Answer
$-\infty \lt x \lt \infty$ or, $x \in (-\infty, \infty)$
Work Step by Step
As we know that $\Sigma_{k=0}^{\infty} \dfrac{x^k}{k!}=e^x$
In the given series we will replace $x$ by $-3x$.
Therefore, the given series can be written as:
$x^2 e^x=x^2 \Sigma_{k=0}^{\infty} \dfrac{(x)^k}{k!}=\Sigma_{k=0}^{\infty} \dfrac{(x)^{k+2}}{k!}$
The given series will converge when $-\infty \lt x \lt \infty$
This implies that $-\infty \lt x \lt \infty$ or, $x \in (-\infty, \infty)$
