3. The graph of the quadratic function $y=a x^{2}+b x+c$ is shown in Figure 2. Let
$$
k=\frac{b(2 a+b)(a+b+c)}{c(2 a-b)(a-b+c)} .
$$

Then the sign of $k$ is
3. $k>0$.

From Figure 2, we can see that the intersection point $(0, c)$ of the function $y=a x^{2}+b x+c$ with the $y$-axis is below the $x$-axis. Therefore, $c<0$;
When $x=-1$, $y=a-b+c<0$.