TITLE: What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.