Please help with synthetic division?

I get 0 as the final remainder, simply by plugging x=-2 into the original polynomial. Did you forget to put a 0x^2 placeholder term in the dividend when you set up your "long division" layout? You should be dividing:

(x^3 + 0x^2 - 3x + 2) / (x + 2)

...if you want the multiples of the divisor (x + 2) to line up with terms of the dividend.

I won't try to do that kind of layout here, but the algebra isn't all that hard.

On the first step, you get (x^3)/(x) = x^2 as the first quotient term. Multiply that by the whole divisor to get x^3 + 2x^2 and subtract from the dividend:

(x^3 - 3x + 2) - (x^3 + 2x^2) = -2x^2 - 3x + 2

Now divide (-2x^2)/(x) to get -2x as the next quotient term. Multiply that by (x+2) and subtract from the current dividend:

(-2x^2 - 3x + 2) - (-2x^2 - 4x) = x + 2

Finally (x/x) = 1 and (x + 2) - (1)(x + 2) = 0 as a remainder.

The quotient terms were x^2 - 2x + 1, with a remainder of 0.

-2| 1+0.-3+2

.......-2..4.-2

....1.-2..1..0

This leaves x² -2x+1 or (x -1)²

x= -2,1,1