### 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 need to factor it into a product of linear and/or quadratic factors.
One possible way to find a linear factor is to 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 (24) and $q$ is a factor of the leading coefficient (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 use synthetic division or long division to test each of these candidates and see if they give a remainder of zero, which means they are a root.
Alternatively, I can use a graphing calculator or software to plot the polynomial and look for the x-intercepts visually.
Either way, I find that one of the roots is $x=4$, which means $(x-4)$ is a factor of the polynomial.
Dividing the polynomial by $(x-4)$, I get $x^3-3x^2-10x+24=(x-4)(x^2+x-6)$.
Now I can factor the quadratic term further by looking for two numbers that multiply to -6 and add to 1.
They are 3 and -2, so I get $x^3-3x^2-10x+24=(x-4)(x+3)(x-2)$.
The roots are the values of x that make each factor zero, so they are $x=4, x=-3, x=2$.