When the two polynomials $x^3 + ax^2 -x -2$ and $ax^3 + x^2 -6x -4$ are divided by $x - 2$, the remainder is the same. Find the value of $a$.
Let $\;$ $f \left(x\right) = x^3 + ax^2 - x - 2$ $\;$ and $\;$ $g \left(x\right) = ax^3 + x^2 -6x -4$
By remainder theorem,
when $f \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= f \left(2\right)$
when $g \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= g \left(2\right)$
Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$
Now, $\;$ $f \left(2\right) = 2^3 + a \times 2^2 - 2 - 2 = 4 + 4a$ $\;\;\; \cdots \; (2)$
$g \left(2\right) = a \times 2^3 + 2^2 - \left(6 \times 2\right) - 4 = 8a - 12$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of equations $(2)$ and $(3)$ equation $(1)$ becomes,
$4 + 4a = 8a - 12$
i.e. $\;$ $4a = 16$ $\implies$ $a = 4$