Help with cubic equations...?

Question

the question asks to find a cubic equation with roots 2 + 3i and -2. can someone help me get started on this...any help is greatly appreciated.

Answer 1

I'm guessing the cubic is equation supposed to have real numbers in it.  

This is helpful because the roots of a cubic equation with real number coefficients occur in conjugate pairs: if 2 + 3i is a root, 2 - 3i must also be a root.  So one candidate would be

(x - (2+3i))(x - (2-3i))(x - (-2)).

This does indeed have real coefficients: if you compute the product

(x - (2+3i))(x - (2-3i)) you should get

x^2 - [(2 + 3i) + (2 - 3i)] x + (2+3i)(2-3i)

In the x term notice that the i's cancel and you are left with 4 in the brackets.  And (2+3i)(2-3i) is 4 - 6i + 6i -9i^2 or 4 + 9 or 13.  So those two terms together multiply to give

x^2 - 4x + 13.

This is a quadratic with 2+3i and 2-3i as roots.  If you multiply it by x-(-2) you will get a cubic with 2+3i, 2-3i, and -2 as roots. I'll leave the expansion of that to you.

(If you aren't required to use real number coefficients, the polynomial (x-(2+3i))^2 (x-(-2)) would work fine.  But it has complex numbers in it.)