Answer
$f(x)= (x-1)(x-2)(2x+3)$
Work Step by Step
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this first.
$k=1$
$\left\{\begin{array}{lllllll}
1 & | & 2 & -3 & -5 & 6 & \\
& & & 2 & -1 & -6 & \\
& & -- & -- & -- & -- & --\\
& & 2 & -1 & -6 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x-1)(2x^{2}-x-6)$
We now factor the trinomial $ax^{2}+bx+c$ by searching for
two factors of $ac$ whose sum is $b.$
If they exist, rewrite $bx$ and factor in pairs.
Factors of $2(-6)=-12$ whose sum is $-1$ ... are $-4$ and $+3$:
$2x^{2}-x-6=2x^{2}-4x+3x-6$
$=2x(x-2)+3(x-2)=(x-2)(2x+3)$
So,
$f(x)= (x-1)(x-2)(2x+3)$