15. Let $a, b \in \mathbf{Z}$. If for any $x \leqslant 0$, we have
$$
(a x+2)\left(x^{2}+2 b\right) \leqslant 0 \text {, }
$$

then $a=$ $\qquad$ ,$b=$ $\qquad$ .
15. $a=1, b=-2$.

Let $x=0$, we know $b \leqslant 0$.
Consider the line $y=ax+2$ passing through the fixed point $(0,2)$, and the parabola $y=x^{2}+2b$ opening upwards, satisfying that for any $x \leqslant 0$, the corresponding points on the graphs are not on the same side of the $x$-axis.
Therefore, $-\sqrt{-2b}=\frac{2}{a}$.
Since $a, b \in \mathbf{Z}$, we have $a=1, b=-2$.