With polynomial long division, why are the un-applied divisor terms still multiplied and subtracted?

Question

Given the following for example:

$$\frac{x^2+x+8}{x+2}$$

The first point of confusion:

The divisor $x + 2$ when performing the division operation, only $x$ is applied. While this works with $x$ dividing $x^2$ fully, if the divisor is $x - 2$ or $x$ is a negative (e.g. $-5 + 2$) than it seems there is a logical problem as this modifies the division scalability relationship of $x$ to $x^2$.

Furthermore, dividing just by $x$ violates my understanding of quotative division in that it is returning how many $x + 2$ overlap with $x^2$. Not how many times just $x$ divides it.

The second point of confusion:

After Division by $x$ the partial quotient is then multiplied by the divisor and subtracted:

$$Divide(Partial-quotient):$$ $$\frac{x^2+x+8}{x+2} = x$$

$$(Confusion) Multiply:$$ $$x(x+2) = (x^2+2x)$$

$$Subtract:$$ $$\frac{x^2+x+8}{x+2} - (x^2 + 2x) = \frac{-x+8}{x+2}$$

Why is the partial quotient multiplied by the constant term 2 in "$x$ + 2" and then the product of 2 is removed? The purpose of subtraction is to eliminate the portion of the dividend $x^2 + x + 8$ that has already been divided. Yet it does not appear the $2$ was involved in the division, so it seems we are removing more than necessary.

Answer 1

The point of the first step in the long division of polynomials is to figure out how to reduce the degree of the polynomial you are dividing.

In your example that means writing $$ x^2+x+8 = a(x+2) + b $$ where you choose the value of $a$ so that the $x^2$ term will cancel. Clearly $a$ should be just $x$. Then that choice tells you what $b$ must be. In your example it is $-x + 8$.

Once you've done this, you do it all over again to reduce the degree of $b$. So your new problem is $$ -x+8 = a(x+2) + b . $$ In this problem $a = -1$ and $b$ works out to $10$. That's of degree less than the degree of $x+2$ so it's the remainder and you are done.

(Can you understand why that $10$ is just what you get when you substitute $-2$ for $x$ in the original polynomial?)

You should think of this algorithm as analogous to the long division algorithm for integer arithmetic. There your goal at each step is to deal with the leftmost digit of the dividend. In fact the polynomial algorithm is easier because you never have to "carry" or "borrow". The powers of $x$ never mix in the way the powers of $10$ can in arithmetic.

You will understand this better when you think only about multiplication. Writing $$ \frac{x^2+x+8}{x+2} = x $$ is confusing nonsense: those two expressions are not equal.

Answer 2

The idea behind the inductive step of the polynomial division algorithm for $\,G\div F\,$ is as follows. Write $\, G = ax^{k+j}+g,\,$ $\,F = x^k + f \,$ as its (highest degree) lead_term plus lower degree terms. We scale the divisor $F$ by the quotient of the leading terms $\,\color{#c00}{ax^j} = (ax^{k+j}/x^k)\,$ so that it has the same leading term as $G$, so it kills the lead term of $G$ when subtracted from $\,G,\,$ leaving a new $\,\color{#0a0}{G'}$ of lower degree than $G.\,$ We repeat this process, dividing $\,\color{#0a0}{G'}$ by $F,\,$ etc, iterating until we reach a polynomial of lower degree than $F$ (which is the remainder $\,G\bmod F).\,$ Below are the equations and associated tabular form of this inductive step:

$$\begin{align} G - \color{#c00}{a x^{\large j}} F \,\ = \,\ \overbrace{a x^{\large k+j} + g}^{\large {\rm dividend}\ G}\ -\ &\color{#c00}{a x^{\large j}} (\!\!\overbrace{x^{\large k} + f\!}^{\large {\rm divisor}\ F})\ \!=\!\!\!\! \overbrace{\color{#0a0}{g-ax^j f}}^{\large {\rm deg}\ \color{#0a0}{G'}\ <\ \deg G\!}\\[.6em] \underset{\!\!\!\!\!\!\large \div\ (x^k+f)}\Longrightarrow\ \ \ \dfrac{a x^{\large k+j}+g}{x^{\large k}+f}\, =\ &\color{#c00}{a x^{\large j}}\ \ + \underbrace{\dfrac{\color{#0a0}{g-a x^{\large j} f}}{x^k + f}}_{\large\rm recurse\ on\ this}\end{align}\qquad\qquad$$ $$\require{enclose} \qquad\ \ \ \begin{array}{r} \color{#c00}{a x^j}\phantom{x^{k+n}+f\ \ \,} \\[0pt] {\Large \smash[t]{\overset{\rm\color{#90f}{tabularly}}{\leadsto^{\phantom{.}}}}}\quad\ x^k\!+f\,\ \enclose{longdiv}{\,\ a x^{k+j}\,+\,g\phantom{x^kg}} \\[-3pt] \underline{ax^{k+j}\ +\, a x^j f} \\ \color{#0a0}{g\,-\,a x^j f = G'}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \end{array}\qquad\qquad\qquad\qquad\quad$$

Remark $ $ The same method works if the leading coef of the divisor is not $1$, as long as it is a unit (invertible), so it divides every coefficient, so the above inductive step is always possible - cf. here.