Question 986063
If  a polynomial has real coefficients, then if a complex
number a + bi is a root, then so is its conjugate a - bi is also the root

given roots are -3 and  -4i
conjugate -4i  = +4i
Hence 3 roots are -3,-4i,4i
If a polynomial f(x) has roots r1, r2, r3, ···, rn
and leading coefficient a, then

f(x) = a(x-r1)(x-r2)···(x-rn)
for this problem a=1
f(x)= (x-(-3))(x-(-4i))(x-4i)
f(x)= (x+3)(x+4i)(x-4i)
f(x)= (x+3)(x(x-4i)+4i(x-4i)
{{{f(x)= (x+3)(x^2-4ix+4ix-16i^2)}}}
{{{F(x)= (x+3)(x^2+16)}}}       (Sine i^2=-1)
{{{f(x)= x(x^2+16)+3(x^2+16)}}}
{{{f(x)= x^3+16x+3x^2+48}}}
 {{{f(x)= x^3+3x^2+16x+48}}}