Answer
$1+\dfrac{x^2}{2}-\dfrac{2x^3}{3}; $ Converges absolutely for all values of $x$.
Work Step by Step
We have $f(x)=(1-x-x^2)e^x$
Now, distribute the polynomial $1-x+x^2+x-x^2+x^3+\dfrac{x^2}{2} -\dfrac{x^3}{2} +\dfrac{x^4}{2}+....$
or, $= (1-x-x^2)x+(1-x-x^2)(x^2/2)+(1-x-x^2)(x^3/6)+.....$
Now, we have the Maclaurin's series for $f(x)$ as follows:
$f(x)=1+\dfrac{x^2}{2}-\dfrac{2x^3}{3}; $ Converges absolutely for all values of $x$.
