Answer :
[tex]2x^4=2x^2\cdot x^2[/tex], and [tex]2x^2\cdot(x^2-2x+1)=2x^4-4x^3+2x^2[/tex]. Subtract this from the dividend to get a remainder of
[tex](2x^4+5x^3+x-1)-(2x^4-4x^3+2x^2)=9x^3-2x^2+x-1[/tex]
[tex]9x^3=9x\cdot x^2[/tex], and [tex]9x\cdot(x^2-2x+1)=9x^3-18x^2+9x[/tex]. Subtracting this from the previous remainder gives a new remainder,
[tex](9x^3-2x^2+x-1)-(9x^3-18x^2+9x)=16x^2-8x-1[/tex]
[tex]16x^2=16\cdot x^2[/tex], and [tex]16\cdot(x^2-2x+1)=16x^2-32x+16[/tex]. Subtract this from the previous remainder to get
[tex](16x^2-8x-1)-(16x^2-32x+16)=24x-17[/tex]
[tex]x^2[/tex] does not divide [tex]24x[/tex], and we've found, one step at a time, that
[tex]\dfrac{2x^4+5x^3+x-1}{x^2-2x+1}=2x^2+\dfrac{9x^3-2x^2+x-1}{x^2-2x+1}[/tex]
[tex]\dfrac{2x^4+5x^3+x-1}{x^2-2x+1}=2x^2+9x+\dfrac{16x^2-8x-1}{x^2-2x+1}[/tex]
[tex]\dfrac{2x^4+5x^3+x-1}{x^2-2x+1}=2x^2+9x+16+\dfrac{24x-17}{x^2-2x+1}[/tex]
Another way to do is involves noticing the denominator is
[tex]x^2-2x+1=(x-1)^2[/tex]
so you can instead carry out two rounds of long division, dividing by [tex]x-1[/tex] both times.