Answer
$f(x)= (x+4)(2x+1)(3x-1)$
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.
$k=-4$
$\left\{\begin{array}{lllllll}
-4 & | & 6 & 25 & 3 & -4 & \\
& & & -24 & -4 & 4 & \\
& & -- & -- & -- & -- & --\\
& & 6 & 1 & -1 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+4)(6x^{2}+x-1)$
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 $+1$ ... are $-2$ and $+3$:
$6x^{2}+x-1\\=6x^{2}+3x-2x-1$
$=3x(2x+1)-(2x+1)\\=(2x+1)(3x-1)$
So,
$f(x)= (x+4)(2x+1)(3x-1)$
