Question

Find a polynomial function f(x) with real coefficients that has the given zeros.-1,-1, 2+3i

Answer 1

Expand
\[
(x+1)(x+1) (x- 2 - 3 i)(x-2+3i)
\]

Answer 2

\[
(x+1) (x+1) ((x-2)-3 i) ((x-2)+3 i)=x^4-2 x^3+6 x^2+22 x+13
\]

Answer 3

If 2 + 3i is a root then 2-3i is also a root

Answer 4

The polynomial would be in the form; y = (x^2 - 1)(x^2 + bx + c) find b and c.
Sum of the 2 roots: (2 + 3i) + (2 - 3i) = 4 , then-> b = -4
Product of 2 roots: c / a = (2 + 3i)*(2 - 3i) = 4 - 9i^2) = (4 + 9) = 13 =
Then, c = 13. The trinomial is (x^2 - 4x + 13)
Check: D = b^2 - 4ac = -36 -> VD = 6i and -6i
2 roots: x1 = 2 + 3i, and x2 = 2 - 3i.
Finally: y = (x^2 - 1)(x^2 - 4x + 13)

Answer 5

Correction.
y = (x + 1)^2*(x^2 - 4x + 13)
because the 2 real roots are -1 and -1.