You could rewrite [tex]F(x)[/tex] as
[tex]\dfrac{x^2+3x+2}{x^2+5x+4}=\dfrac{(x+1)(x+2)}{(x+1)(x+4)}[/tex]
and be tempted to cancel out the factors of [tex]x+1[/tex]. But this cancellation is only valid when [tex]x\neq-1[/tex].
When [tex]x=-1[/tex], you end up with the indeterminate form [tex]\dfrac00[/tex], which is why [tex]-1[/tex] is not a zero.
