1. The set of integer solutions to the inequality $\left(x^{2}+6 x+8\right)\left(x^{2}-4 x+3\right)<0$ is $(\quad)$.
A. $\{-3,1\}$
B. $\{1,2\}$
C. $\{-3,2\}$
D. $\{-3,1,2\}$
1. C. $(x+2)(x+4)(x-1)(x-3)<0$, using the sign chart method we get $-4<x<-2$ or $1<x<3$.