Increasing function

$f(x)=e^x(x^2+ax+2a)$ , so, $f'(x)= e^x(x^2+(a+2)x+3a$ . For $f(x)$ to be increasing $f'(x)\ge 0$ . That is $e^x(x^2+(a+2)x+3a) \ge 0$ $\implies x^2 +(a+2)x+3a) \ge 0$ as $e^x \ge 0$ , and $(x+(a+2)/2)^ 2 +(3a-((a+2)/2)^ 2)\ge 0$ , which leads us to $12a \ge ( a+2)^ 2$ $\implies a^2-8a+4 \le 0$ $\implies 0.53 \le a \le 7.46$ , Since $a$ is integer, so number of possible values of $a$ is $\boxed{7}$ .

$\begin{aligned} f(x) & = (x^2+ax+2a)e^x \\ f'(x) & = (x^2+ax+2a)e^x + (2x+a)e^x \\ & = \left(x^2+(a+2)x+3a\right)e^x \end{aligned}$

For $f(x)$ to be increasing, $f'(x) \ge 0$ . Since $e^x > 0$ for all $x$ , we need $g(x) = x^2+(a+2)x + 3a \ge 0$ . That is when $g(x)$ has no real roots or

$\begin{aligned} (a+2)^2 - 12a & \le 0 \\ a^2+4a+4 - 12a & \le 0 \\ a^2 - 8a + 4 & \le 0 \\ (a-4+2\sqrt 3)(a-4-2\sqrt 3) & \le 0 \end{aligned}$

$\implies 4-2\sqrt 3 \le a \le 4+2\sqrt 3 \approx 0.534 \le a \le 7.464$ or for integers $1 \le a \le 7$ , a total of $\boxed{7}$ integral $a$ .