Observe that
[tex]x^2 + 3x + 1 = (x^2 + 4x + 4) - x - 3 \\\\ ~~~~~~~~ = (x + 2)^2 - x - 3 \\\\ ~~~~~~~~ = (x + 2)^2 - (x + 2) - 1[/tex]
so that
[tex]P(x+2) = (x+2)^2 - (x+2) - 1 \implies P(x) = x^2 - x - 1[/tex]
Then
[tex]P(x - 1) = (x-1)^2 - (x-1) - 1 \\\\ ~~~~~~~~ = (x^2 - 2x + 1) - (x - 1) - 1 \\\\ ~~~~~~~~ = \boxed{x^2 - 3x + 1}[/tex]
Alternatively, we can avoid find [tex]P(x)[/tex] altogether and instead first write [tex]x-1[/tex] in terms of [tex]x+2[/tex] :
[tex]x - 1 = (x - 1) + 2 - 2 = (x + 2) - 1 - 2 = (x + 2) - 3[/tex]
Then
[tex]P(x - 1) = P((x + 2) - 3) \\\\ ~~~~~~~~ = (x - 3)^2 + 3 (x - 3) + 1 \\\\ ~~~~~~~~ = x^2 - 3x + 1[/tex]
