how would i find a 4th degree polynomial that has zeros of -1, -1, and 3i ??

Question

i am assuming that 3i means +/- 3i...which means there must be an (x^2+9) somewhere...but im jsut guessing lol

Answer 1

Yes if 3i is a zero, -3i will also be.

So it's (x+1)^2 * (x^2 + 9)

Answer 2

If you want a real 4th degree polynomial, with four roots, the 4 roots would be
-1, -1, 3i and the conjugate, -3i (to get real polynomial)
so the polynomial is
(x -3i)(x+3i)(x+1)(x+1)
= (x^2 + 9)(x+1)^2
= (x^2 + 9)(x^2 + 2x + 1)
= x^4 + 2x^3 + x^2 + 9x^2 + 18x + 9
= x^4 + 2x^3 +10x^2 + 18x + 9

Answer 3

p(x) = (x^2 + 9)*(x+1)^2
if your assumption of +/- 3i is correct if not
p(x) =(x-a)(x-3i)*(x+1)^2
where a is the fourth zero which obviously reduces to the above equation in the case a=-3i

Answer 4

In general, to find a polynomial with any given roots, simply multiply binomials together made up of (X - one root)
In your problem:
Multiply (x +1)*(x+1)*(x - 3i)*(x + 3i)