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
 A: 
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 $a$.
A: You can do the division between $x^4-6x^3+16x^2-25x+10$ and $x^2-2x+k$ following the polynomial long division, getting:
$$R=(2k-9)x+(k^2-8k+10)$$
but $R=x+a$, so $2k-9=1\longrightarrow k=5$ and $k^2-8k+10=a\longrightarrow a=-5$.
A: HINT
You can write:
$$x^4-6x^3+16x^2-25x+10= (x^2-2x+k)(x^2+bx+c)+x+a=\\
=x^4+(b-2)x^3+(c-2b+k)x^2+(-2c+bk+1)x+(a+kc)$$
So,
$$b-2=-6→b=-4\\
c-2b+k=16\\
-2c+bk+1=-25\\
a+kc=10$$
A: To find the remainder of the long division by $x^2-2x+k$ you can keep replacing $x^2$ with $2x-k$ repeatedly, until getting the remainder of degree $1\,$:
$$
\begin{align}
x^4-6x^3+16x^2-25x+10 & = (2x-k)^2 - 6x(2x-k)+16(2x-k)-25x+ 10 \\
 & = -8x^2 + (-4k +6k +32-25)x+k^2-16k+10 \\
 & = -8(2x-k) + (2k+7)x+k^2-16k+10 \\
 & = (2k-9)x + k^2-8k+10
\end{align}
$$
Identifying coefficients between the calculated remainder and $x+a$ it follows that $2k-9=1$ so $k=5\,$, and $a=5^2-8 \cdot 5+10=-5\,$.
