factor: $x^{\alpha-2}~(1-x)^{\beta-2}~\bigg((\alpha-1)(1-x)-(\beta-1)x\bigg)=0$

Question

I started with a proof to find the maxima of a beta distribution $Beta(x: \alpha, \beta)$. This involves finding the derivative of the beta distribution and equating it with zero to yields the following root equation:

$$x^{\alpha-2}~(1-x)^{\beta-2}~\bigg((\alpha-1)(1-x)-(\beta-1)x\bigg)=0$$

Now i'm wondering, what's the technique to factor the above expression into this expression???

$$(\alpha-1)-x(\alpha + \beta -2) = 0$$

Which leads to the proof's conclusion for the maxima of a beta distribution:

$$x = \frac{\alpha-1}{\alpha+\beta-2}$$

Answer 1

Roots at x=0:

$x^{\alpha-2}=0$

Roots at x=1:

$(1-x)^{\beta-2}=0$

Root between 0 and 1:

$((\alpha-1)(1-x)-(\beta-1)x)=0$

Solve the latter expression for x.

Answer 2

$$y=x^{\alpha-2}~(1-x)^{\beta-2}~\bigg((\alpha-1)(1-x)-(\beta-1)x\bigg)=0 \tag1$$

$$y=x^{\alpha-2}~(1-x)^{\beta-2}(r)=0$$

Expand $r$:

$$r=\alpha-\alpha x-1+x-\beta x +x$$

Group for $x$:

$$r=\alpha -1 + x(-\alpha +1 -\beta+1))$$

$$r=\alpha -1 + x(2-\alpha -\beta))$$

When $r=0$, $$x(2-\alpha -\beta))=1-\alpha$$

$$x=\frac{1-\alpha}{2-\alpha -\beta}$$

Multiply the numerator and denominator by $-1$, you get: $$x = \frac{\alpha-1}{\alpha+\beta-2}$$

I guess this answers your question.

However, I don't see this as the approach to solve (1)!