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.
 A: In this case, you are missing out a term, that is remainder. $$(x+1)^2 = Q(x) (x-1) + R(x)$$
A: Write $(x+1)^2 = (x-1)(ax+b)+c$ and plug in $x=1$, you get $4=0+c$ so $c=4$. Now 
$$ (x+1)^2-4 = x^2+2x-3 = (x-1)(x+3)$$ so $a= 1$ and $b=3$. 
A: As per this equation in your effort: $$x^2\color{red}{+2}x+1=ax^2\color{red}{+(b-a)}x-b$$
Hence, it should be $\,b-a=2$ which is not compatible with $a=1 \, , \,b=-1$.
So, $(x-1)$ is not a divisor/factor of $(x+1)^2$. Rather, you can try adding a remainder, say $\mathrm{c}$, and then solving as follows:
$$\frac{(x+1)^2-c}{x-1}=ax+b$$
A: $x-1$ doesn't divide $(x+1)^2$ exactly, so there must also be a remainder term of lower degree than $x-1$, i.e.
$$ \frac{(x+1)^2}{x-1} = ax+b + \frac{c}{x-1}. $$
Then you find
$$ x^2+2x+1 = ax^2+(b-a)x+(c-b), $$
giving three equations in three unknowns (you expect a unique solution anyway, so you should have the same number of equations as unknowns).
A: $(x+1)^2$ is not divisible by $(x-1)$ and hence you can't say that the result is a polynomial of degree one. You also can see this fact by seeing that how you proceed you can't solve $a=1 ~~b-a=2$ and $b=1$, therefore doesn't exist a polynomial of degree one which satisfy your result (hence its not divisible by $(x-1)$)
A: $$(x+1)^2=x^2+2x+1=x^2-x+3x -3+4=(x-1)(\underbrace{x+3}_\text{quotient})+\underset{\strut{\rlap{\text{remainder}}\qquad}}{4}$$
