### 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 can try to factor it into a product of a linear term and a quadratic term, or use the rational root theorem to test possible candidates.
The rational root theorem says that if the polynomial has a rational root $\frac{p}{q}$, then $p$ must be a factor of the constant term and $q$ must be 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 factors are $\pm 1$.
This means that the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.
I can use synthetic division to test each of these candidates until I find one that gives a remainder of zero, meaning that it is a root and a factor of the polynomial.
I'll start with 1.
Synthetic division involves writing the coefficients of the polynomial in a row and bringing down the first one.
Then I multiply 1 by the first coefficient and write the result below the second coefficient, and add them.
I repeat this process until I reach the last coefficient, which will be the remainder.
Here is what the synthetic division looks like:

\begin{array}{c|cccc}
1 & 1 & -3 & -10 & 24 \\
 & & 1 & -2 & -12 \\
\hline
 & 1 & -2 & -12 & 12
\end{array}
The remainder is 12, which is not zero, so 1 is not a root of the polynomial.
I can try -1 next.
Here is the synthetic division:

\begin{array}{c|cccc}
-1 & 1 & -3 & -10 & 24 \\
 & & -1 & 4 & 6 \\
\hline
 & 1 & -4 & -6 & 30
\end{array}
The remainder is 30, which is also not zero, so -1 is not a root of the polynomial.
I can try 2 next.
Here is the synthetic division:

\begin{array}{c|cccc}
2 & 1 & -3 & -10 & 24 \\
 & & 2 & -2 & -24 \\
\hline
 & 1 & -1 & -12 & 0
\end{array}
The remainder is zero, which means that 2 is a root and a factor of the polynomial.
The other factor is the quadratic polynomial given by the coefficients in the bottom row, which is $x^2-x-12$.
I can factor this quadratic by looking for two numbers that multiply to -12 and add to -1.
These are -4 and 3.
So the quadratic factor is $(x-4)(x+3)$.
This means that the cubic 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.