Complex Roots of a Quartic Equation

Question

Here's the question I'm currently working on:

Suppose the roots of $x^4 + ax^3 + bx^2 +cx + d = 0$ are such that the product of any two equals the product of the other two. Show that $x^4 + ax^3 +bx^2 +cx +d$ factors into either $(x+r)^4$ or $(x^2 - r^2)^2$ for some complex number $r$.

I just finished reading Lagrange's method for solving quartic equations and my textbook really didn't go into detail much for this section. All that he talked about was how permuting the roots can lead to forming new expressions that enables us to solve for the other roots. How should I approach this problem, noting that were working with complex numbers here?

Answer 1

You can ignore the polynomial and just focus on the roots. Call them $p,q,r,s$. First they must all have the same magnitude. Assume $|p| \gt |q|$. Then since $pr=qs$ we must have $|s| \gt |r|$ but then $|ps| \gt |qr|$, a contradiction. Now argue that the angles must all be the same or they must come in two opposite pairs. The two cases have the polynomial factoring as $(x-p)^4$ or as $(x-p)^2(x+p)^2$

Answer 2

Let $P(x)=x^4+ax^3+bx^2+cx+d=(x-A)(x-B)(x-C)(x-D)$ for all $x.$

You have overlooked another case.

(1). If at least one of $A,B,C,D$ is $0$, say $A=0$. Then $0=AB=CD$ so one of $C,D$ is $0.$ So say $C=0.$ Then $0=AC=BD$ so one of $B,D$ is $0$.So say $B=0$. Then $$P(x)=x^3(x-D).$$ Note that $D$ can be anything.

(2). If none of $A,B,C,D$ is $0$ we have $$A=CD/B=BD/C$$ which implies $$A/D=C/B=B/C$$ which implies $$B^2=C^2.$$ By permuting the letters $A,B,C,D$ in every manner in the above argument we will conclude that $$A^2=B^2=C^2=D^2.$$ So at least $2$ of $A,B,C,D$ are equal because $\{A,B,C,D\}\subset \{A,-A\}.$ So say $A=B.$

If $C=A$ then $D=A$ because $0\ne A^2=AB=CD=AD.$

If $C=-A$ then $D=-A$ because $0\ne A^2=AB=CD=-AD.$

So $$P(x)=(x-A)^4 \text { with } A\ne 0$$ $$\text { or } P(x)=(x^2-A^2)^2 \text { with } A\ne 0.$$

(3). Because there are "ifs" in (1) and (2) that are not "iffs" we must confirm that the formulas for $P(x)$ in (1) and (2) do indeed meet the initial conditions.

(4). We can re-arrange the conclusions of (1) and (2) as equivalent to $$ P(x)=(x-A)^4 $$ $$\text {or}\quad P(x)=(x^2-A^2)^2 $$ $$ \text {or}\quad P(x)=x(x-D)^3 \;\text { with } D\ne 0.$$