Answer
$f(x)= (x+5)(x-2)(6x-1)$
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}{lllllll}
-5 & | & 6 & 17 & -63 & 10 & \\
& & & -30 & 65 & -10 & \\
& & -- & -- & -- & -- & --\\
& & 6 & -13 & 2 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+5)(6x^{2}-13x+2)$
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 $6(2)=12$ whose sum is $-13$ ... are $-12$ and $-1$:
$6x^{2}-13x+2\\=6x^{2}-12x-x+2$
$=6x(x-2)-(x-2)\\=(x-2)(6x-1)$
So,
$f(x)= (x+5)(x-2)(6x-1)$
