Roots of Polynomials
Section I: Linear Polynomials
Consider the polynomial equation ax + b = 0, symbolically solve for the value of
x.
Of course the value of x is called the root of the linear
polynomial ax + b. The TI86
has a built in polynomial solver package accessed via 2nd POLY on the key
board.
To find the roots of a polynomial using MAPLE we can use the command
solve(our polynomial=0);
Find the roots of the polynomials in each of three ways: first "by hand", then
via the
TI86, and lastly using MAPLE. Record the answers in order left to right.

P&B 
SGC 
CAS 
Comments 

Try #5 using the command solve(a*x + b=0, x); in
MAPLE.
Why do you think the information ",x" must be put into the MAPLE command in
#5 when we didn't need it in the others?
What conclusion do you make with regard to using the TI86
directly to solve linear
polynomials?
In using MAPLE in #5 we have encountered our first
instance of using the amazing
feature that distinguishes a Computer Algebra System (CAS) from the previous
generation of mathematics software packages: it can do Symbolic Manipulation!
Section II. Quadratic Polynomials
Next let's solve some quadratics; that is, find the roots of polynomials of the
form
ax^{2}+bx+c. You have done this once, but do it again. The quadratic formula tells
us
the two roots of ax^{2}+bx+c are
As above, find the roots of the following "by hand", with
the TI86,(it should work
here) and with MAPLE.(Can you guess which problems will require insertion of ,x
into the solve command, and which will not?)

P&B 
SGC 
CAS 
Comments 

What's special about #8 and #10?
Notice the difference in the way the TI85 expresses
complex numbers and the way
MAPLE expresses them.

P&B 
SGC 
CAS 
Comments 

Now see what MAPLE gives when you ask it to solve for the
roots of
16. ax^{2} + bx +c.
This should look familiar to you, what is it?
Exercise: In each of the above, 615, try to use the evalf
command in MAPLE to
obtain the TI86 approximate answer. You will first need to define something to
be
the roots, eg,
r:=expression;
defines r to be whatever is the expression. So
r:=solve(2x^{2}3x+2=0,x);
represents the two roots of 2x^23x+2=0.
Since r represents two roots evalf(r[1]); will evaluate the floating point
approximation
to the first root and evalf(r[2]); will approximate the second. So try this for
each of
the above. See if you can guess in advance which work and which don't.
MAPLE
Approximation 
TI ans 

evalf 


Next we will play with two MAPLE commands that further
illustrate the nice features
of symbolic manipulation:
expand(algebraic expression); and factor(algebraic expression);
In each of the following first do the required calculation "by hand", then let
MAPLE
do it.
by hand 

MAPLE 

16. expand((x+3)*(x4)); 


17. factor(%); 


18. factor(2x^{2} +3x 2); 


19. expand((ax+1)*(bx+1)); 


20. factor(x^{2} +bx +ax +ab); 


21. expand((x+I)*(xI)); 


22. factor(%); 

You should now be able to "build" quadratics with any
roots you like using expand
and factor. Make some interesting ones.