Multiplying is easier than dividing, so we let the quotient have undetermined coef's and multiply.
Suppose $\rm\ x^4\!-\!6x^3\!+\!16x^2\!-\!25x\!+\!10\, =\, (x^2\!+cx+b)\,(x^2\!-2x+k)\, +\, x+ a.\, $ Comparing coef's
$\rm\ \ x^3\:$ coef $\rm\: \Rightarrow\: c-2 = -6\:\Rightarrow\: c = -4$
$\rm\ \ x^2\:$ coef $\rm\:\Rightarrow\ b+8+k\, =\, 16\ \ \Rightarrow\ \ \ \ b\ +\ k\, =\, 8$
$\rm\ \ x^1\:$ coef $\rm\:\Rightarrow\: -2b\!-\!4k = -26\:\Rightarrow\:-b-2k\, =-13$
Adding the prior two equations yields $\rm\,\ -k = -5,\:$ so $\rm\:k = 5,\:$ so $\rm\:b = 8-k = 3,\:$ so $\rm\:a =\, \ldots$