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Solving Cubic and Quartic Polynomials


Date: 04/30/98 at 17:53:13
From: E. Aevan vonBerwick, IV
Subject: Cubic, Quartic Formulas

Hi,

I know you have received numerous questions about the Cubic and 
Quartic Formulas, and I know they are very long, so I will not ask you 
to give them to me. But, I was wondering if you could tell me where I 
can find these formulas in their entirety, whether they be web pages 
or other math publications.

Thank you,
-Aevan


Date: 04/30/98 at 19:02:58
From: Doctor Barney
Subject: Re: Cubic, Quartic Formulas

You're right, it would be difficult to write them in a text message.  
There is a very common handbook called the CRC Standard Mathematical 
Tables. It should be available in the reference section of any college 
library. Look in the index under "cubic."

-Doctor Barney,  The Math Forum
Check out our web sites! http://mathforum.org/dr.math/   


Date: 04/30/98 at 19:34:15
From: Doctor Anthony
Subject: Re: Cubic, Quartic Formulas

What you are wanting is Tartaglia's Solution (also known as Cardan's 
Solution) to the general cubic. It was invented by Tartaglia but was 
first published in 1545 by Cardan, who had obtained it from Tartaglia 
after giving a solemn promise not to reveal it.

The general cubic ax^3 + 3bx^2 + 3cx + d = 0 is changed to a simpler 
cubic without a term in x^2 by the substitution:

     y = ax+b  (or x = (y-b)/a)

to get:

     y^3 + 3Hy + G = 0

where:

     H = ac-b^2    and    G = a^2*d - 3abc + 2b^3

We now make use of the identity shown below, where w is a complex cube 
root of unity:

     x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x+wy+w^2*z)(x+w^2*y +wz)

This is easily proved if you know how to factorize determinants, but 
if not, you can prove it the long way by multiplying out the factors, 
and remembering that w^3 = 1.

The cubic is now expressed as: 

     y^3 - 3pqy + p^3 + q^3 = 0

with:

     H = -pq    and    G = p^3 + q^3    

So y^3 - 3pqy + p^3+q^3 = (y+p+q)(y+wp+w^2*q)(y+w^2p+wq) = 0

and the three roots will be:

     (1) y = -p - q
     (2) y = -wp - w^2q
     (3) y = -w^2p - wq

So all that is required is that we have the values of p and q from the 
two equations: 

     H = -pq    and    G = p^3 + q^3

Here, think in terms of p^3 and q^3 rather than p and q, and consider 
instead the two equations:

     p^3 + q^3 = G    and    p^3*q^3 = -H^3

Now we can see that p^3 and q^3 are the roots of the quadratic:

     t^2 - Gt - H^3 = 0

and their cube roots must be chosen so that pq = -H.

The roots of this quadratic are:

         (1/2)[G +-sqrt(G^2 + 4H^3)]

If G^2 + 4H^3 > 0   p and q are real and distinct

If G^2 + 4H^3 = 0   p and q are real and equal. The corresponding 
                    roots of the cubic are  -2p, p, p.

If G^2 + 4H^3 < 0   p and q are complex. The cubic then, ironically, 
                    has 3 real roots. In this situation it is better 
                    to use a trigonometrical method.

Trig Solution of Cubic Equation
-------------------------------

Having expressed the equation in the form  y^3 + 3Hy + G = 0  and 
finding that G^2 + 4H^3 < 0, we know that there are three real roots.

For this method we use the trig identity:

     cos(3*theta) =  4*cos^3(theta) - 3*cos(theta)

The method is best illustrated by an example:

     x^3 - 6x - 4 = 0

We note that: 

     3H = -6,  H = -2,  and  G = -4

and: 

     G^2 + 4H^3 =  16 - 32 = -16

so the condition for 3 real roots applies.

Substitute in x = k*cos(theta) and choose k so that the left hand side 
is a multiple of 4*cos^3(theta) - 3*cos(theta):

     k^3.cos^3(theta) - 6k.cos(theta) = 4

This means that:

     k^3     4
    ----- = ---
     6k  =   3

So k^2/6 =  4/3, which implies k^2 = 8. Thus, k = 2*sqrt(2).

Solving for theta:

     16*sqrt(2)cos^3(theta) - 12*sqrt(2)cos(theta) = 4

     4*cos^3(theta) - 3*cos(theta) = 1/sqrt(2)

     cos(3*theta) = 1/sqrt(2)
  
     3*theta = pi/4, 7pi/4,  9pi/4 

     theta = pi/12,  7pi/12,  3pi/4  

Thus:

     x = 2*sqrt(2)cos(pi/12)     or
     x = 2*sqrt(2)cos(7*pi/12)   or
     x = 2*sqrt(2)cos(3*pi/4) = 2*sqrt(2)(-1/sqrt(2)) = -2

Having found that x = -2 is a root we could, if we wished, divide out 
by x+2 and then solve the resulting quadratic, to get the two other 
roots already found above.

In fact:

     x^3 - 6x - 4 =  (x+2)(x^2 - 2x - 2) = 0

So x = -2, and  x = 1 +-sqrt(3)

-------------------------------------------------------------------
QUARTIC

I will outline the method for solving the general quartic:

     z^4 + 4pz^3 + 6qz^2 + 4rz + s = 0

As for the cubic, the first step is to reduce the equation to one with 
no term in z^3. So we get:
 
     w^4 + aw^2 + bw + c = 0

Use the identity:

     (w^2 + t)^2 = u(w + v)^2          (Equation 1)

and show that t is a root of the CUBIC equation (known as the 
reducing cubic):

     8t^3 - 4at^2 - 8ct + 4ac - b^2 = 0

If one of the roots is t and the corresponding values are then u and v 
from the identity (1) above, then:

     w^4 + aw^2 + bw + c =
              [w^2-u^(1/2)*w+t-vu^(1/2)][w^2+u^(1/2)*w+t+vu^(1/2)] 

putting each quadratic factor separately equal to 0 gives the 
values of w.

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
College Number Theory
High School Number Theory

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