If the polynomial $x^{4}-6x^{3}+16x^{2}-25x+10$ is divided by another polynomial $x^{2}-2x+k$,the remainder is $x+a$. Find the values of $k$ and $a$.

(A) By the division algorithm,we have:
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
Therefore,Dividend $-$ Remainder $=$ Divisor $\times$ Quotient.
Subtracting the remainder $(x+a)$ from the dividend:
$(x^{4}-6x^{3}+16x^{2}-25x+10) - (x+a) = x^{4}-6x^{3}+16x^{2}-26x+10-a$.
This resulting polynomial must be perfectly divisible by $x^{2}-2x+k$.
Performing long division of $(x^{4}-6x^{3}+16x^{2}-26x+10-a)$ by $(x^{2}-2x+k)$:
$1$. Dividing $x^{4}$ by $x^{2}$ gives $x^{2}$. Multiplying $(x^{2}-2x+k)$ by $x^{2}$ gives $x^{4}-2x^{3}+kx^{2}$. Subtracting this from the dividend leaves $-4x^{3}+(16-k)x^{2}-26x$.
$2$. Dividing $-4x^{3}$ by $x^{2}$ gives $-4x$. Multiplying $(x^{2}-2x+k)$ by $-4x$ gives $-4x^{3}+8x^{2}-4kx$. Subtracting this leaves $(8-k)x^{2}+(4k-26)x+(10-a)$.
$3$. Dividing $(8-k)x^{2}$ by $x^{2}$ gives $(8-k)$. Multiplying $(x^{2}-2x+k)$ by $(8-k)$ gives $(8-k)x^{2}-2(8-k)x+k(8-k)$.
Subtracting this from the previous remainder,we get the final remainder as $[(4k-26) + 2(8-k)]x + [10-a - k(8-k)] = 0$.
Simplifying the coefficient of $x$: $4k-26+16-2k = 2k-10$. Setting $2k-10=0$ gives $k=5$.
Simplifying the constant term: $10-a-8k+k^{2} = 0$. Substituting $k=5$: $10-a-8(5)+25 = 10-a-40+25 = -5-a = 0$,which gives $a=-5$.
Thus,$k=5$ and $a=-5$.