jmc

The long division is shown below.

$$\begin{array}{cccccc}\text{multicolumn}2r{x}^2& -4x& +(8-k)& & & \\ \text{cline}2-6x^2-2x+k& x^4& -6x^3& +16x^2& -25x& +10\\ \text{multicolumn}2r{x}^2& -2x^3& +k{x}^2& & & \\ \text{cline}2-4\text{multicolumn}2r0& -4x^3& +(16-k)x^2& & & \\ \text{multicolumn}2r& -4x^3& +8x^2& -4kx& & \\ \text{cline}3-5\text{multicolumn}2r& 0& +(8-k)x^2& +(4k-25)x& & \\ \text{multicolumn}2r& & +(8-k)x^2& +(2k-16)x& +k(8-k)& \\ \text{cline}4-6\text{multicolumn}2r& & 0& +(2k-9)x& +(k^2-8k+10)& \end{array}$$Thus, the remainder is $(2k-9)x+(k^2-8k+10).$ We want this to be $x+a,$ so $2k-9=1$ and $k^2-8k+10=a.$ Solving, we find $(k,a)=\boxed{(5,-5)}.$