Solving The Quasi-Symmetric Quartic Equation

Question

This is a self-answer question of the following problem:

Solve the equation:

$$4x^4 - 36x^3 + 61x^2 + 90x + 25 = 0.$$

See my answer.

Answer 1

The rational root theorem gives $$ 4x^4 - 36x^3 + 61x^2 + 90x + 25=(2x + 1)^2(x - 5)^2. $$ Of course, there is some work to do for it, but it seems to me the first thing to try. So the polynomial factors into linear factors.

One could also compare coefficients with $(2x+a)(2x+b)(x+c)(x+d)$ and would find this factorization, or first consider $(4x^2+ax+b)(x^2+cx+d)$.

Answer 2

Solve the equation:

$$4x^4 - 36x^3 + 61x^2 + 90x + 25 = 0.$$

This problem was lifted from this Youtube Video. I didn't particularly like the video's approach, so I used my own approach, which I have never seen discussed anywhere. To understand the approach, first see this answer.

Assuming that $~u = x - \dfrac{1}{x},~$ the foundation of the linked answer is that in a symmetric quartic equation, you can (for example) express $~\displaystyle x^2 + \frac{1}{x^2}~$ as $~u^2 + 2.~$

This implies that a symmetric quartic equation like

$$a_4x^4 + a_3x^3 + a_2x^2 - a_3x + a_4 = 0$$

can be converted into a quadratic equation in $~u.$

The point of this self-answer posting is that this method may be extended beyond symmetric quartic equations to quasi-symmetric quartic equations (a description that I made up).

Suppose (for example) you have the quartic equation

$$a_4 + a_3x^3 + a_2x^2 - a_1x + a_0 = 0$$

where

$$\left| ~\frac{a_1}{a_3} ~\right|^2 = \frac{a_0}{a_4}.$$

I refer to such a quartic equation as quasi-symmetric, because you can set $~\displaystyle k = \left| ~\frac{a_1}{a_3} ~\right|,~$ and then attack the quartic equation with the substitution

$$~u = x - \frac{k}{x} \implies u^2 + 2k = x^2 + \frac{k^2}{x^2}.$$

In the posted problem, you have that

$$\left| ~\frac{a_1}{a_3} ~\right|^2 = \left| ~\frac{-5}{2} ~\right|^2 = \frac{25}{4} = \frac{a_0}{a_4}.$$

The remainder of this posting works through the $~\displaystyle u = x - \frac{5}{2x}~$ solution.

---

First, construct the equation

$$4x^2 - 36x + 61 + \frac{90}{x} + \frac{25}{x^2} = 0.$$

Then, since $~\displaystyle u^2 + 5 = x^2 + \frac{25}{4x^2},$

you have that

$$4(u^2 + 5) - 36u + 61 = 0 \implies 4u^2 - 36u + 81 = 0 \implies $$

$$(2u - 9)^2 = 0 \implies x - \frac{5}{2x} = u = \frac{9}{2} \implies $$

$$2x^2 - 9x - 5 = 0. \tag1 $$

So, (1) above yields two of the roots, and the other two roots can be derived by polynomial long division:

                2x^2 -  9x   -  5
               --------------------------------
2x^2 - 9x - 5 | 4x^4 - 36x^3 + 61x^2 + 90x + 25
                4x^4 - 18x^3 - 10x^2
               ---------------------
                     - 18x^3 + 71x^2 + 90x
                     - 18x^3 + 81x^2 + 45x
                     ---------------------
                             - 10x^2 + 45x + 25
                             - 10x^2 + 45x + 25
                             ------------------

Note
I am deliberately ducking a question that I am not really qualified to answer: was the polynomial long division avoidable? That is, since the quadratic equation in $~u~$ gives a repeated root, does this imply that each of the roots of the original quartic equation is repeated?

Answer 3

I am aware of a technique of reducing quartic polynomials using substitution $y=x+\frac{a_2}{4a_1}$. Applying this substitution leads to $4y^4-\frac{121}{2}y^2+\frac{121^2}{64}$ which despite somewhat scary coefficients is easy to solve directly or factor. Reference: Factoring Quartic Polynomials