Question 7732
  4 and 3 + i among its roots

 Another complex root must be the conjugate of 3+i, ie 3-i.
 The corresponding quadratic monic equation is x^2  -6 x + 10 = 0 (why?)
 [Hint: in one step sum and product of 3+i & 3-i]
 Hence, f(x) = (x-4)(x^2  -6 x + 10)


 Kenny