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\noindent \textsc{Alec Johnson / Math 217} \hfill \textsc{1 May 2007}\\
\noindent \textsc{Wisconsin Emerging Scholars} \hfill \textsc{Worksheet 25}
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{\bf Review: division, factoring, and root-finding.}
(This is a brush-up from last semester.)
The division algorithm allows us to write any two polynomials
$f(x)$ and $p(x)\ne 0$ uniquely in the form $f(x)=p(x)\cdot
q(x)+r(x)$, where $q(x)$ is called the \emph{quotient} and $r(x)$
is called the \emph{remainder}.
In lecture we used the division algorithm to show that factoring
is the same as root-finding. In particular, the remainder
theorem tells you that $f(c)$ is the remainder when you divide
$f(x)$ by $(x-c)$. (To see this, write $f(x)=(x-c)\cdot q(x)+r$.
Then $f(c)=r$.)
{\bf Problems.}
\begin{enumerate}
\item {\bf Constructing a polynomial with given roots.}
Construct a polynomial with real coefficients
that has exactly the given zeros and degree.
\begin{enumerate}
\item $3-4i$, degree 2
\item $3-4i,7$, degree 3
\item $3-4i,7$, degree 4
\end{enumerate}
\item {\bf Rational roots.}
Completely factor the following polynomials.
HINT: The rational roots theorem and synthetic
division can be a big help.
\begin{enumerate}
\item $x^3-x^2+4x-4$
\item $2x^3-5x^2-11x-4$
\end{enumerate}
\item {\bf Complex arithmetic.}
Let $z_1=2\cis(150^\circ)$
and $z_2=3\cis(-60^\circ)$.
Find:
\begin{enumerate}
\item $z_1$ and $z_2$ in $a+bi$ form
\item $z_2/z_1$
\item $z_2-z_1$
\item $z_2^5$
\item All solutions $z$ to $z^2=z_1$.
\end{enumerate}
\item {\bf Derivative of a polynomial with a root of multiplicity
greater than 1.}
Have each person in your group
invent a polynomial $p$ that has a root of multiplicity 2.
Now differentiate it to get the polynomial $p'$.
Factor $p'$. Do you notice a common pattern? Do you think that
this happens in general? Can you prove it?
\item {\bf Factoring a quadratic by depressing it.}
Suppose you want to factor a quadratic polynomial,
i.e. a polynomial of the form $Ax^2+Bx+C$.
If you divide by $A$, you will get an equation of the
form $p=x^2+\beta x+\gamma$.
We wish to put this equation in the ``depressed'' form
$y^2+c$. To accomplish this, substitute
$x=y-\alpha$ and expand out to get an equation of the
form $y^2+by+c$. What should $\alpha$ be
in order to make the $y$-term zero? Choose this
$\alpha$, solve for $x$, and expand your solution
in terms of the original parameters of the
quadratic polynomial ($A$,$B$, and $C$). Do you get
a familiar formula?
\item {\bf Factoring cubics part I: depressing a cubic.}
Ever wonder how to get a general formula for the factors of a
cubic polynomial? Begin with a general cubic, $Ax^3+Bx^2+Cx+D$.
Divide by $A$ to get a cubic of the form $x^3+\beta x^2+\gamma
x+\delta$. We wish to put this equation in the ``depressed''
form $y^3+cy+d$. The trick is to make a substitution of the form
$x=y-\alpha$. What should $\alpha$ be in order to make the $y^2$
term zero?
\item {\bf Factoring cubics part II: solving a depressed cubic.}
Suppose you are trying to factor a depressed cubic,
$p=y^3+cy+d$. So you are trying to solve the
equation $y^3+cy+d=0$.
The trick is to look for a solution
of the form $y=s-t$. Plug this in. Show that
we've got a solution if:
\begin{gather*}
c=t^3-s^3,\ \ d=3st
\end{gather*}
So if we can solve this system for $s$ and $t$
in terms of $c$ and $d$, we're done.
Do you see how to solve it?
\item {\bf Factoring cubics part III: try an example!}
Now let's do an example. To have a nice example
to test out our method, let's make up a cubic
where we already know the roots. I don't know,
how about $p(x)=(x-3)(x^2+1)$? Or maybe you
would prefer real roots, e.g. $p(x)=(x-5)(x+1)(x+4)$?
%I picked those fairly randomly, so I don't have
%any reason to think that they're particularly nice.
Multiply out the factored polynomial that you pick
and then use the methods in the previous two sections
to factor it. Does it work?
You also might try starting with a random depressed cubic,
such as $p(x)=x^3-3x+1$. Compute the roots. Then find
the zeros by graphing or using Newton's method. Do your
answers agree?
\item {\bf Factoring a quartic.}
To factor a quartic equation, first figure out a substitution that
``depresses'' it (i.e. gets rid of the cube term). Then assume
that you can write it as a product of quadratic polynomials with
generic coefficients. Multiply the quadratics to get a quartic
and equate the coefficients with the coefficients of the depressed
quartic. This gives a system of equations which, with some
strategic elimination of variables, you can reduce to the problem
of finding the roots of a cubic. Voil\`a!
\item {\bf Factoring a quintic.}
Show that it is impossible to represent the roots of
the polynomial $f(x)=x^5-x-1$ using addition, subtraction,
multiplication, division, or extraction of roots. Note that
this result shows that it is impossible to find a general
formula for the roots of a polynomial of degree 5 or higher.
(Hint: read up on Galois theory! :)
\end{enumerate}
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