\input common.tex
\centerline{ \bigbf Notes on Problem 4, September 13, 1989 }
\bs
\centerline { \bigbf Math 401 -- Second Course in Abstract Algebra, I}
\ms
\centerline { \bigbf Prof. Daniel Grayson }
\bs

Problem 4 was this: Try to find a formula for the roots of a cubic polynomial $x^3 + a x^2 + b x +
c$.

I would like you to first think about the quadratic formula in a novel way, in
order to see what to do with the cubic equation.  To be precise, let $\alpha$
and $\beta$ be the roots of $x^2-b x+c$.  Then from $x^2-b x+c =
(x-\alpha)(x-\beta)$ we deduce the pair of equations
$$
\align
	\alpha + \beta &= b 
\\
	\alpha \beta &= c
\endalign
$$
Our problem now is to solve this system of two equations in two variables.  
They aren't linear equations -- we like linear equations better, of course,
because they are easier to solve.
Now the trick is to observe that $ (\alpha - \beta)^2 $ is a symmetric
polynomial in the roots, so can be expressed in terms of $b$ and $c$; in fact,
$ (\alpha - \beta)^2  = (\alpha + \beta)^2 - 4 \alpha \beta = b^2 - 4 c$.  So
now we consider instead the system
$$
\align
	\alpha + \beta &= b
\\
	\alpha - \beta &= \sqrt{b^2-4c}
\endalign
$$
This system of equations is linear, and the matrix of coefficients,
$\pmatrix1&1\\1&-1\endpmatrix$, has nonzero determinant as long as the field
over which we are working has characteristic not $2$.  
We can use the inverse matrix to solve for the roots.
$$
\pmatrix\alpha \\ \beta\endpmatrix = \pmatrix 1/2&1/2\\1/2&-1/2 \endpmatrix 
\pmatrix b\\\sqrt{b^2-4c} \endpmatrix
$$

Try to do something similar with three roots.


\vfill
\centerline{\psfig{figure=/usr/users/dan/lib/ps/dodec.ps,width=.5in}}
\vfill\eject

\end
