What must be added to

Let $k$ be added to $2 x^{4}-5 x^{3}+2 x^{2}-x-3$ so that the result is exactly divisible by $(x-2)$. Here, $k$ is a constant.

$\therefore f(x)=2 x^{4}-5 x^{3}+2 x^{2}-x-3+k$ is exactly divisible by $(x-2)$.

Using factor theorem, we have

$f(2)=0$

$\Rightarrow 2 \times 2^{4}-5 \times 2^{3}+2 \times 2^{2}-2-3+k=0$

$\Rightarrow 32-40+8-5+k=0$

$\Rightarrow-5+k=0$

$\Rightarrow k=5$

Thus, 5 must be added to $2 x^{4}-5 x^{3}+2 x^{2}-x-3$ so that the result is exactly divisible by $(x-2)$.