Question

what is polynomial with real coefficients and given zeros -1, 4-5i,4+5i

Answer 1

The polynomial has 3 x- intercepts. It would be a third degree function.
Since it has (-1) a x-intercept, then the polynomial can be factored by (x + 1). It would be under this form:
y = (x + 1)(ax^2 + bx + c ). Find b and c.
Sum of the 2 real roots: (4 + 5i) + (4 - 5i) = 8, then b = -8
Product of the 2 real roots: (4 + 5i)*(4 - 5i) = 16 - 25*i^2 = 16 + 25 = 41 = c/a. Then, c = 41.
Finally: y = (x + 1)*(x^2 - 8x + 41)
Check, by finding the 2 real roots of the trinomial.
Discriminant D = b^2 - 4ac = 64 - 164 = -100 -> VD = 10i and -10i
The real roots are: x1 = 8/2 + 10i/2 and x2 = 8/2 - 10i/2 -->
x1 = (4 + 5i) and x2 = (4 - 5i). Correct.

Answer 2

A different route to the same result:

a polynomial with zeros \(r_1,r_2,...,r_n\) can be written in factored form as
\[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is a constant chosen either for convenience or to make the polynomial fit a specified point. You don't have a point to hit, so \[P(x) = 1(x-(-1))(x-(4-5i))(x-(4+5i)) \]\[\qquad= (x+1)(x-4+5i)(x-4-5i)\]Expand that out to get the polynomial. Hint: multiply the two terms with the conjugate complex zeros first.