Factor the polynomial as $\left(x^{2}-4\right)\left(x^{2}-3\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-4\textless0\textless x^{2}-3$. Solving this inequality, we find $3\textless x^{2}\textless4$. There is no integer $x$ that satisfies this inequality. 
For how many integers $x$ is the number $x^{4}-7x^{2}+12$ negative? (  2014 AMC 10B Problems, Question \#20) 
1: $$2$$ 
2: $$4$$ 
3: $$0$$ 
4: $$3$$ 
5: $$1$$ 