If $y = a \log x + b x^2 + x$ has its extremum value at $x = -1$ and $x = 2$, then

$y' = \frac{a}{x} + 2bx + 1$ Since $y'=0$ at $x=-1$ and $2$: At $x= -1$: $-!a + (-2b) + 1 = 0$ $\Rightarrow -a -2b + 1 = 0$ … (i) At $x= 2$: $\frac{a}{2} + 4b + 1 = 0$ $\Rightarrow a + 8b + 2 = 0$ … (ii) Solve (i) and (ii): From (i): $a = 1 - 2b$ Put into (ii): $(1 - 2b) + 8b + 2 = 0$ $1 - 2b + 8b + 2 = 0$ $6b + 3 = 0$ $b = -\frac12$ Then $a = 1 - 2b = 1 - 2(-\frac12) = 1 + 1 = 2$ So $a = 2,\ b = -\frac12$