Question

Find the cubic polynomial in standard form with real coefficients, having the given zeros. Let the leading coefficient be 1.

4 and 2+i

Answer 1

To get real coefficients, you need to multiply the zero 2+i with its complex conjugate, 2-i:
$$
\large{
(x-4)(x-(2+i))(x - (2-i))=0\\
\implies (x^2-x(2+i)-4x+4(2+i))(x - (2-i))=0\\
\implies x^3-8 x^2+21 x-20=0
}
$$
This give zeros at 4 and 2+i. But also, a necessary zero at 2-i to make the coefficients real.

Answer 2

Make sense?