When $\left(3+2 x+x^{2}\right)\left(1+m x+m^{2} x^{2}\right)$ is expanded and fully simplified, the coefficient of $x^{2}$ is equal to 1 . What is the sum of all possible values of $m$ ?
(A) $-\frac{4}{3}$
(B) $-\frac{2}{3}$
(C) 0
(D) $\frac{2}{3}$
(E) $\frac{4}{3}$
When $\left(3+2 x+x^{2}\right)\left(1+m x+m^{2} x^{2}\right)$ is expanded, the terms that include an $x^{2}$ will come from multiplying a constant with a term that includes $x^{2}$ or multiplying two terms that includes $x$. In other words, the term that includes $x^{2}$ will be

$$
3 \cdot m^{2} x^{2}+2 x \cdot m x+x^{2} \cdot 1=3 m^{2} x^{2}+2 m x^{2}+x^{2}=\left(3 m^{2}+2 m+1\right) x^{2}
$$

From the condition that the coefficient of this term equals 1 , we see that $3 m^{2}+2 m+1=1$ which gives $3 m^{2}+2 m=0$ or $m(3 m+2)=0$, which means that $m=0$ or $m=-\frac{2}{3}$.

The sum of these possible values of $m$ is $-\frac{2}{3}$.

ANswer: (B)