Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder is linear? Given question:

If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
  what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?

Solution:

We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
  the division statement becomes
  $p(x) = (x^2 − 2x − 3)q(x) + ax + b $

Can someone please explain at a PRE-CALCULUS level? Thanks

While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)
If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.
For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:


*

*Note that the ratio of the leading terms is $\frac {3x^3}{x^2} = 3x$

*Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$

*Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $\frac {-2x^2}{x^2} = -2$.

*Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$

*Note that the ratio of the leading terms is now $\frac {-2x}{x^2} = \frac {-2}x$, which is not a polynomial, so we cannot continue.


Combining the results from steps 2 and 4:
$$\begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\\&=(3x - 2)(x^2 + 1) + (-2x-3)\end{align}$$
If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.
