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### Cubic Polynomial

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Date: 6/27/96 at 14:23:46
From: Anonymous
Subject: Cubic Polynomial

Sketch the graphs of y=ax^3, y=ax^3+mx, y=ax^3-mx for some a,m>0. Note
that all of these graphs have a point of symmetry at the origin. Show
that every cubic polynomial function y=ax^3+bx^2+cx+d can be expressed
in the form y=a(x-h)^3+m(x-h)+k for some h,k; thus the graph of every
cubic polynomial has a point of symmetry.
```

```
Date: 9/12/96 at 11:26:24
From: Doctor Ken
Subject: Re: Cubic Polynomial

Hello!

This is our standard answer for the question "What are the general
solutions to cubic and quartic polynomial equations?"  This actually
gives a description of how to solve the cubic, but we don't describe
how to do the quartic.  It's just too darn messy.  Contained in the
method for solving the cubic equation is a transformation (a variable
substitution, which is essentially the step you ask for too -
substitute x-h for x) that will eliminate the squared term.

Okay, here's how you do the cubic.  Let's say you have the equation
ax^3 + bx^2 + cx + d = 0.  The first thing you do is to get rid of the
"a" out in front by dividing the whole equation by it.  Then we get
something in the form of x^3 + ex^2 + fx + g = 0.  The next thing we
do is to get rid of the x^2 term by replacing x with (x - e/3).  That
will give us something of the form of x^3 + px + q = 0.  This is much
easier to solve, although it's still kind of hard and it took a lot of
really smart people a really long time to do.

Let's introduce two new variables, t and u, defined by the equations
u - t = q               and
tu = (p/3)^3.

If you're really weird, you might want to know what t and u actually
are, in terms of p and q.  Well, here they are:

-3q +- Sqrt{4p^3 + 27 q^2}         27q +- Sqrt{108p^3 + 729q^2)
t = -------------------------  ,   u = ----------------------------
6 Sqrt{3}                                54

Where that +- is either + or - for BOTH of the equations at the same
time. You can't have one as a + and the other as a -.

Anyway, x = CubeRoot{t} - CubeRoot{u} will be a solution of x^3 + px +
q = 0. Verify this result now (DON'T use the explicit formulas for t
and u, but just plug CubeRoot{t} - CubeRoot{u} into the cubic
polynomial), and make sure you see why it works.  To find the other
two solutions of our cubic polynomial (if there are any real ones) we
could divide x^3 + px + q by its known factor (x - CubeRoot{t} +
CubeRoot{u}), getting a quadratic equation that we could solve by the

As you can see the complexity is much greater than it is in the
Quadratic Formula.  To try to go backwards and come up with a closed
form for the Cubic Formula in terms of the original a, b, c, d would
be a real pain.

stuff about the Quartic, see if you can lay your hands on the 3-volume
set called "Mathematical Thought from Ancient to Modern Times" by
Morris Kline.  It gives the math and the history together, which is
pretty cool.

-Doctor Ken,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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