Answer
$f(x)=(x-2)(x+1)(x-7)$
Work Step by Step
We are given the polynomial function:
$$f(x)=x^3-8x^2+5x+14.$$
$\bf{Step\text{ }1}$
First we will list the possible rational zeros. The leading coefficient is $1$ and the constant term is $18$. So the possible rational zeros are:
$$\pm \pm 1,\pm2,\pm 7,\pm 14.$$
$\bf{Step\text{ }2}$
Test these zeros using synthetic division:
Test $x=1$:
Because we got that the remainder is $12\not=0$, $1$ is not a zero of $f$.
Test $x=2$:
Since the remainder is $0$ we can write:
$$f(x)=(x-2)(x^2-6x-7).$$
$\bf{Step\text{ }3}$
Factor the polynomial using the Factor Theorem:
$$\begin{align*}
f(x)&=(x-2)((x^2+x)-(7x+7))\\
&=(x-2)(x(x+1)-7(x+1)\\
&=(x-2)(x+1)(x-7).
\end{align*}$$
