If dividing by $x+a$ or $x-2a$ gives the same remainder, what's $a$?

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As \(x-2\) is a factor of both of these polynomials, by the factor theorem, both expressions are zero when evaluated at \(x = 2\) .

We thus have the simultaneous equations \[\begin{align*} p \times 2^4 - 5 \times 2 + q &= 0, \\ 2^4 - 2 \times 2^3 - p \times 2^2 - q \times 2 - 8 &= 0. \end{align*}\] That is, \[\begin{align*} 16p + q &= 10, \\ -4p - 2q &= 8. \end{align*}\]

The second equation becomes \(q = -4-2p\) when dividing through by \(2\). By substituting this into the first equation, we have that \[ 16p + (-4-2p) = 10. \] That is, \(14p = 14\) or \(p = 1\). Consequently, \(q = -4-2 = -6\).

The remainder theorem tells us that the expression evaluated at \(x=-a\) is equal to the expression evaluated at \(x=2a\). This gives us \[\begin{equation} -a^3 - a^2 + 2a - 3 = 8a^3 - 4a^2 - 4a - 3. \label{eq:1} \end{equation}\]

That is, \(a\) is such that \[ 9a^3 - 3a^2 - 6a = 0 \] or \[ 3a(3a^2 - a - 2) = 0. \] Factorising, we see that \[ 3a^2 - a - 2 = (3a + 2)(a - 1) \] and so the non-zero values of \(a\) that satisfy \(\eqref{eq:1}\) are \(a = 1\) and \(a = -\dfrac{2}{3}\).

By substituting into either side of \(\eqref{eq:1}\), the remainder in the case when \(a=1\) is given by \[ -1^3 - 1^2 + 2 \times 1 - 3 = -3, \] and the remainder in the case when \(a=-\dfrac{2}{3}\) is given by \[ -\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 + 2 \times \left( -\frac{2}{3} \right) - 3 = \frac{8 - 12 - 36 - 81}{27} = -\frac{121}{27}. \]