APL(NARS), 92 chars

{{⍬≡↑⌽⍵:(⊂¯1↓⍵),⊂,0⋄0=≢1↓⍵:(⊂,0),⍵⋄(⊂¯1↓⍵),¯1↑⍵}⍺{0>k←-/≢¨⍵⍺:⊂,⍵⋄d,⍺∇1↓⍵-(⍺,k⍴0)×d←÷/↑¨⍵⍺}⍵}

Input: 1 2 3 4 means x^3+2x^2+3x^1+4

Input: divisor function dividend

Output: quotient, remainder

Im sorry for the one has upvote this, but i find some bugs... Possible there are other bugs, not too much test... last few test:

  f←{{⍬≡↑⌽⍵:(⊂¯1↓⍵),⊂,0⋄0=≢1↓⍵:(⊂,0),⍵⋄(⊂¯1↓⍵),¯1↑⍵}⍺{0>k←-/≢¨⍵⍺:⊂,⍵⋄d,⍺∇1↓⍵-(⍺,k⍴0)×d←÷/↑¨⍵⍺}⍵}

 2 5   f 1 3
┌2────────────┐
│┌1───┐ ┌1───┐│
││ 0.5│ │ 0.5││
│└~───┘ └~───┘2
└∊────────────┘
  1 0 ¯5 f 12 ¯5 3 ¯1
┌2─────────────────┐
│┌2─────┐ ┌2──────┐│
││ 12 ¯5│ │ 63 ¯26││
│└~─────┘ └~──────┘2
└∊─────────────────┘
  2 0 ¯5 f 12 ¯5 3 ¯1
┌2────────────────────┐
│┌2──────┐ ┌2────────┐│
││ 6 ¯2.5│ │ 33 ¯13.5││
│└~──────┘ └~────────┘2
└∊────────────────────┘
  1 2 f 1 2 3 4
┌2─────────────┐
│┌3─────┐ ┌1──┐│
││ 1 0 3│ │ ¯2││
│└~─────┘ └~──┘2
└∊─────────────┘
  1 1 f 1 6 12 7
┌2────────────┐
│┌3─────┐ ┌1─┐│
││ 1 5 7│ │ 0││
│└~─────┘ └~─┘2
└∊────────────┘
  1 3 ¯1 f 5 ¯2 ¯3 171 4
┌2───────────────────┐
│┌3────────┐ ┌2─────┐│
││ 5 ¯17 53│ │ ¯5 57││
│└~────────┘ └~─────┘2
└∊───────────────────┘
  1  f 12 ¯5 3 ¯1
┌2─────────────────┐
│┌4──────────┐ ┌1─┐│
││ 12 ¯5 3 ¯1│ │ 0││
│└~──────────┘ └~─┘2
└∊─────────────────┘
  2  f 12 ¯5 3 ¯1
┌2──────────────────────┐
│┌4───────────────┐ ┌1─┐│
││ 6 ¯2.5 1.5 ¯0.5│ │ 0││
│└~───────────────┘ └~─┘2
└∊──────────────────────┘
  1 1 1 1 1 1  f 12 ¯5 3 ¯1
┌2─────────────────┐
│┌1─┐ ┌4──────────┐│
││ 0│ │ 12 ¯5 3 ¯1││
│└~─┘ └~──────────┘2
└∊─────────────────┘
3 f 4
┌2──────────────────┐
│┌1───────────┐ ┌1─┐│
││ 1.333333333│ │ 0││
│└~───────────┘ └~─┘2
└∊──────────────────┘
  1 1 1 1 1 1  f 0
┌2────────┐
│┌1─┐ ┌1─┐│
││ 0│ │ 0││
│└~─┘ └~─┘2
└∊────────┘

this below is a little more long of the recursive one, but i like it more

r←a D w;q;k;d
r←,w⋄q←⍬
→3×⍳0>k←-/≢¨r a⋄q←q,d←÷/↑¨r a⋄r←1↓r-(a,k⍴0)×d⋄→2
→4×⍳q≢⍬⋄q←,0
→5×⍳r≢⍬⋄r←,0
r←(⊂q),⊂r

//14+9+49+13+13+10= 108 it would returns the same of the recursive one. It is need a function remove the more 0 in the remainder. For example in

  1 1 1 1 f 1 0 0 0 ¯2
┌2────────────────┐
│┌2────┐ ┌3──────┐│
││ 1 ¯1│ │ 0 0 ¯1││
│└~────┘ └~──────┘2
└∊────────────────┘

it is better that is just -1 and not 0x^2+0x-1.

Python, 123 bytes

f=lambda x,y,z=[],l=len:z+x if l(x)<l(y)else f([X[0]-X[1]*x[0]/y[0]for X in zip(x[:l(y)],y)][1:]+x[l(y):],y,z+[x[0]/y[0]])

Divides term by term. Run with print(f(dividend, divisor))

Try it online

APL (Dyalog Unicode) 18.0, 63 bytes

{0≥d←1+⍺-⍥≢⍵:⍬⍺⋄q(r↓⍨⊥⍨0=⌽r←⍺-m+.×q←(d↑⍺)⌹d↑m←⍉{⍵,⍨⍺⍴0}∘⍵⍤0⍳d)}

Try it online!

The TIO link (which uses 17.1) has a polyfill for ⍥.

Well, it uses ⌹ to solve a system of equations, but technically it's not a library, so...

How it works

Illustration for the inputs x←12 ¯5 3 ¯1 and y←1 0 ¯5:

• We know that, from the lengths of x and y, the quotient will have two terms.
• Construct a matrix which, when matrix-multiplied with the quotient, will give the polynomial product of quotient with y:

 1  0
 0  1
¯5  0
 0 ¯5

• Calculate the quotient q so that first two terms of x agree with those of y×q, using matrix division ⌹.
• Calculate the remainder r using x - y×q, then remove leading zeros of r. (q is guaranteed to have no leading zeros.)

{ ⍝ x←dividend, y←divisor
  d←1+⍺-⍥≢⍵  ⍝ Degree of the quotient; 1 + length of x - length of y
  0≥d: ⍬⍺    ⍝ If degree is zero or lower: quotient is zero, remainder is x

  m←⍉{⍵,⍨⍺⍴0}∘⍵⍤0⍳d  ⍝ Construct the matrix
                 ⍳d  ⍝ 0..d-1
     {      }∘⍵⍤0    ⍝ For each number i, yielding a matrix row
      ⍵,⍨⍺⍴0         ⍝ Make i copies of 0, and append y
                     ⍝ (Implicitly right-pad with zeros for short rows)
    ⍉                ⍝ Transpose

  q←(d↑⍺)⌹d↑m  ⍝ Get the quotient
          d↑m  ⍝ Take first d rows of the matrix
    (d↑⍺)      ⍝ Take first d numbers of x
         ⌹     ⍝ Matrix divide aka. solve linear system of equations

  r←⍺-m+.×q  ⍝ Get the remainder

  q(r↓⍨⊥⍨0=⌽r)  ⍝ Remove leading zeros from r, and join with q
       ⊥⍨0=⌽    ⍝ Reverse r, test equal to zero, and count trailing ones (⊥⍨)
    r↓⍨         ⍝ Drop that many numbers from start of r
  q(         )  ⍝ Join with q
}

Javascript with lambdas, 108

f=(a,b)=>{for(n=b.length;a.length>=n;a.shift())for(b.push(k=a[q=0]/b[0]);q<n;++q)a[q]-=k*b[q];b.splice(0,n)}

It replaces first argument by reminder and second by result.

Example of usage in Firefox:

f(x=[12,-5,3,-1], y=[1,0,-5]), console.log(x, y)
// Array [ 63, -26 ] Array [ 12, -5 ]

Sorry for the bug. Already fixed.

Python 2, 260 258 257 255 bytes

exec'''def d(p,q):
 R=range;D=len(p);F=len(q)-1;d=q[0];q=[q[i]/-d@R(1,F+1)];r=[0@R(D)];a=[[0@R(F)]@R(D)]
@R(D):
  p[i]/=d;r[i]=sum(a[i])+p[i]
  for j in R(F):
   if i<D-F:a[i+j+1][F-j-1]=r[i]*q[j]
 return r[:D-F],[d*i@r[D-F:]]'''.replace('@',' for i in ')

This executes:

def d(p,q):
 R=range;D=len(p);F=len(q)-1;d=q[0];q=[q[i]/-d for i in R(1,F+1)];r=[0 for i in R(D)];a=[[0 for i in R(F)] for i in R(D)]
 for i in R(D):
  p[i]/=d;r[i]=sum(a[i])+p[i]
  for j in R(F):
   if i<D-F:a[i+j+1][F-j-1]=r[i]*q[j]
 return r[:D-F],[d*i for i in r[D-F:]]

Use like so:

>>>d([12., -5., 3., -1.],[1.,0.,-5.])
([12.0, -5.0], [63.0, -26.0])

J, 94

f=:>@(0&{)
d=:0{[%~0{[:f]
D=:4 :'x((1}.([:f])-((#@]{.[)f)*d);([:>1{]),d)^:(>:((#f y)-(#x)))y'

eg.

(1 0 _5) D (12 _5 3 _1;'')
63 _26 | 12  _5

Explanation of some snippets, given that a: (12 -5 3 -1) and b: (1 0 -5)

length of a:

#a
4

make a and b same order by appending zeroes to b:

(#a) {. b
1 0 -5 0

divide higher powers (first elements) of a, b:

(0{a) % (0{b)
12

multiply b by that and subtract it from a:

a - 12*b
12 0 _60

repeat n times b = f(a,b):

a f^:n b

Haskell, 126

For a start:

l s _ 0=s
l(x:s)(y:t)n=x/y:l(zipWith(-)s$map(*(x/y))t++repeat 0)(y:t)(n-1)
d s t=splitAt n$l s t n where n=length s-length t+1

Sample use:

*Main> d [12, -5, 3, -1] [1, 0, -5]
([12.0,-5.0],[63.0,-26.0])