Polynomial Division: dividing by a double root I have this fairly interesting problem that is based on Polynomial Division and/or factor/remainder theorem.

Determine the values of $a$ and $b$ such that $ax^4 + bx^3 -3$ is divisible by $(x-1)^2$.

This is interesting because the root we are dividing by is a double root, so its difficult to get $2$ equations and $2$ unknowns.
(note that one cannot use ideas from Calculus)
The only approach I have been able to come up with is to expand the perfect square divisor to a full quadratic and perform a brute force division, but it did not really lead me to a solution.

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...  
