Answer
$f(x)= (x+5)(2x+3)(4x-1)$
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.
$k=-5$
$\left\{\begin{array}{rrrrrrr}
-5 & | & 8 & 50 & 47 & -15 & \\
& & & -40 & -50 & 15 & \\
& & -- & -- & -- & -- & --\\
& & 8 & 10 & -3 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+5)(8x^{2}+10x-3)$
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 $8(-3)=-24$ whose sum is $+10$ ... are $+12$ and $-2$:
$8x^{2}+10x-3\\=8x^{2}+12x-2x-3$
$=4x(2x+3)-(2x+3)\\=(2x+3)(4x-1)$
So,
$f(x)= (x+5)(2x+3)(4x-1)$