complex roots always come in conjugate pairs
so roots are 9, 3 ± 5i, ± 9i
fo f(x) = (x-9)(x-3-5i)(x-3+5i)(x+9i)x-9i)
= (x-9)(x^2 - 6x + 34)(x^2 + 91)
expand if need be
Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f.
Degree 5; zeros: 9, 3+5i, -9i
1 answer