The polynomials $2x^3 - 7x^2 + ax - 6$ and $x^3 - 8x^2 + \left(2a + 1 \right)x - 16$ leave the same remainder when divided by $\left(x - 2\right)$. Find the value of $a$.
Let $\;$ $f \left(x\right) = 2x^3 - 7x^2 + ax - 6$, $\;$ $g \left(x\right) = x^3 - 8x^2 + \left(2a + 1\right)x - 16$
By remainder theorem,
remainder obtained when $f \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = f \left(2\right)$
remainder obtained when $g \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = g \left(2\right)$
Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$
Now, $\;$ $f \left(2\right) = 2 \times 2^3 - 7 \times 2^2 + 2a - 6$
i.e. $\;$ $f \left(2\right) = 2a - 18$ $\;\;\; \cdots \; (2a)$
and $\;$ $g \left(2\right) = 2^3 - 8 \times 2^2 + \left(2a + 1\right) \times 2 - 16$
i.e. $\;$ $g \left(x\right) = 4a - 38$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,
$2a - 18 = 4a - 38$
i.e. $\;$ $2a = 20$ $\implies$ $a = 10$