Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac{(x+1)^2}{x-1}=x-1.$$ Which is far from the correct answer. I can't spot my mistake. I feel that the more I sit and study, the worse at math I become.

$$(x+1)^2=x^2+2x+1=x^2-x+3x -3+4=(x-1)(\underbrace{x+3}_\text{quotient})+\underset{\strut{\rlap{\text{remainder}}\qquad}}{4}$$
