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. 
 A: Polynomial divisibility follows many of the same rules as number divisibility.  In particular, it is possible to perform "long division" with polynomials the same as with integers:
$$x^4+2x^3-2x^2+x-1 = x^2(x^2+2x-3)+x^2+x-1$$
This means there is remainder $x^2+x-1$ after dividing by $x^2+2x^3-3$.  If we then subtract these two (i.e., divide and look for the remainder), we get $-x+2$ as the final remainder, which means that
$$x^4+2x^3-2x^2+x-1=(x^2+1)(x^2+2x-3)-x+2$$
So it should be fairly clear what you need to add to make the original expression divisible by $x^2+2x-3$.
In general, if your divisor is of degree $n$, then the remainder after division can be at most degree $n-1$, just like the remainder after integer division by $k$ will be at most $k-1$.  This is why the book would tell you to look for a degree $1$ polynomial offset for a divisor of degree $2$.
A: Hint: $x^4 + 2x^3 - 2x^2 + x - 1 = x^4 + 2x^3 - 3x^2 + x^2 + 2x  - 3- x+2 = (x^2+1)(x^2 + 2x - 3) -(x-2).$
A: Hint: break $-2x^2=-3x^2+x^2$ to get two sorta similar trinomials (one should look a lot like your $x^2+2x-3$).. which coefficients prevent you from successfully factoring $x^2+2x-3$ out of both parts? How can you 'fix' said coefficients?
A: If you don't want to use polynomial long division, you can factor $x^2+2x-3$ by seeing it has roots at $1$ and $-3$,  Then, you need to add a degree 1 polynomial to your polynomial such that the sum is divisible by $x-1$ and $x+3$.  Those are relatively prime, so the sum will then be divisible by the product, which is $x^2+2x-3$. So, 
$$
\begin{aligned}
x^2 + 2x -3 &\mid x^4+2x^3−2x^2+x−1+ax+b\ \mbox{if}\\
(x-1)(x+3)  &\mid x^4+2x^3−2x^2+x−1+ax+b\\
P(1) &= P(-3) = 0\ \mbox{where}\\
P(x) &= x^4+2x^3−2x^2+x−1+ax+b\\ 
P(1) &= a + b + 1 = 0,\\
P(-3) &= -3a  + b + 5 = 0\\
a + b &= -1\\
-3a + b &= -5\\
a&=1\\
b&=-2 
\end{aligned}
$$
Added
The poster asked why I needed a degree $1$ polynomial.  Remember that integers have this property: when I divide integer $a$ by integer $b$, I get an equation:
$$
a = bq + r,\quad 0\le r < |b|.
$$
The absolute value is what we call a Euclidean norm, and division has the remainder have its norm smaller than the dividend.  The integers $\mathbb{Z}$ are called a Euclidean ring.
Polynomials are also a Euclidean ring, where the norm is just the degree:
$$
a(x) = b(x)q(x) + r(x),\quad 0\le \deg r < \deg b.
$$
So, if you had found an arbitrary solution $p(x)$ to your problem, then you could divide $p$ by $x^2+2x-3$ and get a remainder of degree less than $2$, and that would be unique.  So, use $ax+b$ so you have only two integers to find.
A: Imagine that we add $ax+b$ to the given polynomial, obtaining a new polynomial 
$$P(x)= x^4 +2x^3-2x^2+x-1+ax+b.$$
Note that $x^2+2x-3=(x+3)(x-1)$. So if $x^2+2x-3$ divides our new polynomial $P(x)$, then $P(1)=0$ and $P(-3)=0$. 
We have 
$$P(1)=a+b+1, \qquad\text{and}\qquad P(-3)=-3a+b+5.$$
Solve the system of linear equations $a+b+1=0$, $-3a+b+5=0$ for $a$ and $b$. 
Remark: We have skipped a logical step that in principle should not be skipped. If $ax+b$ is to work, our argument shows that $a$ and $b$ must satisfy the two equations. We have not shown that if $a$ and $b$  satisfy the two equations, then $ax+b$ automatically works. This is in fact true, but requires some theory. 
A: 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
$$\text{something}=(2b-3a-1)x+(1-3b)= x-2$$
