Question 977052
Since it has rational coefficients, the complex conjugate of {{{3i}}} is also a root.
So {{{(x-3i)(x+3i)=x^2+9}}} is a factor of f(x).
Divide the factor from the polynomial to find the remaining factor.
{{{(4x^3+5x^2+36x+45)/(x^2+9)}}}
You know that {{{4x}}} is the leading term of the factor because of the {{{4x^3}}} leading term in f(x).
{{{4x(x^2+9)=4x^2+36x}}}
Which leaves,
{{{4x^3+5x^2+36x+45-(4x^2+36x)=5x^2+45}}}
Then the constant term of the factor would be {{{5}}} since,
{{{5(x^2+9)=5x^2+45}}}
So the remaining factor is, {{{4x+5}}}
{{{(4x^3+5x^2+36x+45)=(x^2+9)(4x+5)}}}
{{{(4x^3+5x^2+36x+45)=(x+3i)(x-3i)(4x+5)}}}
{{{4x+5=0}}}
{{{4x=-5}}}
{{{x=-5/4}}}
So the zeros are 3i,-3i, and -5/4.