write a polynomial with real,rational coefficients whose zeros are 4i and -3?

Answer 1

complex roots occur in pairs.

So the roots of the required polynomial are 4i, -4i and -3

so the factors are (x+3), (x+4i) and (x-4i)

So the required polynomial = (x+3)*(x+4i)*(x-4i)

=>(x+3)(x^2 - 16i^2) = (x+3)(x^2 - 16(-1))

=>(x+3)(x^2 + 16)

=>x^3 + 16x + 3x^2 + 48

=>x^3 + 3x^2 + 16x + 48

Answer 2

since the roots of polynomials with real coeff must have conjugate pairs, the eq have three roots
p(x) = (x+4i)(x-4i)(x+3)
= (x^2 +16)(x+3)
=x^3 + 3x^2 + 16x + 48