Question 764825
Avoiding to give intricately fancy reasoning, {{{highlight(2+3i)}}} is one other zero because complex solutions to polynomial equations occur in conjugate pairs.  Knowing that both 2-3i and 2+3i are zeros of f(x), you should be able to form the resulting quadratic factor which those two zeros form.  You might then want to divide f(x) by this quadratic factor and possibly find the other two zeros of f(x).


Note that you have {{{(x-(2-3i))(x-(2+3i))=(x-2+3i)(x-2-3i)=((x-2)+3i)((x-2)-3i)}}}
={{{(x-2)^2-(3i)^2=(x-2)^2+9=x^2-4x+4+9=highlight(x^2-4x+13)}}}