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if -6i is a root, then so is 6i since complex roots come in complex conjugate pairs
so you really have these 3 roots: -4, -6i, 6i
ok but how do you make that into an equation...its cubic because there are three roots correct?
which means x = -4, x = -6i, x = 6i x+4 = 0, x^2 = 36i^2, or x^2 = 36i^2 x+4=0 or x^2 = -36 or x^2 = -36 x+4=0 or x^2 = -36 x+4=0 or x^2+36 = 0 (x+4)(x^2+36) = 0 keep going...
and yes, all cubic equations have 3 complex roots
any nth degree polynomial has n complex roots (fundamental theorem of algebra)
ok I think I understand now. thank you
so is this the final answer? (x+4) (x+6)(x+6)=0
(x+4) (x+6)(x+6)=0 implies that x+6=0 ---> x =-6 is a root, but -6 is NOT a root
Too bad the question does not ask for what was provided. If we want the Complex Conjugate Pairs, the question MUST state "with real coefficients". Otherwise, (x+4)(x+6i)(x-2) = 0 would also be valid, or (x+4)(x+6i)^2 = 0 or infinitely many other options. Please do not try to factor x^2 + 36. That makes no sense.
ok so just leave it as (x+4)(x^2+36)
no you have to expand it out
Make sure the question says "with Real Coefficients". If it doesn't, you might want to have a chat with the author of the question.
(x+4)(x^2+36) x(x^2+36) + 4(x^2+36) what's next?
you somehow turned +4x^2 into -4x^2
otherwise, it's perfect
so y = x^3+4x^2+36x+144 is your cubic equation
oh ok it just clicked in my head. thanks I really appreciate your help
ok great, yw