The remainder when the polynomial x4−6x3+16x2−25x+10 is divided by x2−2x+k is (2k−9)x+(k2−8k+10) But, the remainder is x+a Then, (2k−9)x+(k2−8k+10)=x+a Compare the coefficients of x and constant term on both the sides, Then 2k−9=1 and k2−8k+10=a 2k=10 k=5 Now a=k2−8k+10 =25−40+10=35−40=−5 Hence, the required value of a is – 5.