Complex Numbers: Roots of a cubic formula help

Let P(z)=z^3+az^2+bz+c, where a, b, and c E R. Two of the roots of P(z)=0 are -2 and (-3+2i). Find the value of a, of b and of c.
Please explain if you can, and if you have a link to a website that can help with these problems I would appreciate if you could give that too.

Since coefficients are real, and since one of the roots is not, then the last root must be complex conjugate of −3+2i, i.e. −3−2i

Therefore, factors of P(z) are (z + 2), (z + 3 − 2i), (z + 3 + 2i)

P(z) = (z + 2) (z + 3 - 2i) (z + 3 + 2i)
P(z) = (z + 2) ((z + 3)² - (2i)²))
P(z) = (z + 2) (z² + 6z + 9 + 4)
P(z) = z (z² + 6z + 13) + 2 (z² + 6z + 13)
P(z) = z³ + 6z² + 13z + 2z² + 12z + 26
P(z) = z³ + 8z² + 25z + 26

a = 8, b = 25, c = 26

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