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