### 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 a cubic polynomial, I need to find at least one factor of the polynomial.
One way to do that is to use the rational root theorem, which says that any rational root of the polynomial must be of the form $\pm p/q$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.
In this case, the constant term is 24 and the leading coefficient is 1, so the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.
I can try plugging in these values into the polynomial and see if any of them make it zero.
If I try $x=1$, I get $1-3-10+24=12$, not zero.
If I try $x=-1$, I get $-1+3+10+24=36$, not zero.
If I try $x=2$, I get $8-12-20+24=0$, zero!
So $x=2$ is a root of the polynomial, and $(x-2)$ is a factor of the polynomial.
To find the other factor, I can use long division or synthetic division to divide the polynomial by $(x-2)$.
I prefer synthetic division, so I write the coefficients of the polynomial in a row and bring down the first one: 

\begin{array}{c|ccccc}
2 & 1 & -3 & -10 & 24 \\ 
  &   &  2 &  -2 & -24 \\ \hline
  & 1 & -1 & -12 & 0  \\
\end{array}
The last row represents the coefficients of the quotient, which is a quadratic polynomial.
So the original polynomial can be written as $(x-2)(x^2-x-12)$.
To find the other roots, I need to solve $x^2-x-12=0$.
I can use the quadratic formula, or I can try to factor the quadratic into two binomials.
I notice that $-12$ can be written as $-4 \times 3$, and $-4+3=-1$, so I can factor the quadratic as $(x-4)(x+3)$.
This means that the other roots are $x=4$ and $x=-3$.