Doesn't polynomial long division violate the general rules of division? Doesn't polynomial long division violate the general rules of division? As we don't divide each term in the numerator by the whole denominator (like with the division of the real numbers), what we do is dividing each term in the numerator (starting from the highest power term till the lowest power term) by the highest power term only in the denominator ignoring the other terms. How does this style of division succeed?
I searched many websites like Quora and even this site but no one has discussed this point which I consider it as a violation of division rules. Could anyone clarify this point to me? Thanks in advance.
 A: Dividing $3x^4 + 2x^3 + 6x^2 + 8x + 4$ by $x^3+2x+1$ to obtain $3x+2$ remainder $x+2$ works exactly the same way you do long division with numbers to find that $32684$ divided by $1021$ is $32$ remainder $12$. Just compare the steps in detail and there is no real difference.

You can view it this way:
We have polynomials $f(x)$ and $g(x)$. We want to find polynomials $q(x)$ and $r(x)$ such that 
$$\tag1 f(x)=q(x)g(x)+r(x).$$
Well, that's trivial: Just take $q(x)=0$ and $r(x)=f(x)$. Okay, so we want something better: We want $\deg r$ to be as small as possible. 
Can we improve upon the trivial solution $q=0, r=f$? We can replace $q(x)$ with $q(x)+cx^d$ and $r(x)$ with $r(x)-cx^dg(x)$ without destroying the equality $(1)$. And as long as $\deg r\ge deg g$, we can let $d=\deg r-\deg g$ and $c$ the quotient of their leading coefficients and - voila! - we have another solution to $(1)$ but the new $r$ has lower degree. If we continue this, we will ultimately obtain a solution where $\deg r<\deg g$ and $(1)$ still holds.
A: You need to understand that when we do polynomial division, we mean exactly what we mean when we do integer division. Suppose you want to divide $384$ by $25.$ Then this is the same as subtracting as many copies of $25$ as you can from $384.$ Whatever is left is the remainder. You can see that this is simply a problem in multiplication and subtraction. Now, you can do this anyhow, but we usually split our integers in digital multiples of nonnegative integer powers of $10,$ like $300+80+4.$ Then you can see that $25$ goes $12$ times in the first part, and $3$ times in the second. The remainder is $9.$ This is what we do with the long division method usually taught to schoolchildren.
Now come to polynomials. Given two polynomials $p$ and $d,$ so that $p$ has a higher degree, we want to see how many times $d$ divides into $p,$ and then note the remainder, which will be necessarily of lower degree than $d.$ We're concerned with degrees here because it indicates the rough size of the polynomial when the variable is large enough. Now consider the example of trying to divide $x^2+2x-4$ by $x+2.$ Then we want to subtract as many multiples of the divisor from the dividend as we can. You can see in this case that no constant multiple will be large enough. So we take the next higher multiple of the divisor, namely $x(x+2),$ which works since it reduces the dividend by one degree. Then we subtract this from $x^2+2x-4$ to see what remains, which is $-4$ in this case, which is of lesser degree than the divisor, so that our work is complete. So the quotient is $x.$
In general we subtract as many copies of $d$ as needed from $p$ until we can no longer do so. So, if $p$ is of degree $m$ and $d$ of degree $n,$ with $n\le m,$ then we see that we can take as many as $x^{m-n}$ copies of $d$ from $p,$ provided all polynomials here are monic (otherwise use an appropriate constant coefficients). Then we now have the remainder $p-x^{m-n}d,$ whose degree is now less than that of $p.$ If its degree is greater than that of $d,$ we continue in this way until the degree of the remainder is less than that of $d.$ Let this remainder of degree $<n$ be denoted by $r,$ then we eventually have something of the form $$p-x^{m-n}d-c_1x^{m-k}d-\cdots-c_kx^{n}d=r,$$ so that factoring $d$ out and rearranging, we have $$p=dq+r,$$ where $q=x^M+\sum c_kx^i,$ with $M=m-n.$
Hope this helps!
