### Instruction: Find all roots of the polynomial $x^4+2x^3-7x^2-8x+12$. Enter your answer as a list of numbers separated by commas.### Response: I know that if a polynomial has a root $r$, then $x-r$ is a factor of the polynomial.
So I want to find numbers $r$ that make the polynomial zero.
One way to do that is to try some possible rational roots, which are of the form $\pm p/q$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.
The constant term is 12, so its factors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
The leading coefficient is 1, so its factors are $\pm 1$.
That gives me a lot of possible rational roots to try.
I can use synthetic division to quickly check if a possible root is actually a root.
For example, let me try $r=1$.
I write 1 in the left corner, and the coefficients of the polynomial in a row: 1, 2, -7, -8, 12.
Then I bring down the first coefficient, which is 1, and multiply it by the root, which is also 1, and write the product below the second coefficient.
That gives me 1.
I add 1 and 2 and get 3, and write it below the line.
I repeat this process for the rest of the coefficients, multiplying by the root and adding to the next coefficient.
I get 1, 3, -4, -12, 0.
The last number is zero, which means that $r=1$ is indeed a root, and $x-1$ is a factor of the polynomial.
Now I have a simpler polynomial, which is the quotient of the original polynomial by $x-1$.
It is $x^3+3x^2-4x-12$.
I can repeat the process of synthetic division to find another root.
I notice that $r=-3$ seems like a good candidate, because it would make the last term of the quotient zero.
So I try $r=-3$ using synthetic division.
I write -3 in the left corner, and the coefficients of the quotient in a row: 1, 3, -4, -12.
I bring down the first coefficient, which is 1, and multiply it by the root, which is -3, and write the product below the second coefficient.
That gives me -3.
I add -3 and 3 and get 0, and write it below the line.
I repeat this process for the rest of the coefficients, multiplying by the root and adding to the next coefficient.
I get 1, 0, -4, 0.
The last number is zero, which means that $r=-3$ is indeed a root, and $x+3$ is a factor of the polynomial.
Now I have an even simpler polynomial, which is the quotient of the previous polynomial by $x+3$.
It is $x^2-4$.
This is a difference of squares, so I can easily factor it as $(x-2)(x+2)$.
That means that $r=2$ and $r=-2$ are also roots of the original polynomial.
So I have found all four roots of the polynomial: 1, -3, 2, and -2.
To write them as a list of numbers separated by commas, I just put them in ascending order: -3, -2, 1, 2.