n=3 4 and 5i are zeros f(2) =58

Answer 1

I assume your question is to find the polynomial with the given properties. When asking a question, please provide the full question.

Assuming integer coefficients, if 5i is a zero so is -5i. A degree 3 polynomial has three zeros and we now have all three: 4, 5i, -5i. Use the factored form:

f(x) = a(x-4)(x-5i)(x+5i)

To find the value of the constant a, plug in the given point f(2) = 58 and solve for a:

f(2) = a(2-4)(2-5i)(2+5i)

58 = a(-2)(4-25i^2)

58 = a(-58)

-1 = a

f(x) = -(x-4)(x-5i)(x+5i)

Multiply it out if you want it in standard form.