Question

find a second degree polynomial with real coefficients that has a root at x=2i

Answer 1

If \(2i\) is a root then so is \(-2i\).

Answer 2

Remember that if such a polynomial has root 2i then it also has -2i as a root.
Now you can write a factored form of it as follows:

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

Expand (FOIL) and there is your polynomial.

In fact, any number a (nonzero) can be put in front of the brackets:

a(x-2i)(x+2i). There are infinitely many functions. Leaving a=1 gives you the simplest one...

Answer 3

I cant foil an x w 2i

Answer 4

Yes you can! it is of the form (a+b)(a-b)=a²-b² (remember that one?)
So (x-2i)(x+2i) = x² - (4i²) = x²-4*-1=x²+4

Answer 5

so thatd b my final answer?

Answer 6

Yes, that is the polynomial of degree 2 with 2i as a root (and also -2i).

Answer 7

so my final answer is x^2+4

Answer 8

Yes! Final, definite, 100%!