Solving the inequality $x^2 + 4x + 3 > 0$ So here is the question:

For what values of $x$ is $x^2 + 4x + 3 > 0$?

So I decided to factor the right-hand side into:
$$(x+1)(x+3) > 0$$
getting:
$$x+1>0 \text{ or } x+3 > 0 \implies x > -1 \text{ or } x > -3$$
But this is incorrect because I graphed the function and I saw $x\not>$ $-3$ for some values of $x$ as the graph dips below the $x$-axis. Where I am going wrong? Could someone help?

Option:
$(x+2)^2>1;$
$\Rightarrow:$ $|x+2|>1$, i. e.
$x+2>1$, or $x+2<-1.$
