Originally Posted by

**Archie Meade** No, one of the roots is $\displaystyle x=-2$

and it's the only real root.

There are also two complex roots, which are the solution to the 2nd factor being zero.

$\displaystyle x^3+2x^2+x+2=(x+2)\left(x^2+1\right)$

$\displaystyle x^3+2x^2+x+2=0\Rightarrow\ (x+2)\left(x^2+1\right)=0\Rightarrow\ x+2=0\;\;\;or\;\;\;x^2+1=0$

$\displaystyle (x+2)=0\Rightarrow\ x=-2$

If you are familiar with complex numbers, then you need to solve $\displaystyle x^2+1=0$

Otherwise, you need to be aware that $\displaystyle x=-2$ is the only real root

since there is no real solution to $\displaystyle x^2+1=0$

since $\displaystyle x^2=-1$ is the solution to that.