How to find roots of $4t^3-18t^2+24t-10=0$ Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.
Position is given as $s=t^4-6t^3+12t^2-10t+3$
I'd like to know when the particle is moving to the right.
ie:  Velocity is positive.
I took the derivative:
$v=s'=4t^3-18t^2+24t-10$
What's the best way to find the roots/zeros of this?
(In order to solve the inequality $4t^3-18t^2+24t-10>0$)
Using tech/calc, I can use TI83+ Solver or just graph it and find zeros. (x = 1 & 2.5)
But, what's the easiest/best way to solve without using a calculator?  Factoring?
 A: Generally the best method here is the rational roots test. The rational roots theorem states that if an integer polynomial has rational roots, then they must be of the form $\frac{a}{b}$ where $a$ divides the last coefficient and $b$ divides the leading coefficient.
In practice, we list all the divisors of the first and last coefficients (positive and negative) then check to see if any are roots. For your polynomial, we would check all the possibilities from
$$ \frac{\left\{\pm 1,\pm 2,\pm 5,\pm 10\right\}}{\left\{\pm 1,\pm 2,\pm 4\right\}}$$
It's usually easiest to check the integers first. If this yeilds a root, then we are hoome free (in your case) because we are left with factoring a quadratic polynomial, which is easy.
If this fails, you could use the formula for the roots of a cubic, but that's horrendous and no one wants to do that.
A: You can try factoring if there are rational solutions using a rational roots test, solving for roots by using depressed cubic equations $(x = y - \frac{b}{3a})$ or approximate a solution using something like Newton's Method. 
A: Hint 1: Divide by $2$ on each side to simplify. 
Hint 2: Use the Rational Roots Theorem to try to find some 'easy' roots. 
Hint 3: You will find some nice roots. Divide the original polynomial by $x-r$ for each of the roots $r$. 
Hint 4: Solving the resulting equation should now be easy. 
A: Hint:
Equation 
$$4t^3-18t^2+24t-10=0$$
is equivalent to
$$2t^3-9t^2 +12t-5=0.$$
But
$$2t^3-9t^2 +12t-5=2t^3-2t^2-7t^2+7t+5t-5 \\=2t^2(t-1)-7t(t-1)+5(t-1)\\
=(t-1)(2t^2-7t+5).$$
A: Notice that 1 is a root of your equation and t-1 is a factor. Clearly the other factor should be a quadratic expression
$(t-1)(At^2+Bt+C)=4t^3-18t^2+24t-10$
Compare the coefficients of comparable terms to get
A=4,B=14,C=10
And you know the roots of a quadratic equation? http://en.wikipedia.org/wiki/Quadratic_equation
A: So we have to solve $$4t^3-18t^2+24t-10=0.\tag{$\star$}$$
We can first note that all the coefficients are even and so we can divide by $2$ to be left with the following simplified equation: $$2t^3-9t^2+12t-5=0.$$
Now, observe that the sum of all of those coefficients is exactly $0$. So $1$ is a solution. Therefore, we can divide the polynomial $2t^3-9t^2+12t-5$ by $t-1$ using Euclidean division and we get: $2t^2-7t+5$ and the remainder is of course $0$. 
The coefficient of the polynomials $2t^2-7t+5$ also add up to $0$ so $1$ is a root. Thus, by Vieta's theorem, the other root (we will call it $\omega$) is $\omega=5/2$. Therefore, we conclude that:
$$\text{The set of solutions to $(\star)$}=\{1,2.5\}$$

We used to derive those solutions the following important property:

Let $a\in\Bbb R$. For each polynomial $\mathcal P(x)$ such that its degree is greater than $1$, there exist one polynomial $\mathcal O(x)$ such that $\mathcal P(x)=(x-a)\cdot\mathcal O(x)+\mathcal P(a)$ such that the degree of $\mathcal P$ is greater than that of $\mathcal O$ by $1$. $\mathcal O(x)$ is called the quotient and $\mathcal P(a)$ is named the remainder.  If $a$ is a root of the polynomial $\mathcal P(x)$ then it is easy to see that $\mathcal P(x)=(x-a)\cdot\mathcal O(x)$. In this case, we say that $\mathcal P(x)$ is divisible by $x-a$.

I hope this helps. 
Best wishes, $\mathcal H$akim.
