Question

Find all the zeros of f(x) help please

Answer 1

Since $f(6)=0$, by the remainder theorem this means that x - 6 divides f(x) exactly. This means there are constants a, b, and c such that

$\dfrac{f(x)}{x-6} = ax^2 + bx + c$

Multiplying both sides by x - 6 gives

$f(x) = (x-6)(ax^2+bx+c) \\\\ 2x^3-19x^2+45x-18 = ax^3 + (b-6a)x^2 + (c-6b)x - 6c$

Then a, b, c satisfy

$\begin{cases}a=2 \\ b-6a=-19 \\ c-6b=45 \\ -6c=-18\end{cases}$

and solving this system gives a = 2, b = -7, and c = 3.

So, we have

$2x^3-19x^2+45x-18 = (x - 6) (2x^2 - 7x + 3)$

The quadratic term can be factored as

$2x^2 - 7x + 3 = (2x - 1)(x - 3)$

which leaves us with

$f(x) = (x-6)(2x-1)(x-3)$

so that the zeros of f(x) are 6, 1/2, and 3.