Note that 4+1-(2+3)=0, so if x=-1, 4x4+3x3+2x+1=4-3-2+1=0, and x+1 must be a factor. Also note that there is no x2 term and the denominator (divisor) cannot be factorised.
Using synthetic division we can find the other factor of the dividend (numerator):
-1 | 4 3 0 2 1 (0 is required for the missing x2 term)
4 -4 1 -1 | -1
4 -1 1 1 | 0 = 4x3-x2+x+1.
There are no common factors between the denominator and numerator. [However, if the divisor had been x2-x-2=(x+1)(x-2), there would have been the common factor x+1.]
Using algebraic long division we get:
4x -5
x2+x+2)4x3 -x2 +x+1
4x3+4x2+8x
-5x2-7x+1
-5x2-5x-10
-2x+11 remainder
Therefore (4x4+3x3+2x+1)/(x2+x+2)=(x+1)(4x3-x2+x+1)/(x2+x+2)=
(x+1)[4x-5 rem -2x+11].