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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 74 "Dr. Sarah's Maple Demo on
Non-Solvability of Quintic Equations by Radicals" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 215 "The Quadratic, Cubic and
Quartic solution formulae are built up from the coefficients of the e
quation by repeated addition, subraction, multiplication, division and
taking roots. Expressions of this kind are called " }{TEXT 257 19 "ra
dical expressions" }{TEXT -1 3 ". " }}}{SECT 0 {PARA 3 "" 0 ""
{XPPEDIT 18 0 "x^5+1/3*x^4*sqrt(3)-3*x^3-x^2*sqrt(3)-4*x-4/3*sqrt(3);
" "6#,.*$)%\"xG\"\"&\"\"\"F(*(#F(\"\"$F()F&\"\"%F(-%%sqrtG6#F+F(F(*&F+
F()F&F+F(!\"\"*&)F&\"\"#F(F.F(F3*&F-F(F&F(F3*&#F-F+F(*$F.F(F(F3" }
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Let's look at a \+
plot of this quintic polynomail and then have Maple solve for the root
s, where the function is 0 and intercepts the x-axis." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "plot(x^5-3*x^3+1/3*x^4*sqrt(3)-x^2*
sqrt(3)-4*x-4/3*sqrt(3),x=-3..3);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 63 "solve(x^5-3*x^3+1/3*x^4*sqrt(3)-x^2*sqrt(3)-4*x-4/3*s
qrt(3),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "Notice that these \+
roots can be expressed nicely in terms of the coefficents of the polyn
omial. For example, the first coefficient (of " }{XPPEDIT 18 0 "x^5"
"6#*$%\"xG\"\"&" }{TEXT -1 143 ") is a 1 and doubling it yields 2, one
of the roots. Taking the square root of negative the first coefficen
t gives I, the square root of -1. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT
258 8 "Exercise" }{TEXT -1 41 ": Give instructions in order to expres
s " }{XPPEDIT 18 0 "-1/3*sqrt(3);" "6#,$*(\"\"\"F%\"\"$!\"\"-%%sqrtG6#
F&F%F'" }{TEXT -1 31 " in terms of the coefficients." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {XPPEDIT 18 0 "x^5-2*x^3-8*x-
2" "6#,**$%\"xG\"\"&\"\"\"*&\"\"#F'*$F%\"\"$F'!\"\"*&\"\")F'F%F'F,F)F,
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "Some fifth de
gree polynomials can be solved by writing the roots as nice expression
s of the coeffients, but the following quintic polynomial is not solva
ble by radicals: " }{XPPEDIT 18 0 "x^5-2*x^3-8*x-2;" "6#,**$%\"xG\"\"
&\"\"\"*&\"\"#F'*$F%\"\"$F'!\"\"*&\"\")F'F%F'F,F)F," }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 260 "It does certainly have roots that we can find,
but we cannot express these roots only in terms of the combinations o
f the coefficients of the polynomial. So, this quintic is not solvable
by radicals.\n\nLet's first plot the function so that we can see the \+
zeros:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot(x^5-2*x^3-8*
x-2,x=-3..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 243 "Notice that the
graph of this polynomial looks just like our first example. We can s
ee that there are three intervals which have roots on this graph. Cal
culus techniques would show us that these are the only three intervals
we need to check." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "fsol
ve(x^5-2*x^3-8*x-2=0,x=-2..-1); \nfsolve(x^5-2*x^3-8*x
-2=0,x=-1..0); \nfsolve(x^5-2*x^3-8*x-2=0,x=2.
.3);\n \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "On the
surface, these roots don't look like nice numbers which could be expr
essed as combinations of the integer coefficients of " }{XPPEDIT 18 0
"x^5-2*x^3-8*x-2" "6#,**$%\"xG\"\"&\"\"\"*&\"\"#F'*$F%\"\"$F'!\"\"*&\"
\")F'F%F'F,F)F," }{TEXT -1 419 ", we are allowed to take square roots,
cube roots, etc. of the coefficients, which would themselves look qui
te messy (recall that the square root of 2 is irrational, for example,
and so we cannot find a repeating pattern to its decimal expansion). \+
In addition, Maple's fsolve command rounds, so these are just an esti
mation of the roots. Let's try and have Maple solve for the exact ex
pression of the roots directly." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "solve(x^5-2*x^3-8*x-2=0,x);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 539 "Maple cannot solve for exact
expressions of the roots, but this is still not a proof that the poly
nomial cannot be solved by radicals (why?). Other techniques, like th
ose used by Galois, can prove that we can not write the roots nicely i
n terms of the coefficients of the polynomial, like we do for the lowe
r degree polynomials. Abel began to examine which polynomial's have s
olutions by radicals, but did not complete his work before his death. \+
Galois was the first to determine a condition for a polynomial to be \+
solvable by radicals." }}}{SECT 0 {PARA 3 "" 0 "" {XPPEDIT 18 0 "x^5-1
" "6#,&*$%\"xG\"\"&\"\"\"F'!\"\"" }{TEXT -1 12 " Exploration" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 259 0 "" }{TEXT 260 8 "Exercise" }{TEXT
261 0 "" }{TEXT -1 5 " Is " }{XPPEDIT 18 0 "x^5-1;" "6#,&*$%\"xG\"\"&
\"\"\"F'!\"\"" }{TEXT -1 24 " solvable by radicals? " }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }