Polynomials/ remainder theorem What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?
The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor has roots 1 and -3. We can solve by putting p(1) and p(-3) equal to zero and finding the values of a and b. I know how to solve this but I am confused as to why we say that the expression to be added is of the form ax+b. How do we know this. Essentially what the book seems to be saying is add the remainder. But if we take the example7/3, the remainder is 1 but adding the remainder will not make 7 exactly divisible by 3. Here we subtract 1 from 3 and add the result to make 7 exactly divisible by 3. I believe the same principle must hold in the case of polynomials and the explanation in the book that we add the remainder to the dividend must be wrong. Please tell me the reason why we add ax+b. Please explain.
Also, can it be solved through polynomial long division? If so, how?
Are there any other easier ways to solve this? Feel free to ignore the last question and even the second last question if it will make the answer too long.

The division algorithm says that if $p(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m\le 0$ then there is a unique quotient polynomial $q(x)$ of degree $n-m$ and a unique remainder polynomial of degree $< m$ so that:  $p(x) = q(x)g(x) + r(x)$.
So if $p(x) = x^4+2x^3-2x^2+x-1$ is of degree $4$ and $g(x) = x^2+2x-3$ is of degree $2$ then there is a unique polynomial of degree $4-2 = 2$ called $q(x)$ and a unique polynomial $r(x)$ of degree less than $2$ so or degree of at most $1$ so that
$p(x) = q(x)g(x) + r(x)$.
Now as $r(x)$ is at most of degree $1$ it is of the form $ax + b$ (where $a$ could be $0$).
And

$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma) + (ax+b)$

so to make $p(x)$ exactly divisible by $g(x)$ we must add the NEGATIVE of the remainder (I don't understand why the book says to add rather than subtract the remainder) to get:

$x^4 + 2x^3 - 2x^2 + x-1 +(-ax -b) = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma)$

And the most straightforwar way to do this is long division.
But typing long division on a keyboard is too hard for me so I'll talk us through:
$(x^2+2x-3)(\alpha x^2 + \beta x + \gamma)= \alpha x^4 + (2\alpha + \beta)x^3 +(\gamma+2\beta -3\alpha) x^2 + (2\gamma -3\beta)x- 3\gamma$.
We want that to be as close to $x^4 + 2x^3 -2x^2 + x-1$ as possible.
So $\alpha = 1$ and $2\alpha + \beta = 2$ and $\gamma +2\beta -3\alpha = -2$.  And that's as fair as we will be able to go.
$\alpha = 1$ and $\beta = 0$ and $\gamma = 1$.
And so we have

$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(x^2 +1) + (ax+b)$


$x^4+2x^3-2x^2+x-1 = x^4 +2x^3 - 2x^2 +2x -3 +(ax+b)$.

And so the remainder is $ax + b = -x +2$
so what we must add is $-(-x+2) = x -2$.
That is to say
$(x^4+2x^3-2x^2+x-1) + (x-2) = (x^2+2x-3)(x^2 +1)$
.......
But that is the most straightforward way.
The easiest way is to note:
If $p(x) + (mx+n) = (x^4 + 2x^3 - 2x^2 + x-1) + (mx + n) = (x^2 + 2x -3)q(x)=g(x)q(x)$.
And the roots of $g(x) = x^2 + 2x -3)$ are $x = 1$ and $x=-3$ then
$p(1) + (m*1 + n) = g(1)q(1) = 0*q(1) = 0$ so
$(1^4 + 2*1^3 -2*1^2 + 1-1) + (m+n) = 0$
$1+m+n = 0$
And $p(-3) + (-3m + n) = 0$ and
$(3^4 - 2*3^3 - 2*3^2 - 3-1) -3m +n =0$
$81 - 54 -18 -4-3m+n = 0$
$5- 3m + n=0$
So $5-3m + n = 1+m+n$
$5-3m = 1+m$
$4=4m $ and $m =1$ and $1+m+n = 0$ so $1+1+n =0$ so $n=-2$.
