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

Solution 1:

Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that

1. $p_1(x)=p_2(x)q(x)+r(x)$;
2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.

In particular, if $p_2(x)$ is a quadratic polynomial, then the degree of $r(x)$ will be at most $1$.

Solution 2:

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:

1. Note that the ratio of the leading terms is $\frac {3x^3}{x^2} = 3x$
2. 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.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$
3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $\frac {-2x^2}{x^2} = -2$.
4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$
5. 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.

Solution 3:

The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can then replace $x^2$ as above.

Solution 4:

An explanation rather than a full proof. Trying to describe it more simply:

Suppose your first polynomial is $$P=ax^4+bx^3+cx^2+dx+e$$ and you want to divide it by $$Q=fx^2+gx+h$$

The first step is to multiply $Q$ by something which will turn its first term into $ax^4$. You've now got a polynomial which, when you subtract it from $P$, will make $ax^4$ disappear.

Then you do the same again, except you arrange for the $x^3$ term to disappear.

Now you have two quadratics: what's left of $P$, and $Q$.

So, multiply $Q$ by a suitable constant to make the $x^2$ terms match, and subtract it. Now all that's left is a multiple of $x$ and a constant, since you've got rid of all the higher order terms—and that's your linear remainder.