What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms . 
I cant fully understand the explanation in my book .
It just assumes that the expression that should be added has a degree of 1. 
I apologize if this question is too simple or just stupid but this is a genuine doubt. 

You can get to a quartic divisible by $x^2+2x-3$ by writing
$$\begin{align}
x^4+2x^3-2x^2+x-1+\text{something}&=(x^2+2x-3)(x^2+ax+b)\cr
&=x^4+(a+2)x^3+(2a+b-3)x^2+(2b-3a)x-3b\cr
\end{align}$$
which leads to
$$\text{something} = ax^3+(2a+b-1)x^2+(2b-3a-1)x+(1-3b)$$
for any coefficients $a$ and $b$ (of the quotient) that your heart desires.  What the book presumably has in mind is to make the "something" of degree as small as possible.  You can obviously get rid of the $ax^3$ by setting $a=0$ and then the $(2a+b-1)x^2$ by setting $b=1$.  This leaves
$$\text{something}=(2b-3a-1)x+(1-3b)= x-2$$
