If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadratic form. so how do I  use the synthetic division


Question: A polynomial $f(x)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $g(x)=x^2-2x+k$. Their remainder is $x+a$. Find $k$.

We're given a polynomial, and given that $\frac {f(x)}{g(x)}=p(x)+(x+a)$ where $p(x)$ is a polynomial. Therefore, if we divide $\frac {f(x)}{g(x)}$ and equate the remainder, we should be able to find $k$.
The process for extended synthetic division for higher powers is described below.
\begin{array}{c |c c} & 1 & & -6 & & +16 & & -25 & & +10\\ 2 & & & 2 & & -8 & & 16-2k & & \\-k & & & & & -k & & 4k & & k^2-8k\\\hline\\ &1 & & -4 & & 8-k & & \color{red}{2k-9} & & \color{blue}{k^2-8k+10}\end{array}
Where the top row of numbers are the coefficients of $f(x)$, and the vertical numbers are the negated coefficients of $g(x)$. And the last two numbers being the remainder $\color{red} px+\color{blue} q$.
Comparing the last two numbers to the remainder $x+a$, we have$$(\color{red}{2k-9})x+(\color{blue}{k^2-8k+10})=\color{red}{1}x+\color{blue}{a}\tag1$$$$\implies\begin{cases}2k-9=1\\k^2-8k+10=a\tag2\end{cases}$$
From the first equation of $(2)$, we have$$2k-9=1\implies 2k=10\implies k=5$$And. if necessary, plug in $k=5$ into the second equation of $(2)$ to find $a$.
