Since [tex]f(6)=0[/tex], by the remainder theorem this means that x - 6 divides f(x) exactly. This means there are constants a, b, and c such that
[tex]\dfrac{f(x)}{x-6} = ax^2 + bx + c[/tex]
Multiplying both sides by x - 6 gives
[tex]f(x) = (x-6)(ax^2+bx+c) \\\\ 2x^3-19x^2+45x-18 = ax^3 + (b-6a)x^2 + (c-6b)x - 6c[/tex]
Then a, b, c satisfy
[tex]\begin{cases}a=2 \\ b-6a=-19 \\ c-6b=45 \\ -6c=-18\end{cases}[/tex]
and solving this system gives a = 2, b = -7, and c = 3.
So, we have
[tex]2x^3-19x^2+45x-18 = (x - 6) (2x^2 - 7x + 3)[/tex]
The quadratic term can be factored as
[tex]2x^2 - 7x + 3 = (2x - 1)(x - 3)[/tex]
which leaves us with
[tex]f(x) = (x-6)(2x-1)(x-3)[/tex]
so that the zeros of f(x) are 6, 1/2, and 3.
