We are given that the remainder when $x^4 + 3x^2 - 2x + 2$ is divided by $x+a$ is the square of the remainder when $x^2-3$ is divided by $x+a$. We need to find the possible values of $a$.

1. Problem Description

We are given that the remainder when $x^4 + 3x^2 - 2x + 2$ is divided by $x+a$ is the square of the remainder when $x^2-3$ is divided by $x+a$. We need to find the possible values of $a$.

2. Solution Steps

Let $P(x) = x^4 + 3x^2 - 2x + 2$ and $Q(x) = x^2 - 3$.

By the Remainder Theorem, the remainder when $P(x)$ is divided by $x+a$ is $P(-a)$, and the remainder when $Q(x)$ is divided by $x+a$ is $Q(-a)$.

Therefore, we have

$P(-a) = (-a)^4 + 3(-a)^2 - 2(-a) + 2 = a^4 + 3a^2 + 2a + 2$

$Q(-a) = (-a)^2 - 3 = a^2 - 3$

We are given that $P(-a) = [Q(-a)]^2$, so we have

$a^4 + 3a^2 + 2a + 2 = (a^2 - 3)^2$

$a^4 + 3a^2 + 2a + 2 = a^4 - 6a^2 + 9$

$0 = -9a^2 - 2a + 7$

$9a^2 + 2a - 7 = 0$

We can use the quadratic formula to solve for $a$:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$a = \frac{-2 \pm \sqrt{2^2 - 4(9)(-7)}}{2(9)}$

$a = \frac{-2 \pm \sqrt{4 + 252}}{18}$

$a = \frac{-2 \pm \sqrt{256}}{18}$

$a = \frac{-2 \pm 16}{18}$

$a = \frac{-2 + 16}{18} = \frac{14}{18} = \frac{7}{9}$

$a = \frac{-2 - 16}{18} = \frac{-18}{18} = -1$

Therefore, the possible values of $a$ are $\frac{7}{9}$ and $-1$.

3. Final Answer

The possible values of $a$ are $\frac{7}{9}$ and $-1$.