Answer
$f(x)= (x-2)(x+3)(2x-5)$
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.
$k=2$
$\left\{\begin{array}{rrrrrrr}
2 & | & 2 & -3 & -17 & 30 & \\
& & & 4 & 2 & -30 & \\
& & -- & -- & -- & -- & --\\
& & 2 & 1 & -15 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x-2)(2x^{2}+x-15)$
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 $2(-15)=-30$ whose sum is $1$ ... are $+6$ and $-5$:
$2x^{2}+x-15\\
=2x^{2}+6x-5x-15$
$=2x(x+3)-5(x+3)\\=(x+3)(2x-5)$
So,
$f(x)= (x-2)(x+3)(2x-5)$