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.

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:
  $\lfrac{4x^3-7x^2-6}{2x^2+5}$
  $= -1.2 + \lfrac{-4.6x^2+4x^3}{5+2x^2}$
  $= -1.2 + 0.2·\lfrac{-4.6x^2+4x^3}{1+0.4x^2}$
  $∈ -1.2 + 0.2·(-4.6x^2+4x^3)(1-0.4x^2+O(x^4))$
  $⊆ -1.2 + 0.2·(-4.6x^2+4x^3+O(x^4))$
  $= -1.2-0.92x^2+0.8x^3+O(x^4)$.
