### Instruction: Find the roots of $x^3-3x^2-10x+24$. Enter your answer as a list of numbers separated by commas.### Response: I need to find the values of $x$ that make this polynomial equal to zero.
A possible strategy is to factor the polynomial and use the zero product property.
To factor the polynomial, I can try to find a rational root first, using the rational root theorem.
The rational root theorem says that if $p/q$ is a rational root of the polynomial, where $p$ and $q$ are relatively prime integers, then $p$ must divide the constant term and $q$ must divide the leading coefficient.
In this case, the constant term is $24$ and the leading coefficient is $1$, so the possible values of $p$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$ and the possible values of $q$ 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 try plugging in these values into the polynomial and see if any of them make it zero.
I start with $x=1$.
I get $1^3-3(1)^2-10(1)+24 = 1-3-10+24 = 12 \neq 0$, so $x=1$ is not a root.
I try $x=-1$.
I get $(-1)^3-3(-1)^2-10(-1)+24 = -1-3+10+24 = 30 \neq 0$, so $x=-1$ is not a root either.
I try $x=2$.
I get $2^3-3(2)^2-10(2)+24 = 8-12-20+24 = 0$, so $x=2$ is a root.
Since $x=2$ is a root, I can use synthetic division to divide the polynomial by $x-2$ and get a quadratic quotient.
I write the coefficients of the polynomial in a row: $1, -3, -10, 24$.
I write $2$ below the first coefficient and draw a vertical line after it.
I bring down the first coefficient and write it below the line: $1$.
I multiply $2$ by $1$ and write the result below the second coefficient: $2$.
I add $-3$ and $2$ and write the result below the line: $-1$.
I multiply $2$ by $-1$ and write the result below the third coefficient: $-2$.
I add $-10$ and $-2$ and write the result below the line: $-12$.
I multiply $2$ by $-12$ and write the result below the fourth coefficient: $-24$.
I add $24$ and $-24$ and write the result below the line: $0$.
The numbers below the line are the coefficients of the quotient polynomial, which is $x^2-x-12$.
To find the other roots, I need to solve $x^2-x-12=0$.
I can factor this quadratic equation as $(x-4)(x+3)=0$.
Using the zero product property, I get $x-4=0$ or $x+3=0$.
Solving for $x$, I get $x=4$ or $x=-3$.
Therefore, the roots of the original polynomial are $x=2, x=4, x=-3$.