### Instruction: Find the four roots of
\[2x^4 + x^3 - 6x^2 + x + 2 = 0.\]Enter the four roots (counting multiplicity), separated by commas.### Response: I notice that this is a quartic polynomial equation, which means it could have up to four complex roots.
I also notice that the coefficients add up to zero, which means that x = 1 is a root by the rational root theorem.
I can use synthetic division to divide the polynomial by x - 1 and get a cubic quotient.
I write the polynomial as 2x^4 + x^3 - 6x^2 + x + 2 = (x - 1)(2x^3 + 3x^2 - 3x - 2).
To find the roots of the cubic factor, I can use the rational root theorem again and look for possible rational roots of the form p/q, where p is a factor of -2 and q is a factor of 2.
The possible values of p/q are ±1, ±1/2, ±2, and ±2/2.
I can use synthetic division again to test these values and see if any of them are roots.
I find that x = -1/2 is a root, and the cubic factor becomes (2x^3 + 3x^2 - 3x - 2) = (2x + 1)(x^2 + x - 2).
To find the roots of the quadratic factor, I can use the quadratic formula or factor it further.
I see that the quadratic factor can be factored as (x^2 + x - 2) = (x + 2)(x - 1).
Therefore, the quartic polynomial has four roots: x = 1, x = -1/2, x = -2, and x = 1.
I check that these roots satisfy the original equation by plugging them in and verifying that they make it zero.
I write the final answer as 1, -1/2, -2, 1.