If we have a degree $n$ polynomial $$p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0$$with coefficients in a field, say $\Bbb C$, for concreteness, it is well known that the substitution $y= x + \frac{a_{n-1}}{na_n}$ "kills" the $a_{n-1}x^{n-1}$ term, that is, we get something of the form $$b_ny^n +b_{n-2}y^{n-2}+\cdots+ b_1y+b_0.$$ I think of that substitution as "translating" the polynomial such that the mean of the roots "in the new variable $y$" is zero. Everything seems to work fine since the arithmetic mean of $n$ numbers is a linear function of them, in some sense.
Question: is there a suitable substitution that eliminates an arbitrary term $a_kx^k$, $0\leq k < n $, of our choice?
Working with $n=3$ around, it seems that there are two possibilities for "depressing" the $a_1x$ term (given by the quadratic formula), and these expressions are very close to the closed expressions for the critical points of a third degree polynomial. I don't know if this is a coincidence or not, but I digress.
Also, thinking of Vieta's formulae doesn't seem to lead anywhere, because of the non-linearity of the expressions involved. I googled around a bit, and this might be related to Tschirnhaus transformations, but my background in algebra is weak and I don't know enough Galois theory to digest some of that information (and that's probably overkill).
I would like some intuition behind such substitutions instead of the obvious "write $y = x+t$, plug in, put the desired coefficient equal to zero and solve for $t$", if said substitutions do exist - I'd like some intuition that avoids computations as most as possible (something geometric, maybe?)
Thanks, and sorry for the rant. (You can edit in tags that might be also relevant, I'm a bit lost here)
Yes, if you want to eliminate only one term, then the linear substitution $x = y+t$ is enough. For example, given the cubic,
$$P(x): = ax^3+bx^2+cx+d\tag1$$
Let $x = y+t$, collect the new variable $y$, and you get,
$$P(y): = a y^3 + (b + 3 a t) y^2 + (c + 2 b t + 3 a t^2) y + (d + c t + b t^2 + a t^3)$$
You can eliminate the non-leading terms by equating any of the coefficients to zero and solving for $t$. I'm sure you can see this easily generalizes to higher powers. However, once you go above the quintic, getting rid of some terms may involve $t$ that is no longer in radicals.
P.S. If you want to get rid of two or three terms simultaneously, then you need a Tschirnhausen transformation which is just a generalization, though you have to equate your polynomial to zero. See this post for an easy explanation on how to do this as applied to the quintic.