### 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: To find the roots of a polynomial, I want to factor it as much as possible.
A good strategy is to look for possible rational roots using the rational root theorem, which says that any rational root of the form $\frac{p}{q}$ must satisfy $p$ divides the constant term and $q$ divides the leading coefficient.
In this case, the constant term is 12 and the leading coefficient is 1, so the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
I can test each of these candidates using synthetic division or the remainder theorem, which says that $f(a)$ is the remainder when $f(x)$ is divided by $x-a$.
I start with 1, and find that $f(1) = 1^4+2(1)^3-7(1)^2-8(1)+12 = 0$, so 1 is a root and $x-1$ is a factor.
I can divide $x^4+2x^3-7x^2-8x+12$ by $x-1$ using synthetic division or long division, and I get $x^3+3x^2-4x-12$.
I repeat the process with the other possible rational roots, and I find that -2 is also a root, since $f(-2) = 0$.
I divide $x^3+3x^2-4x-12$ by $x+2$, and I get $x^2+x-6$.
This is a quadratic, so I can use the quadratic formula or factoring to find the remaining roots.
I notice that $x^2+x-6$ can be factored as $(x+3)(x-2)$, so the roots are -3 and 2.
I have found all four roots of the polynomial, and they are 1, -2, -3, and 2.