Is there a substitute that will eliminate terms of the degree $3$ and $2$ from the quartic polynomial?

Question

The standart form of the quartic equation:

$$ax^4+bx^3+cx^2+dx+e=0$$ where $$a\neq 0.$$

The purpose of the question is not to solve quartic. Resources related to the solution are sufficiently available. Just about special curiosity, I am looking for a special substitute that will transform the following:

$$ax^4+bx^3+cx^2+dx+e=0 \Longrightarrow z^4+pz+q=0$$ where $$x=f(z)$$

Here $f(z)$ can be a linear or nonlinear complex function.

This substitute is known:

$$x\longmapsto x-\frac b{4a}$$

Which gives us

$$x^4+px^2+qx+r=0$$

Is a substitution known to eliminate term of degree $3$ and $2$ together?

Answer 1

We can assume that $a=1$ by dividing by it, so start with $$ x^4 + bx^3 + cx^2 + dx + e = 0 $$ (there's no need for $f$, either).

We will use a more general substitution known as a Tschirnhaus substitution: put $$ y = x^2 + \lambda x + \mu . $$ Then $$ y^2 = x^4 + 2\lambda x^3 + 2(\mu+\lambda^2) x^2 + 2\lambda\mu x + \mu^2 = (2\lambda-b)x^3 + (\lambda^2+2\mu-c)x^2 + (2\lambda\mu-d)x + (\mu^2-e) \\ y^3 = \dotsb = (-b^3 + 2 b c - d + 3 b^2 \lambda - 3 c \lambda - 3 b \lambda^2 + \lambda^3 - 3 b \mu + 6 \lambda \mu) x^3 + (-b^2 c + c^2 + b d - e + 3 b c \lambda - 3 d \lambda - 3 c \lambda^2 - 3 c \mu + 3 \lambda^2 \mu + 3 \mu^2) + \dotsb \\ y^4 = \dotsb , $$ and then since $1,y,y^2,y^3,y^4$ give five equations with four coefficients, we can find a linear combination so that they add up to zero, leaving us with the equation $$ y^4 + (-b^2 + 2 c + b \lambda - 4 \mu) y^3 + (c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2) y^2 + ()y + () = 0 , $$ where we're not interested in the last two coefficients. (This effectively calculates the resultant of the two polynomials $x^4+bx^3+cx^2+dx+e$ and $x^2+\lambda x-\mu-y$, which is generally best done with a computer.)

To eliminate the first two coefficients, we need $$ -b^2 + 2 c + b \lambda - 4 \mu = 0 \\ c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2 = 0 . $$ The first of these is linear, the second quadratic, so it is easy to solve both simultaneously: we find $$ \lambda = \frac{3 b^3 - 10 b c + 12 d \pm 2 \Delta}{3b^2-8c}, \qquad \mu = \frac{2 b^2 c - 8 c^2 + 6 b d \pm b \Delta}{6b^2-16c} , $$ where $\Delta^2 = 6 b^3 d - 28 b c d - 2 b^2 (c^2 - 6 e) + 4 (2 c^3 + 9 d^2 - 8 c e) $. I've left out a few of the details of the calculation, but they should be easy, if tedious, to fill in.

Answer 2

If you mean linear substitutions $x\mapsto \alpha x+\beta$, then the answer is no. Note that $x^4+qx+r$ has positive second derivative everywhere (equivalently, its graph is convex), and this property is preserved by linear substitutions of this type. On the other hand, not every quartic polynomial has this property—for example, $x^4-x^3$.