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I'm a self-learner trying to solve polynomial equations.

This equation $$x^4-6x^3+\frac{31} 4x^2-\frac 3 2x-2=0$$ has $4$ irrational roots (if I'm right.) I found the solution by on-line solver. HOWEVER, I want to solve this by hand, without graphing calculator, and without using quartic formula.

Can it be done? If so, how? Is the procedure efficient, or should I rely on electronic devices?

Brian
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  • It's not clear to me what you mean by "solve" - do you want to calculate the roots, or just prove that they're all irrational? (I would normally assume the former) – Zev Chonoles Sep 08 '13 at 03:26
  • There are roots in $(0,1)$ and $(-1,0)$ – lab bhattacharjee Sep 08 '13 at 03:27
  • Yes Zev, the former is what I meant. Lab, I don't understand your comment. As far as I know any root of a polynomial should be a number(be it rational or irrational) not an interval. – Brian Sep 17 '13 at 16:40

2 Answers2

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Great question I believe! You probably know this but this is said in MacTutor's Quadratic, cubic and quartic equation.

"After Tartaglia had shown Cardan how to solve cubics, Cardan encouraged his own student, Lodovico Ferrari, to examine quartic equations. Ferrari managed to solve the quartic with perhaps the most elegant of all the methods that were found to solve this type of problem. Cardan published all 20 cases of quartic equations in Ars Magna. Here, again in modern notation, is Ferrari's solution of the case: x4 + px2 + qx + r = 0. First complete the square to obtain

[ ... ]

In the years after Cardan's Ars Magna many mathematicians contributed to the solution of cubic and quartic equations. Viète, Harriot, Tschirnhaus, Euler, Bezout and Descartes all devised methods. Tschirnhaus's methods were extended by the Swedish mathematician E S Bring near the end of the 18th Century."

I just wanted to put it here to emphasize how "non-trivial" the problem of solving quartic equations can be. (I believe that you are aware of methods of finding rational roots, Uspensky's Theory of Equations is a great reference for that; but the general problem is definitely non trivial).

Here are many of this methods explained. I also found some of them here.

Mauricio Tec
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One method to solving the general quartic entails factoring it into two quadratics, but you will need to solve an intermediate cubic. First, given,

$$x^4+Ax^3+Bx^2+Cx+D = 0$$

let $x = y+t$ for some $t$ and new variable $y$,

$$(y+t)^4+A(y+t)^3+B(y+t)^2+C(y+t)+D = 0$$

Expand and collect the variable $y$,

$$y^4+(A+4t)y^3+\dots = 0$$

Let $t = -A/4$ to get rid of its $y^3$ term to get the depressed form,

$$y^4+py^2+qy+r=0\tag{1}$$

Assume it can be factored as,

$$(y^2+ay+b)(y^2-ay+c) = y^4+(-a^2+b+c)y^2-a(b-c)y+bc = 0\tag{2}$$

Equate coefficients of $(1)$ and $(2)$,

$$\begin{aligned} p &= -a^2+b+c\\ q &= -a(b-c)\\ r &= bc \end{aligned}\tag3$$

Solving for $b,c$, you will end up with a cubic in $a^2$. Once you have $a,b,c$, you can then solve,

$$y^2+ay+b = 0$$ $$y^2-ay+c = 0$$

which should give you the 4 roots of $(1)$.

Tedious, yes, but at least now you know how to solve the general quartic. (Though it means first knowing how to solve the general cubic.)

Tito Piezas III
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