Let $$f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$$

Differentiating both sides,

$$f'\left( x \right) = {\alpha \over x} + 2\beta x + 1$$

Since $$x=-1$$ and $$x=2$$ are extreme points therefore

$$f'\left( x \right) = 0$$ at these points.

Put $$x = - 1$$ and $$x = 2$$ in $$f'\left( x \right),$$

we get $$ - \alpha - 2\beta + 1 = 0$$

$$ \Rightarrow \alpha + 2\beta = 1\,\,...\left( i \right)$$

$${\alpha \over 2} + 4\beta + 1 = 0$$

$$ \Rightarrow \alpha + 8\beta = - 2\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$, we get

$$6\beta = - 3 \Rightarrow \beta = - {1 \over 2}$$

$$\therefore$$ $$\,\,\,\,\alpha = 2$$