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### Using Substitution

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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/
```
Associated Topics:
High School Basic Algebra

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