Ray B. answered 08/12/20
Passionate about tutoring Mathematics
The 3 roots of f(x) are 3, 3+5i and 3-5i. By the fundamental theorem of algebra, for some constant k,
f(x)= k (x+3)(x-(3+5i))(x-(3-5i)) = k (x+3)(x^2-6x+34)= k (x^3-3x^2+16x+102). We calculate that
f(1)=k(1-3+16+102)=116k and since it is given that f(1)=116, then k=1 and
f(x) = x^3 - 3x^2 + 16x + 102