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Cubic Polynomial

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


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 
quadratic formula.

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.

If you'd like a nice reference about this problem, as well as some 
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!   
Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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