Question

If P(x + 2) = x² + 3x + 1, then find P(x-1)?

A) x²+3x+1
B) x²-3x+1
C) x-2
D) X-3

Answer 1

Observe that

`x^2 + 3x + 1 = (x^2 + 4x + 4) - x - 3 \\\\ ~~~~~~~~ = (x + 2)^2 - x - 3 \\\\ ~~~~~~~~ = (x + 2)^2 - (x + 2) - 1`

so that

`P(x+2) = (x+2)^2 - (x+2) - 1 \implies P(x) = x^2 - x - 1`

Then

`P(x - 1) = (x-1)^2 - (x-1) - 1 \\\\ ~~~~~~~~ = (x^2 - 2x + 1) - (x - 1) - 1 \\\\ ~~~~~~~~ = \boxed{x^2 - 3x + 1}`

Alternatively, we can avoid find
`P(x)` altogether and instead first write
`x-1` in terms of
`x+2` :

`x - 1 = (x - 1) + 2 - 2 = (x + 2) - 1 - 2 = (x + 2) - 3`

Then

`P(x - 1) = P((x + 2) - 3) \\\\ ~~~~~~~~ = (x - 3)^2 + 3 (x - 3) + 1 \\\\ ~~~~~~~~ = x^2 - 3x + 1`