### Instruction: Find the roots of $x^3-3x^2-10x+24$. Enter your answer as a list of numbers separated by commas.### Response: To find the roots of this cubic polynomial, I will use the rational root theorem, which says that any rational root of the polynomial must be of the form $\pm \frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.
The constant term is 24, so its factors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.
The leading coefficient is 1, so its only factor is $\pm 1$.
Therefore, the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.
I will try these candidates one by one, using synthetic division or the remainder theorem to see if they are actually roots.
If I find a root, I will also get a quadratic factor of the polynomial, which I can then solve by the quadratic formula or factoring.
I will start with 1, since it is the easiest to check.
If I plug in 1 into the polynomial, I get $1-3-10+24=12$, which is not zero.
So 1 is not a root.
Next, I will try -1.
If I plug in -1 into the polynomial, I get $-1-3+10+24=30$, which is also not zero.
So -1 is not a root.
Next, I will try 2.
If I plug in 2 into the polynomial, I get $8-12-20+24=0$, which is zero.
So 2 is a root.
That means I can divide the polynomial by $x-2$ and get a quadratic factor.
To do the synthetic division, I write the coefficients of the polynomial in a row, and write 2 under the first coefficient.
Then I bring down the first coefficient, multiply it by 2, and add it to the second coefficient.
I repeat this process until I reach the end of the row.
The synthetic division looks like this:

\begin{array}{c|cccc}
  2 & 1 & -3 & -10 & 24 \\
    &   & 2  & -2  & -24 \\
  \hline
    & 1 & -1 & -12 & 0
\end{array}
The last row gives me the coefficients of the quadratic factor, which is $x^2-x-12$.
I can factor this by finding two numbers that multiply to -12 and add to -1.
These are -4 and 3.
So the quadratic factor is $(x-4)(x+3)$.
Therefore, the original polynomial can be written as $(x-2)(x-4)(x+3)$.
The roots are the values of x that make each factor zero, which are 2, 4, and -3.