### 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 equation, I could try to factor it into a product of linear and quadratic terms.
I notice that the constant term is 24, which has a lot of possible factors.
I also notice that the coefficient of $x^3$ is 1, which makes factoring easier.
I can use the rational root theorem to list all the possible rational roots of the equation, which 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 and see if any of them are actually roots.
I start with 1, and write the coefficients of the equation in a row: 1, -3, -10, 24.
I bring down the first coefficient, 1, and write it below the row.
I multiply 1 by 1, and write the result, 1, above the second coefficient.
I add -3 and 1, and write the result, -2, below the row.
I multiply 1 by -2, and write the result, -2, above the third coefficient.
I add -10 and -2, and write the result, -12, below the row.
I multiply 1 by -12, and write the result, -12, above the fourth coefficient.
I add 24 and -12, and write the result, 12, below the row.
The last number in the bottom row is 12, which is not zero, so 1 is not a root of the equation.
I repeat this process with -1, and write the coefficients in a row: 1, -3, -10, 24.
I bring down the first coefficient, 1, and write it below the row.
I multiply -1 by 1, and write the result, -1, above the second coefficient.
I add -3 and -1, and write the result, -4, below the row.
I multiply -1 by -4, and write the result, 4, above the third coefficient.
I add -10 and 4, and write the result, -6, below the row.
I multiply -1 by -6, and write the result, 6, above the fourth coefficient.
I add 24 and 6, and write the result, 30, below the row.
The last number in the bottom row is 30, which is not zero, so -1 is not a root of the equation.
I continue this process with 2, and write the coefficients in a row: 1, -3, -10, 24.
I bring down the first coefficient, 1, and write it below the row.
I multiply 2 by 1, and write the result, 2, above the second coefficient.
I add -3 and 2, and write the result, -1, below the row.
I multiply 2 by -1, and write the result, -2, above the third coefficient.
I add -10 and -2, and write the result, -12, below the row.
I multiply 2 by -12, and write the result, -24, above the fourth coefficient.
I add 24 and -24, and write the result, 0, below the row.
The last number in the bottom row is 0, which means 2 is a root of the equation.
I also notice that the other numbers in the bottom row, 1, -1, -12, are the coefficients of a quadratic equation: $x^2-x-12$.
I can factor this equation into $(x-4)(x+3)$.
This means that the original equation can be written as $(x-2)(x-4)(x+3)$.
Therefore, the roots of the equation are 2, 4, and -3.