Let's forget the word "division" for a moment.
The claim is that if $f$ and $g$ are polynomials in one variable (i.e., in one indeterminate) with coefficients in a field (e.g., with rational coefficients), and if $g$ is not the zero polynomial, then there exist unique polynomials $q$ and $r$, with $\deg r < \deg g$, such that
$$
f = qg + r.
$$
The preceding equation expresses equality of polynomials as abstract algebraic entities. If you're working with coefficients in a field of characteristic zero (e.g., rational, real, or complex coefficients), it's safe to reformulate the preceding in terms of values of polynomial functions:
$$
f(x) = q(x) g(x) + r(x)
\quad\text{for all $x$.}
\tag{1}
$$
The discussion below uses "function value" language.
Uniqueness is easy: if
$$
q_{1}(x) g(x) + r_{1}(x)
= f(x)
= q_{2}(x) g(x) + r_{2}(x)
\quad\text{for all $x$,}
$$
then $\bigl(q_{1}(x) - q_{2}(x)\bigr) g(x) = r_{2}(x) - r_{1}(x)$. Since the left-hand side is a polynomial multiple of $g$ (hence is either $0$ for all $x$, or is a polynomial of degree at least $\deg g$) and the right-hand side is a polynomial of degree strictly smaller than $\deg g$, each side must be $0$. That is, $q_{1}(x) = q_{2}(x)$ for all $x$, and $r_{1}(x) = r_{2}(x)$ for all $x$.
One proof of existence proceeds by successive subtraction of (polynomial multiples of) $g$ from $f$: The preceding conclusion may be written
$$
f(x) - q(x) g(x) = r(x)
\quad\text{for all $x$,}
$$
in which case it's clear the goal is "to subtract a polynomial multiple of $g$ from $f$, obtaining a polynomial of degree strictly smaller than $\deg g$".
To illustrate the main idea, consider the example
$$
f(x) = x^{3} - 12x^{2} - 42,\quad
g(x) = x - 3.
$$
The "game" will be to subtract successive monomial multiples of $g$ from $f$ with the goal of reducing the degree of the difference at each step.
To that end, focus on the highest-degree term of each: $f(x) = x^{3} + \cdots$ and $g(x) = x + \cdots$. Since $x^{3} = x^{2} \cdot x$, we're led to consider
\begin{align*}
f(x) - x^{2}g(x)
&= x^{3} - 12x^{2} - 42 - x^{2}(x - 3) \\
&= -9x^{2} - 42.
\end{align*}
Since $\deg (-9x^{2} - 42) = 2 < 3 = \deg f$, we've succeeded in writing $f$ as a polynomial multiple of $g$ plus a remainder of degree smaller than $\deg f$ [sic., not $\deg g$]. Note that the preceding holds for all $x$.
Continue in this vein. By similar consideration of the highest-degree terms, $-9x^{2} = -9x\cdot x$, we're led to write
\begin{align*}
\bigl[f(x) - x^{2}g(x)\bigr] - (-9x)\, g(x)
&= \bigl[-9x^{2} - 42\bigr] + 9x(x - 3) \\
&= -27x - 42.
\end{align*}
Combining the two preceding steps, we have
$$
f(x) - (x^{2} - 9x)\, g(x) = -27x - 42\quad\text{for all $x$.}
$$
The remainder still has degree greater than or equal to the degree of $g$, so we can perform another step, obtaining
$$
f(x) - (x^{2} - 9x - 27)\, g(x) = -123\quad\text{for all $x$.}
$$
Here the process ends: Since $\deg(-123) = 0 < \deg g$, we cannot reduce the degree of the remainder further by subtracting monomial multiples of $g$. We have established a decomposition of the form (1) in this example:
$$
f(x) - \underbrace{(x^{2} - 9x - 27)}_{q(x)}\, g(x) = \underbrace{-123}_{r(x)}\quad\text{for all $x$.}
$$
The close analogy with Euclid's algorithm should be clear. The customary "polynomial long division" notation expresses this process more concisely.
Writing out a general proof is straightforward. The key technical step is the following observation: If $f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots$ is a polynomial of degree $n$ (i.e., if $a_{n} \neq 0$), if $g(x) = b_{m} x^{m} + b_{m-1}x^{m-1} + \cdots$ has degree $m$ (i.e., if $b_{m} \neq 0$), and if $\deg g = m \leq n = \deg f$ (i.e., if we do not already have a "remainder" of degree smaller than $\deg g$), then
$$
f(x) - \frac{a_{n}}{b_{m}} x^{n - m}\, g(x)
= \underbrace{\left(a_{n} - \frac{a_{n}}{b_{m}}\, b_{m}\right)}_{= 0} x^{n}
+ \left(a_{n-1} - \frac{a_{n}}{b_{m}}\, b_{m-1}\right) x^{n-1} + \cdots
$$
has degree at most $n - 1$. (The second coefficient in parentheses might be zero, so the degree of this "partial remainder" might be strictly smaller than $n - 1$.)
Now do induction on the following statement $P_{n}$:
If $f$ is a polynomial of degree an most $n$ in one variable, and if $g$ is a polynomial in one variable, there exist polynomials $q$ and $r$ such that $f(x) = q(x) g(x) + r(x)$ and $\deg r < \deg g$.
As base case, the induction starts with constant polynomials, for which the conclusion is obvious.
The "key technical step" above shows that if $\deg f = k + 1$, then $f$ may be written as a monomial multiple of $g$ plus a polynomial of degree at most $k$, to which the inductive hypothesis may be applied.
Method one (compare the coefficients):
$$\begin{eqnarray}ax^4+bx^3-3 &=& (x-1)^2(cx^2+dx+e)\\
&=& (x^2-2x+1)(cx^2+dx+e)\\
&=& cx^4+(d-2c)x^3+(e-2d+c)x^2+(-2e+d)x+e\end{eqnarray}$$
From here we see that:
$$
\begin{eqnarray*}
c&=&a\\
e&=& -3\\
-2e+d&=& 0\implies d=-6\\
e-2d+c&=&0 \implies c=-9\implies a=-9\\
d-2c&=&b\implies b=-12
\end{eqnarray*}$$
Method two (Vieta formulas), $x_1=x_2=1$:
$$ 2+x_3+x_4 = -{b\over a}$$
$$ 1+2(x_3+x_4)+x_3x_4 = 0$$
$$ 2x_3x_4+ x_3+ x_4 =0$$
$$ x_3x_4 = -{3\over a}$$
from 2. and 3. equation we get
$$x_3+x_4 = -{2\over 3}\;\;\;{\rm and}\;\;\;x_3x_4 ={1\over 3}$$
From 4. equation we get $a=-9$ and from 1. equation we get $b=-12$.
Method three Since $1$ is root of a polynomial $p(x)=ax^4+bx^3-3$ we get $b=3-a$ so we have
$$p(x)=ax^4+3x^3 -ax^3-3 $$
$$= ax^3(x-1)+3(x-1)(x^2+x+1) $$
$$= (x-1)\underbrace{\Big(ax^3+3(x^2+x+1)\Big)}_{q(x)} $$
Now since $1$ double root we have also $q(1)=0$ so $a+9=0$.
Method four: (Horner schema)
$$\begin{array}{cccccc}
& a & b & 0 & 0 & -3 \\ \hline
1 & & a & a+b & a+b & a+b \\ \hline
& a & a+b & a+b & a+b & \color{red}{a+b-3} \\ \hline
1 & & a & 2a+b & 3a+2b & \\ \hline
& a & 2a+b & 3a+2b & \color{red}{4a+3b} & \\
\end{array}$$
Both red expressions must be zero...
Method five: Direct (long) division. What you are left ($1$. degree polynomial ) must be identical to $0$, so you get $2$ equations again...
Best Answer
Let $f(x)$ be your polynomial. It has a number of symmetries. Dietrich Burde's answer exhibits one. We see another by calculating
$$g(u):=2^{-9}f(\sqrt{2u})=32 u^{14}+144 u^{13}+312 u^{12}+380 u^{11}+152 u^{10}-384 u^9-964 u^8-1217 u^7-964 u^6-384 u^5+152 u^4+380 u^3+312 u^2+144 u+32.$$
A most obvious feature of this is that it is palindromic. Meaning that $u=\xi$ is a zero of $g(u)$ if and only if $u=1/\xi$ is. A common trick taking advantage of this is to write everything in terms of the new variable $v=u+1/u$. It is simple to verify that $$ u^{-7}g(u)=P(v), $$ with $$ P(v)=32 v^7+144 v^6+88 v^5-484 v^4-960 v^3-608 v^2-84 v+23. $$ Mathematica thinks that $P(v)$ is irreducible over $\Bbb{Q}$. Judging from the plot it has five zeros in the interval $(-2,0)$ and two positive zeros approximately $0.1$ and $2.1$. If $u$ is real, then $v=u+1/u$ has absolute value $\ge2$. This would yield four real zeros of $f(x)$ subject to symmetries: $x\mapsto -x$ and $x\mapsto 2/x$.