Factoring polynomials and equating coefficients I am studying Factorising polynomials and equating coefficients at the moment, and up until now all has been well. I have no problem working these out up as high as the third degree as there are plenty of examples. 
Questions like this:
Factorise the expression $x^3 - 17x^2 + 54x - 8$ given $(x-4)$ is a factor.
I can work this out using $(x-4)(\alpha x^2 + \beta x + \gamma)$ and then equating the coefficients.
However, in the end of block exercises some of the questions are 4th degree polynomials and I don't know where to begin. 
With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a factor for example.
Do I use the same procedure but start a degree higher, like $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$  and move on down to figure out the 2nd degree after that? 
 A: 
With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a
  factor for example.
Do I use the same procedure but start a degree higher, like
  $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$  and move on down
  to figure out the 2nd degree after that?

There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to
$$\begin{eqnarray*}
P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$
Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality
$$\begin{eqnarray*}
P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right)  \\
&=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right)
x^{2}+\left( \gamma +\delta \right) x+\delta .
\end{eqnarray*}$$
Equating coefficients you get the simple system of four linear equations
$$
\begin{equation*}
\left\{ 
\begin{array}{c}
\delta =20 \\ 
\gamma +\delta =31 \\ 
\beta +\gamma =3 \\ 
\beta +1=-7,
\end{array}
\right. 
\end{equation*}$$
whose solution is $\delta =20,\beta =-8,\gamma =11$. So
$$\begin{equation*}
x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left(
x^{3}-8x^{2}+11x+20\right). 
\end{equation*}\tag{2}$$
Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that
$$\begin{equation*}
Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left(
x-5\right) .
\end{equation*}$$
Consequently, 
$$\begin{equation*}
x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left(
x-5\right). 
\end{equation*}\tag{3}$$
A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, )  is shown bellow, where the remainder is $P(r)=s$.

