The Third Root

By the Complex Conjugate Root Theorem, if $f(3+4i) = 0$ , then also $f(3-4i)=0$ , which gives us a bigger clue of what $f(x)$ looks like.

Let by now $f(x) = k \times (x^2 - 6x + 25) \times (x - R)$ , where $k$ is a non-zero constant and $R$ is the real root of $f(x)$ . By plucking in values, we get that $20 = 20k \times (1 - R)$ and $51 = 17k \times (2-R)$ . Isolating k, we find that $k = \frac{1}{1-R} = \frac{3}{2-R}$ . Thus $R = \frac{1}{2}$ and $k =2$ .

Now we have $f(x)$ 's full form, which is $f(x) = 2x^3 - 13x^2 + 56x - 25$ . And now it's easy: for $x=3$ , $\boxed{f(3) = 80}$ .