If the polynomial $f(x)=x^4- 6x^3+16x^2-25x+10$ is divided by another polynomial $q(x) = x^2-2x+k$, the reminder is $x+a$ . Find $k$ and $a$.
please don't solve it by long division as i am searching for some other approaches
my attempt : i thought $x^2-2x + k$ has two roots let say $m$ and $2-m$ then i tried to equate both sides by putting $m$ and $2-m$ in f(x) but i am not arriving at any answer please don't solve it by long division.
It should be $$x^4-6x^3+16x^2-25x+10=(x^2-2x+k)(x^2-4x+n)+x+a,$$ which gives $$n+k+8=16,$$ $$-2n-4k+1=-25$$ and $$kn+a=10,$$ which gives $k=5$,$n=3$ and $a=-5.$
Since the divisor is of first rank, it follows that the quotient is also of second order. That is, the given polynomial can be written as $$f(x)=q(x)\cdot p(x) + x+a,$$ where $p(x)$ is a quadratic polynomial in $x.$ Indeed, we nust have $p(x)=(x-m)(x-n),$ since $f(x)$ is monic, so that with well-chosen substitutions for $x$ you should be able to get a system of equations in $a,k,m,n.$ You may use $x=-1,0,1,1/2,$ for example.
