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### Solving Cubic and Quartic Polynomials

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

Hi,

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

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

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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