Question 731291
TWO of the roots are -2i and +2i.  The -2i was given.  The +2i is a root because complex roots to polynomial equations occur in conjugate pairs.



Using those two roots, find the corresponding quadratic factors of f(x), and use polynomial division to find the other quadratic factor.



{{{(x+2i)(x-2i)=x^2-(2i)^2}}}=x^2-(-1)*4={{{highlight(x^2+4)}}}
You want to divide {{{x^4-3x^3-12x-16}}} by {{{x^2+4}}}.  You should use the divisor in the form, {{{x^2+0*x+4}}}.



The polynomial division, {{{(x^4-3x^3+0*x^2-12x-16)/(x^2+0*x+4)=x^2-3x-4}}}
and that quadratic polynomial is factorable into {{{(x+1)(x-4)}}}.



So the function is factorable into {{{highlight(f(x)=(x^2+4)(x+1)(x-4))}}}.  The roots found which were not given in the description  but were asked for are -1 and +4.