Question

Find a cubic equation that has -2 and 1-2i as roots.

Answer 1

Okay, so we have a cubic equation at hand.
Do you know how many solutions a cubic equations has?
Three, it has three real or a mix of real and imaginary solutions.

Now, there is one very neat thing about this.
Whenever you have the imaginary solution, its conjugate must also be a root. That means, imaginary roots occur in pairs. That means in a cubic polynomial if one root is real and the other is imaginary the third and final root of the cubic polynomial MUST be imaginary.

And what is the third root? Well, it is the conjugate of the root you are already given.
So find that root and then tell me!

Getting this till here? :)

Answer 2

So the third root would be 1+2i?

Answer 3

yes :)
Do you understand how to write the equation now that you have your roots?

Answer 4

i tried to do it out again when i thought i did and it didn't come out right

Answer 5

If a,b are roots of f(x), then f(x) = (x-a)(x-b)
So, if -2, 1-2i, 1+2i are roots of any polynomial p(x) = {x - (-2)}{x - (1-2i)}{x - (1+2i)}
Expand it and you'll have the polynomial you need.

Answer 6

Thank you so much!!!

Answer 7

You're welcome. :)