What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$?

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.

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 $2$.

Hint: $x^4 + 2x^3 - 2x^2 + x - 1 = x^4 + 2x^3 - 3x^2 + x^2 + 2x - 3- x+2 = (x^2+1)(x^2 + 2x - 3) -(x-2).$