The polynomial #p(x)=x^4-6x^3+16x^2-25x+10,# when
divided by #(x^2-2x+k),# leaves the remainder # r(x)=(x+a).#
If the quotient polynomial is #q(x),# then, knowing that,
#p(x)=(x^2-2x+k)q(x)+r(x),# we have,
#p(x)-r(x)=(x^2-2x+k)q(x).#
This means that, #p(x)-r(x)# is divisible by #(x^2-2x+k).#
So, when #p(x)-q(x)# is divided by #(x^2-2x+k),# the Remainder
must be #0.#
Now,
#p(x)-r(x)=(x^4-6x^3+16x^2-25x+10)-(x+a), i.e.,#
#p(x)-r(x)=x^4-6x^3+16x^2-26x+(10-a).#
By Long Division of #p(x)-r(x)# by #(x^2-2x+k),# we get,
#p(x)-r(x)=x^4-6x^3+16x^2-26x+(10-a),#
#=(x^2-2x+k){x^2-4x+(8-k)}+{2(k-5)x+k^2-8k+10-a}.#
#:."The Remainder="0 rArr {2(k-5)x+k^2-8k+10-a}=0.#
#:. (k-5)=0, and, k^2-8k+10-a=0.#
#:. k=5, and, a=k^2-8k+10=25-40+10=-5.#
Enjoy Maths.!