Finding a new polynomial with roots $k$ greater than a given polynomial A similar question appears here, though I wish to ask for polynomials of higher degree.
Let's say we have the polynomial $f(x)=x^4-3x^3-3x^2+4x-6$, and we wish to find a polynomial with roots 3 greater than those of $f(x)$.  I can see that putting $g(x)=f(x-3)$ works, but then we have the expression $$(x-3)^4-3(x-3)^3-3(x-3)^2-4(x-3)+6$$ which will take quite a while to evaluate. (It comes out to $\boldsymbol{x^4-15x^3+78x^2-175x+441}$). 
Is there a shortcut to find the polynomial above through some 'trick' (e.g. some form of synthetic division?)
Furthermore, if we wish to find a polynomial with roots $k$ greater than a given polynomial, will this new polynomial have a given 'form' of $k$? E.g. what if I wanted to solve the above question, but with my roots being $4,5,$ or $6$ greater than those of the given polynomial?

Well, your roots will be shifted $3$ units to the right. From there, you can calculate the new roots, using the sum and product of roots formulae to calculate the coefficients of each term. This can be considered a shortcut of sorts. 
For a polynomial $f(x)$ where $$f(x)=a_nx^n+...+a_1x+a_0$$
$$\alpha\beta\gamma...=(-1)^n\frac{a_0}{a_n}$$
$$\alpha+\beta+\gamma+...=-\frac{a_{n-1}}{a_n}$$
The greek letters represent the roots.
There are other formulae for finding $a_{n-2}$ and so forth. I will give $a_{n-2}$ in cyclic sum notation, but the rest are easily accessible and directly quotable.
$$\sum \alpha\beta=\frac{a_{n-2}}{a_n}$$
