### 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 need to factor it into linear and/or irreducible quadratic factors.
I can start by looking for possible rational roots using the rational root theorem, which says that any rational root of the form p/q (where p and q are integers and q is not zero) must be a factor of the constant term (12) divided by a factor of the leading coefficient (1).
So the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
I can try plugging in each of these values into the polynomial and see if any of them make it zero.
If so, then they are roots and I can factor out a linear term of the form (x-r).
I notice that if I plug in x = -2, I get $(-2)^4+2(-2)^3-7(-2)^2-8(-2)+12 = 16-16-28+16+12 = 0$.
So -2 is a root and I can factor out (x+2) from the polynomial using long division or synthetic division.
I get $x^4+2x^3-7x^2-8x+12 = (x+2)(x^3-7x+6)$.
Now I have a cubic factor to deal with.
I can try the same process again and look for rational roots of the cubic.
The possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$.
I see that if I plug in x = 1, I get $(1)^3-7(1)+6 = 0$.
So 1 is another root and I can factor out (x-1) from the cubic using long division or synthetic division.
I get $x^3-7x+6 = (x-1)(x^2+x-6)$.
Now I have a quadratic factor to deal with.
I can use the quadratic formula or try to factor it further.
I notice that the quadratic factor can be factored as $(x+3)(x-2)$ by looking for two numbers that multiply to -6 and add to 1.
So I have $x^4+2x^3-7x^2-8x+12 = (x+2)(x-1)(x+3)(x-2)$.
This means that the roots of the polynomial are the values of x that make any of these factors zero.
Therefore, the roots are -2, 1, -3, and 2.