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 \(x + a\). Calculate the possible values of \(a\).

The remainder theorem tells us that if a polynomial \(p(x)\) is divided by \(x-a\), the remainder is \(p(a)\).

So on dividing \(x^2 - 3\) by \(x + a\), we have a remainder of \(a^2-3\).

We also have that if \(f(x) = x^4 + 3x^2 - 2x + 2\) is divided by \(x + a\), the remainder is \(f(-a)=a^4 + 3a^2 + 2a + 2\).

Thus the question tells us that \((a^2-3)^2 = a^4 + 3a^2 + 2a + 2 \implies 9a^2+2a-7=0.\)

Factorising, we have that \((a+1)(9a-7) = 0\), and so the possible values for \(a\) are \(-1\) and \(\dfrac{7}{9}\).