Substitution to Eliminate Linear Terms

Date: 2/9/96 at 10:51:49
From: Anonymous
Subject: Modern algebra topics

I have a few questions about Modern Algebra:

  1. When finding the roots of a cubic equation we substitute 
x = y - b/3 to eliminate the square term.  Are there any such 
substitutions for eliminating the linear term in both quadratic 
and quartic equations?

  2. Given a set S and a relation R on S, how do I show the 
equivalence classes are disjoint and that their union is S?

Thanks,   Joe D'Amato

Date: 6/27/96 at 12:35:7
From: Doctor Ceeks
Subject: Re: Modern algebra topics

In any polynomial in x of any degree n, substitution of y-k for x 
yields a polynomial in y whose coefficients are polynomials in k.  
In particular, the coefficient of the linear term y will be a 
polynomial in k generally of degree n-1.  Any value of k which 
makes this coefficient 0 will work.  (And there will always be a 
solution...though possibly complex...thanks to Gauss' Ph.D. 
thesis!)

For quadratics, this process is known as completing the square.
(Note that your question asks something different from what you 
did for the cubic, where you eliminated the quadratic term, not 
the linear term.  Also, note that you cannot simultaneously 
eliminate both the linear and quadratic terms in general, although 
in special cases you can.)

For your second question: every element of S belongs to an 
equivalence class, so the union is definitely all of S.  On the 
other hand, if x belongs to the equivalence classes A and B, then 
since all a in A are equivalent to x and all b in B are equivalent 
to x, by transitivity, all a in A are equivalent to all b in B, 
and therefore A = B.

-Doctor Ceeks,  The Math Forum