I'll quote the full question and afterwards my attempt at solution.
$x(x-2)(x-\tau )+\epsilon x + \epsilon^2 = 0 $ has three roots, $x\approx 0,2,\tau$. Find expansions for the roots $x\approx 0$ and $x\approx \tau$ valid when $\tau \approx 0$ with error $\mathcal{O}(\epsilon^2)$. Determine the shape of the regions in the $\epsilon , \tau$ plane for which the estimate holds uniformly.
Here's what I understand from the question. $\tau = \tau_1 \epsilon + \mathcal{O}(\epsilon^2) , \ \ x=x_1\epsilon +\mathcal{O}(\epsilon^2), \ \ x=\tau + x_1 \epsilon + \mathcal{O}(\epsilon^2)$
For the first root I find the following expansion coefficinets: $x_1=-1=\tau_1$ and the same expansion also works for the second root of $x\approx \tau$; perhaps I am doing something wrong here?
You have to take $$ x=x_0+\epsilon x_1+O(\epsilon^2) $$ and, when you assume $\tau\approx 0$, this means that you can neglect higher powers of $\tau$. Then, putting this into your equation you get $$ x_0^3-2x_0^2-\tau x_0^2+2\tau x_0+3\epsilon x_1x_0^2-4\epsilon x_0x_1+\epsilon x_0-2\epsilon\tau x_1x_0+2\epsilon\tau x_1+O(\epsilon^2)=0. $$ This yields two equations $$ x_0^3-2x_0^2-\tau x_0^2+2\tau x_0=0, $$ the leading order one having solutions $x=0,2,\tau$, and $$ 3x_1x_0^2-4x_0x_1+x_0-2\tau x_1x_0+2\tau x_1=0 $$ that has the solution $$ x_1=-\frac{x_0}{3x_0^2-2(\tau+2)x_0+2\tau}. $$ This yields $$ x^{(1)}=O(\epsilon^2), $$ $$ x^{(2)}=2-\epsilon\left[\frac{1}{2}-\frac{\tau}{4}+O(\tau^2)\right]+O(\epsilon^2), $$ $$ x^{(3)}=\tau+\epsilon\left[\frac{1}{2}+O(\tau^2)\right]+O(\epsilon^2). $$