Here is something right away to express: \[ (\alpha + \beta)^n = \sum^{n}_{k=0} {n \choose k} \alpha^{n-k} \beta^k = \sum^{n}_{k=0} {n \choose k} (\alpha + \beta - 1)^k \] My other favourite self-discocery is:

\[ \begin{align} \sum \alpha &= \alpha + \beta + \gamma = -\frac{b}{a} \\ \sum \alpha \beta &= \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \sum \alpha \beta \gamma &= \alpha \beta \gamma = -\frac{d}{a} \end{align} \]

Let \(\gamma = x_1\) which is one root we assume we know, then

\[ \begin{align} \alpha + \beta + x_1 &= -\frac{b}{a} \\ \alpha \beta + \beta x_1 + \alpha x_1 &= \frac{c}{a} \\ \alpha \beta x_1 &= -\frac{d}{a} \end{align} \]

We know from here that:

\[ \begin{equation*} \alpha + \beta = -\frac{b}{a} - x_1 \\ \alpha \beta = -\frac{d}{ax_1} \end{equation*} \]

Plugging this into a quadratic equation:

\[ \begin{align} x^2 - (\alpha + \beta)x + \alpha \beta &= 0 \\ x^2 - \left(-\frac{b}{a} - x_1\right)x + \left(-\frac{d}{ax_1}\right) &= 0 \\ x^2 + \frac{b}{a}x + x_1 x - \frac{d}{ax_1} &= 0 \end{align} \]

\[ \begin{aligned} a x_1 x^2 + b x x_1 + a x_1^2 x - d &= 0 \end{aligned} \]

This means the other two roots are:

\[ \begin{equation} \alpha, \beta = \frac{-(ax_1^2 + bx_1) \pm \sqrt{(ax_1^2 + bx_1)^2 + 4(ax_1)(d)}}{2ax_1} \end{equation} \]