Intuition for polynomial long division So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.
I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials since a number in its decimal expansion can be viewed as a sort of polynomial e.g $a_n10^n+\cdots+a_2100+10a_1+a_0$
What i don't understand is why we only use the term with the highest degree in the denominator as the divisor for the whole polynomial as shown here  Polynomial Long Division whereas with normal division we divide the whole divisor into the dividend for example when calculating $88÷32$ we look at how many times $32$ goes into $88$ and not how many times $30$ does whilst with polynomial division only the highest degree term is considered which corresponds to 30 in this example ignoring 2.
So how are they the same? Some intuition behind why polynomial division works would be greatly appreciated.
Thanks in advance.
 A: In long division, we have to pay attention to the entire divisor because multiples of the divisor can result in carries between the various digits (columns in the usual tableau for long division).  There is no equivalent to carries when multiplying polynomials.
There are no carries between columns in polynomial division, so the non-leading degree terms cannot alter the multiple of the divisor that gives matching leading terms.
Consider $40 \div 13$.  The carry from the $3$ in the units place spoils the candidate quotient, $4$: $4 \times 13 = 52$, so the leading digit is too large.  This can't happen in the analogous $(4x+0) \div(x+3)$ -- when we multiply by $4$, $4(x+3) =4x+12$, there are no carries between terms, so the non-leading terms do not interact with the multiple of the leading term that gives the correct term in the quotient.
In long division of numbers, carries allow different columns to talk to each other.  In long division of polynomials, there is no communication between columns, so it is always sufficient to compare just the leading terms.
A: Lets just forget how regular long division works for a minute and do it the hard way. Say we're dividing $888/17$ then we could repeatedly subtract $17$ from $888$ and count how many times this occurs until the remainder is less than $17$. This count will be the answer we're looking for.
This would however take a long time. Instead we could use groups of $17$ count how many times we subtract them off then multiply by the size of the groups. So for example I could do groups of $170$, count the number of times then multiply it by $10$. This makes the work a lot easier.
Now if we choose the optimal group size to subtract off at each stage you're now doing long division as usual. You find the highest multiple that fits into the existing number, subtract it off, then repeat until the remainder is less than $17$. You then add the group counts up and you're left with the correct answer.
Polynomial long division is the same, but the optimal size is determined by the highest power and not the coefficients. This is because we lose some information when working with polynomials that we have in base $10$, namely that $x=10$ and the coefficients are less than $10$. It limits how much we can subtract off at each stage reliably but the idea is the same, repeatly removing as much as you can until the remainder is smaller than the divisor.
A: The fundamental idea is to replace a "digit" in a Euclidean long division's quotient with a "monomial" in a polynomial long division's quotient. The reason we only care about the term of highest degree is because, at each step, we're trying to find the "biggest" (in terms of degree) multiple of the denominator $B(x)$ that allows us to cancel out a monomial from the numerator $A(x)$.
Here is a worked example, and a table of analogies, to give you a better idea.
    4x^4 + 7x^3 - 4x^2 + 3x + 12 = A(x) | x + 2 = B(x)
  - 4x^4 - 8x^3                         +------- 
  _____________                         | 4x^3 - x^2 - 2x + 7 = Q(x)
      0  -  x^3 - 4x^2 + 3x + 12        |
         +  x^3 + 2x^2                  |
           ___________                  |
              0 - 2x^2 + 3x + 12        |
                + 2x^2 + 4x             |
                 __________             |
                     0 + 7x + 12        |
                       - 7x - 14        |
                        ________        |
                           0 - 2        |

\begin{array}{ |c|c|c|c| } 
 \hline
A(x) & 4x^4 + 7x^3 - 4x^2 + 3x + 12 & a & 51 \\ 
 \hline
B(x) & x + 2 & b & 10 \\ 
 \hline
Q(x) & 4x^3 - x^2 - 2x + 7 & q & 5 \\
 \hline
R(x) & -2 & r & 1 \\
 \hline
A(x) = B(x) \times Q(x) + R(x) & \text{Verify by yourself} & a = b \times q + r & 51 = 10 \times 5 + 1 \\
 \hline
\end{array}
A: But that is how we do long division!
Suppose we are diving $56264377$ by $712$.  I have no idea how many times $712$ goes into $52664377$ so I simply divide it into $52660000$ at first.  But I have no idea how many times $712$ goes into $5266$ either.  So I divide $700$ into $5266$ first and get $700*7$ is $49$ so that's close and $712*7 = 7*(700 + 12) = 4900 + 7*12 = 49+84$ as subtraction I get $300 + (66-84) = 300+(66-86+2)=300 -20 + 2=282$.
Then I need to figure out how many times $712$ divides into $2824$.  Again, no idea but $700$ goes in $4$ times so $4(700 + 12) = 2800 + 48=2824$ is too be big.  SO $3$.  And $3(700+12) = 21 + 36$ so we get remainder $2824 - 2136= 700 + (24-26) = 700 - 12 = 688$
So so fore we have $56264377\div 712 = 74xxxxx$ and we keep going.
A: Hint: in the algebra of polynomials there is no concept of "carrying" or "borrowing" when we add two polynomials, because there is no constraint on the size of the coefficients. In the algebra of integers expressed as decimals, when we add or subtract two decimal representations, we have to make adjustments to make the digits (coefficents) lie between $0$ and $9$. The intuition behind polynomial division is the same as the intuition behind integer division (subtract the biggest allowable multiple on each iteration), but the implementation is simpler - because there is no trial and error required to find the the biggest allowable multiple.
A: If you really want to understand the mathematics, do not do the kind of long division that you were taught. Instead, you must learn to write down true equalities. For example:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
  $\lfrac{4x^3-7x^2-6}{2x^2+5}$
  $ = \lfrac{\color{blue}{2x}(2x^2+5)-10x-7x^2-6}{2x^2+5}$
  $ = 2x + \lfrac{-7x^2-10x-6}{2x^2+5}$
  $ = 2x + \lfrac{\color{blue}{-3.5}(2x^2+5)+17.5-10x-6}{2x^2+5}$
  $ = 2x-3.5 + \lfrac{17.5-10x-6}{2x^2+5}$
  $ = 2x-3.5 + \lfrac{-10x+11.5}{2x^2+5}$.
Where did the blue terms come from? First you need to recognize that what I wrote is true. Next the reason for choosing those blue terms was to eventually get an equal expression where the fractional part is 'simpler'. You should try other terms in place of the blue terms and see why they do not work, and then you will understand why those blue terms work.
Also, there is absolutely nothing forcing you to 'simplify' the given polynomial fraction in that way. The following is also true:
  $\lfrac{4x^3-7x^2-6}{2x^2+5}$
  $= \lfrac{-6-7x^2+4x^3}{5+2x^2}$
  $= \lfrac{\color{blue}{-1.2}(5+2x^2)+2.4x^2-7x^2+4x^3}{5+2x^2}$
  $= -1.2 + \lfrac{2.4x^2-7x^2+4x^3}{5+2x^2}$
  $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$.
Why might you want to do this? Well, if you want to know the behaviour of $\lfrac{4x^3-7x^2-6}{2x^2+5}$ as $x→0$, then you want to extract out the 'big' terms, and leave the 'residual' fraction with a 'small' numerator, so that it is clear what the asymptotic behaviour is. For instance as $x→0$ we have:
  $\lfrac{4x^3-7x^2-6}{2x^2+5}$
  $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$
  $= -1.2 + 0.2·\lfrac{-4.6x^2+4x^3}{1+0.4x^2}$
  $∈ -1.2 + 0.2·(-4.6x^2+4x^3)(1-0.4x^2+O(x^4))$
  $⊆ -1.2 + 0.2·(-4.6x^2+4x^3+O(x^4))$
  $= -1.2-0.92x^2+0.8x^3+O(x^4)$.
