Write a polynomial in standard form , with a zero 3i, an x-intercept 6, a multiplicity 2, and a leading coefficient 1?

Answer 1

Complex roots appear in conjugate form, so we have roots 3i and -3i
giving us two factors of (x-3i)(x+3i) = x^2 + 9
an x-intercept of 6 with multiplicity of 2 yields (x-6)(x-6)

I would state the polynomial as
y = (x^2 + 9)(x-6)^2 , expanding this would be superfluous.

Answer 2

Thanks so much