Answer
$f(x)= (x+3)(2x-1)(3x-1)$
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.
$k=-3$
$\left\{\begin{array}{lllllll}
-3 & | & 6 & 13 & -14 & 3 & \\
& & & -18 & 15 & -3 & \\
& & -- & -- & -- & -- & --\\
& & 6 & -5 & 1 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+3)(6x^{2}-5x+1)$
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 $6(1)=6$ whose sum is $-5$ ... are $-3$ and $-2$:
$6x^{2}-5x+1\\
=6x^{2}-3x-2x+1$
$=3x(2x-1)-(2x-1)\\=(2x-1)(3x-1)$
So,
$f(x)= (x+3)(2x-1)(3x-1)$
