Question 756608
+i and -7-i are also zeros because of -i and -7+i being zeros.  You can begin forming your function using binomial factors this way:


{{{f(x)=(x-2)(x+i)(x-i)(x-(-7+i))(x-(-7-i))}}}


Next perform multiplications to achieve real coefficients when simplified, beginning with the binomial factors containing constant terms of conjugate complex pairs.


{{{(x+i)(x-i)=x^2+1}}}
and 
{{{(x-(-7+i))(x-(-7-i))=(x+7-i)(x+7+i)=(x+7)^2+1=x^2+14x+49+1=x^2+14x+50}}}


Writing the update for the function,
{{{f(x)=(x-2)(x^2+1)(x^2+14x+50)}}} which is still not yet in the general form you want, but is still correct.  Continue multiplying all the way if you want this in general form.