Answer
$\{-16,-8,8,16\}$
Work Step by Step
We are given the trinomial:
$$p(x)=3x^2+bx+5.$$
As the trinomial is a quadratic function, it can be factored if it has real root(s)
$$\begin{align*}
b^2-4(3)(5)&\geq 0\\
b^2-60&\geq 0.
\end{align*}$$
So the trinomial can be factored for any real value of $b$ in the set $(-\infty,4\sqrt{15}]\cup[4\sqrt{15},\infty)$.
$$\begin{align*}
3x^2+bx+5&=(3x+u)(x+v)\\
3x^2+bx+5&=3x^2+(3v+u)x+uv\\
b&=3v+u\\
uv&=5\\
u=1,v=5\text{ or }u=-1,v=-5&\text{ or }u=5,v=1\text{ or }u=-5,v=-1\\
b=3(5)+(1)\text{ or }b=3(-5)+(-1)&\text{ or }b=3(1)+5\text{ or }b=3(-1)+(-5)\\
b=16\text{ or }b=-16&\text{ or }b=8\text{ or }b=-8.
\end{align*}$$
All $4$ values are inside the domain. So the integer values of $b$ are $\{-16,-8,8,16\}$.
