic to have at least one "nice" rational root, which often (but not always) turns out to be an integer.
Let's proceed on the assumption that this is a problem specifically designed to be solvable at an elementary level.
Rearrange the cubic and make it monic (lead coefficient one):
$t^3 - 15t^2 - 200t + 1250 = 0$
By Rational Root Theorem, any rational root (if it exists) will need to divide $1250$. You'll have to consider both positive and negative possibilities for each number.
$1250 = (2^1)(5^4)$
This means it has a total of $(1+1)(4+1)= 10$ factors, considering only the positive ones. Double this for the negatives, and you're looking at $20$.
At this point, you're going to have to "appeal" to the good sense of the question setter and hope the root you're looking for is within that narrow range they stipulated. Which means you'll start by testing $1, 2, 5, 10$. You don't even need to get to the end of that list because you'll hit paydirt at $5$.
Once you know $t=5$ is a root, you can immediately deduce that $(t-5)$ is a factor of the cubic polynomial (this is Factor Theorem). You can then divide the cubic by $(t-5)$ and get a quadratic you can easily solve by any method you've learned (factorisation by inspection, completing the square or quadratic formula).
The division can be simplified by Ruffini's method of synthetic division. The quadratic you get as the result will be $t^2 - 10t - 250$, for which the roots (by formula) are $5(1 \pm \sqrt{11})$. Neither of these values is in the required range, so you can reject them after finding them. But finding them is a must, for rigour.
Now, what would've happened if, in the original application of Rational Root Theorem, you couldn't find a root among the candidates $1,2,5,10$? Technically you have a few choices. You can go on to test the remaining $20-4 = 16$ possibilities. It is possible you'll get a nice rational root outside the stipulated range that, after polynomial division, gets you to a quadratic which does have one or two root(s) in the stipulated range. But that would have to be a pretty sadistic question setter. Not impossible, just unpleasant.
But it's still a more pleasant undertaking than actually solving the cubic using the general solution, which I won't be going into. It's far too tedious and involved.
Anyway, the only solution for your question in the required range is $t=5$.
This is mainly an extended comment on the answers that have already been posted, but it may be of help in approaching problems of this type, where you can assume the problem poser has carefully constructed things to have a nice answer.
As the other answers observe, the cubic simplifies initially to
$$t^3-15t^2-200t+1250=0$$
which you can then attack by hoping it has an integer root. The key observation to make is that since the coefficients $15$, $200$, and $1250$ are all divisible by $5$, then $t^3$, if it's an integer, must also be divisible by $5$. So if we let $t=5u$, the equation becomes $125u^3-15\cdot25u^2-1000u+1250=0$, which simplifies to
$$u^3-3u^2-8u+10=0$$
The smaller coefficients here make this a much easier cubic to think about, and it might even jump right out at you that $u=1$ is a root, since $3+8=1+10$. This leads to the solution $t=5u=5$.
