### 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: This is a quartic equation, so it might have up to four roots.
To find them, I will try to factor the polynomial into simpler expressions.
One way to factor a polynomial is to use the rational root theorem, which says that any rational root of the form p/q, where p and q are integers, must satisfy p is a factor of the constant term and q is a factor of the leading coefficient.
In this case, the constant term is 2 and the leading coefficient is 2, so the possible rational roots are ±1, ±2, ±1/2.
I can test each of these values by plugging them into the polynomial and seeing if they make it zero.
If they do, then they are roots and I can use synthetic division to find the other factors.
I start with 1.
I plug it into the polynomial and get 2(1)^4 + (1)^3 - 6(1)^2 + (1) + 2, which simplifies to 0.
So 1 is a root and a factor of the polynomial.
I use synthetic division to divide the polynomial by x - 1 and get 2x^3 + 3x^2 - 3x - 2 as the quotient.
Now I have 2x^4 + x^3 - 6x^2 + x + 2 = (x - 1)(2x^3 + 3x^2 - 3x - 2).
I repeat the same process with the quotient polynomial.
I try the same possible rational roots as before, starting with 1 again.
I plug 1 into the quotient polynomial and get 2(1)^3 + 3(1)^2 - 3(1) - 2, which simplifies to 0 again.
So 1 is another root and a factor of the quotient polynomial.
I use synthetic division to divide the quotient polynomial by x - 1 and get 2x^2 + 5x + 2 as the new quotient.
Now I have 2x^4 + x^3 - 6x^2 + x + 2 = (x - 1)^2(2x^2 + 5x + 2).
I try to factor the quadratic quotient further.
I look for two numbers that multiply to 4 and add to 5.
I see that 4 and 1 work.
I write 2x^2 + 5x + 2 = 2x^2 + 4x + x + 2 = 2x(x + 2) + (x + 2) = (2x + 1)(x + 2).
Now I have 2x^4 + x^3 - 6x^2 + x + 2 = (x - 1)^2(2x + 1)(x + 2).
I set each factor equal to zero and solve for x. I get x = 1, x = -1/2, and x = -2.
These are the four roots of the polynomial, counting multiplicity.