William W. answered • 02/19/19

Math and science made easy - learn from a retired engineer

Since 9+5i is a zero, then 9-5i must also be a zero. That means you can write the polynomial as a series of binomial expression, each equating to one of the roots: f(n) = (n-3) (n+3) [n-(9+5i)] [n-(9-5i)} to account for the roots but there could also be a multiplier A in front of these meaning the polynomial could be f(n) = A(n-3) (n+3) [n-(9+5i)] [n-(9-5i)}

To find A, you must plug in x=-1 for f(x) = 250 and solve for A:

250 = A(-1-3) (-1+3) [-1-(9+5i)] [-1-(9-5i)}