Using Substitution

Date: 10/05/98 at 21:53:52
From: Joshua Gronsbell
Subject: Explanation of your Explanation of Cubic Polynomials

In your section on solving cubic polynomials, you make a leap of logic 
that I just don't understand. Here's the explanation:

"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."  

This much I'm okay with. Where things get complicated is the next step:  

"The next thing we do is to get rid of  the x^2 term by replacing x 
with (x - e/3)."

How exactly does this work? I don't understand how replacing x with 
(x-e/3) helps anything, or, for that matter, how it gets rid of the 
x^2 term. Maybe it's just me, but if you could please explain this 
further, it would be most appreciated. Thanks.

Date: 10/06/98 at 19:49:46
From: Doctor Peterson
Subject: Re: Explanation of your Explanation of Cubic Polynomials

Hi, Joshua. You are referring to the Dr. Math FAQ on cubic equations, 
which can be found here:

  http://mathforum.org/dr.math/faq/faq.cubic.equations.html   

That explanation obviously assumed you are familiar with some of the 
methods you need to use to solve cubics. Substitution is a useful 
technique that we don't tend to teach in algebra. We often see it in 
calculus instead. The basic idea is that we can change to a new 
variable for which the equation is simpler, because some terms drop 
out. It's easier to follow if we don't reuse x, but call the new 
variable y. So let's let:

    x = y - e/3

and put that in place of x in our equation:

                            x^3 + ex^2 + fx + g = 0

    (y - e/3)^3 + e(y - e/3)^2 + f(y - e/3) + g = 0

                 y^3 - e*y^2 + e^2*y/3  - e^3/27
                     + e*y^2 - 2e^2*y/3 + e^3/9
                             + f*y      - ef/3
                                        + g     = 0

                 y^3    +(f - e^2/3)y +(g - ef/3 + 2e^3/27) = 0

Do you see how the y^2 term dropped out? You might like to think about 
why that happened, and why y - e/3 was chosen as the substitution in 
order to make that happen. Pretty smart, huh? (Of course, e, f, and g 
would be numbers derived from your specific problem, so this would 
look simpler than it does here.)

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/