Solving Third Degree Polynomial
Equations
Michel Beaudin, [email protected]
Frédérick Henri, [email protected]
École de technologie supérieure, Montréal, Québec, Canada
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
2017-01-17
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Abstract
Starting with some concrete examples, we use TI-Nspire
CX CAS to solve third degree polynomial equations with
real coefficients. Then we compare the answers obtained
to the ones provided by other CAS.
Using Cardano’s method or François Viète’s formulas, we
create a homemade function allowing Nspire to produce
clear and compact answers for the solutions of a third
degree polynomial equation: "Everything should be made
as simple as possible, but not simpler".
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Table of contents
1. Two examples showing CAS imperfections
2. An algorithm to beautify the solutions
3. Solving a third degree polynomial equation with 1 real
root
4. Solving a third degree polynomial equation with 3 real
roots
5. Implementation of a new TI-Nspire function
6. Conclusion
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Two examples showing CAS imperfections –
first example
Consider the following third degree polynomial function
Since :
a. the derivative of the function is strictly positive ;
b.
and
;
we infer that the function has a single real root located
and .
between
Let’s look at the graph of the function to confirm this.
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Two examples showing CAS imperfections –
first example
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Two examples showing CAS imperfections –
first example
Let’s compare the solutions we obtain when we solve the
equation
using 4 different CAS:
 Maple ,
TM
 Mathematica ,
TM
 TI-Nspire and
TM
 Derive .
TM
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Two examples showing CAS imperfections –
first example
With Maple :
TM
1
4
2
1
4
4
1
4
4
1
4 5
4
4 5
1
4 5
4
4 5
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2
4 5
4
4 5
1
I 3
2
1
4
2
1
I 3
2
1
4
2
2
4 5
4
4 5
2
4 5
4
4 5
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Two examples showing CAS imperfections –
first example
With Mathematica :
TM
2
5
1
1
1
2
1
1
2
i 3
i 3
1
2
1
2
1
2
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1
1
5
1
5
1
2
1
i 3
1
5
2
5
i 3
1
5
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Two examples showing CAS imperfections –
first example
With TI-Nspire :
TM
2
2
5
5
1
·2
1
·2
3
1
2
5
5
5
5
1
2
5
3
2
5
1
1
2·2
·
5
2·2
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1
·2
2 ·4
· 3
·2
2 ·4
· 3
16
2
5
3
2·2
1
2
5
5
1
·
5
1
16
2017-01-17
i
i
9
Two examples showing CAS imperfections –
first example
With Derive (notice the simplicity of these solutions):
TM
4 5
4
4 5
2
4 5
4
4 5
4
4 5
4
4
4
4
4 5
4
4
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
4
2
i
3 4 5 4
4
3 4 5 4
4
i
3 4 5 4
4
3 4 5 4
4
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Two examples showing CAS imperfections –
second example
Consider now this other simple third degree polynomial
function
Having 2 critical points, we may expect this function to
have 3 real roots.
Let’s look at the graph of the function to confirm this.
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
2017-01-17
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Two examples showing CAS imperfections –
second example
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
2017-01-17
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Two examples showing CAS imperfections –
second example
Again, let’s compare the solutions we obtain when we
solve the equation
using 4 different CAS:
 Maple ,
TM
 Mathematica ,
TM
 TI-Nspire and
TM
 Derive .
TM
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Two examples showing CAS imperfections –
second example
With Maple (it’s hard to believe these are real values)
TM
1
2
1
4
4
4
1
4I 3
4
4I 3
4
4
1
4
2
4I 3
1
4I 3
4
4I 3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
1
1
I 3
2
2
1
1
I 3
2
2
4I 3
4
2
4I 3
4
4
4I 3
2
4I 3
4
4I 3
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Two examples showing CAS imperfections –
second example
With Mathematica :
TM
1
2
1
1
2
1
2
1
1
i 3
1
i 3
1
2
1
i 3
i 3
1 1
2 2
i 3
1
2
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1
1
i 3
1
1
i 3
2
i 3
i 3
1
i 3
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Two examples showing CAS imperfections –
second example
With TI-Nspire :
TM
cos
2·
9
sin
2·
9
· 3
1
cos
2·
9
2·
9
2·
cos
9
cos
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
sin
2·
9
· 3
1
2·
· 3 1
9
2·
sin
· 3
9
sin
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Two examples showing CAS imperfections –
second example
With Derive (again, so much simpler):
TM
2
2 · cos
9
2 · cos
2 · sin
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18
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An algorithm to beautify the solutions
Our goal is to create a new function for TI-Nspire that
finds the roots of a third degree polynomial function
while:
a. avoiding the presence of
denominators of the roots;
radicals
in
the
b. avoiding needless occurrence of in the real roots
and using, instead, trig expressions for the roots;
The next slide presents the algorithm we will implement.
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An algorithm to beautify the solutions
Choose a 3rd degree function
Get rid of the 2nd degree term
1 real root
Ascertain the number of real roots
3 real roots
Transform into a 6th degree polynomial equation
Apply a trigonometric substitution
Transform into a 2nd degree polynomial equation
Solve the equation
Select one solution of the previous equation
Compute the roots of the original
function
Compute the 3 cubic roots of this solution
Compute the roots of the original function
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
2017-01-17
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Solving a third degree polynomial equation
with 1 real root
Let’s use our algorithm to solve a third degree polynomial
equation that possesses a single real root.
To solve the equation, we will need to accomplish 8 steps,
as seen in the algorithm’s organigram.
START
2
3
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5
6
7
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Solving a third degree polynomial equation
with 1 real root
Step #1 – Pick a 3rd degree polynomial function
SELECT
A
3RD
DEGREE POLYNOMIAL FUNCTION IN THE FORM OF
.
Let’s apply our algorithm to solve the equation
As we will see, this function only possess a single real
root.
START
2
3
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5
6
7
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Solving a third degree polynomial equation
with 1 real root
Step #2 – Get rid of the 2nd degree term
DEFINE
BY SUBSTITUTING
AND THEN
.
DIVIDING BY
5
Since
START
FOR
2
3
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, we obtain:
5
6
7
8
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Solving a third degree polynomial equation
with 1 real root
Step #3 – Ascertain the number of real roots
REWRITE
IN THE FORM OF
WHERE
.
AND
Since
, we find that
and
START
.
2
3
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5
6
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Solving a third degree polynomial equation
with 1 real root
Step #3 – Ascertain the number of real roots
IF
,
THERE IS A SINGLE REAL SOLUTION.
THERE ARE 3 REAL SOLUTIONS.
Since
and
,
OTHERWISE,
.
We conclude that our equation only has a single real
solution.
START
2
3
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5
6
7
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Solving a third degree polynomial equation
with 1 real root
Step #4 – Transform the equation into a 6th degree
polynomial equation
DEFINE
IN
BY SUBSTITUTING
.
AND THEN MULTIPLYING BY
, we find:
Since
START
FOR
2
3
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6
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Solving a third degree polynomial equation
with 1 real root
Step #5 – Transform the equation into a 2nd degree
polynomial equation
DEFINE
.
BY SUBSTITUTING
Since
START
FOR
IN
7
8
, we find:
2
3
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5
6
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Solving a third degree polynomial equation
with 1 real root
Step #6 – Select 1 of the 2 solutions of the 2nd degree
polynomial equation
SOLVE
THE EQUATION
AND SELECT
.
THE SOLUTION
After solving
solution
START
·
, we select the
.
2
3
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Solving a third degree polynomial equation
with 1 real root
Step #7 – Compute the cubic roots of the chosen solution
SINCE
,
AND
SOLVE THE EQUATION
.
·
Solving
, and
START
, we find 3 solutions that we note
.
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Solving a third degree polynomial equation
with 1 real root
Step #7 – Compute the cubic roots of the chosen solution
·
The solutions of the equation
9 · 401
191
are:
·2
6
START
2
9 · 401 191
12
·2
9 · 401
191
12
· 3·2
9 · 401 191
12
·2
9 · 401
191
12
· 3·2
3
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5
6
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Solving a third degree polynomial equation
with 1 real root
Step #8 – Compute the roots of the original function
USING
THE
3
ROOTS OF THE FUNCTION
ROOTS OF THE FUNCTION
Since
START
,
COMPUTE THE
.
, we compute the 3 roots of
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
3
4
5
6
:
7
8
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Solving a third degree polynomial equation
with 1 real root
Step #8 – Compute the roots of the original function
USING
THE
3
ROOTS OF THE FUNCTION
COMPUTE THE
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
3
.
, we compute the 3 roots of
Since
START
,
ROOTS OF THE FUNCTION
4
5
6
:
7
8
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Solving a third degree polynomial equation
with 3 real roots
Let’s use our algorithm to solve a third degree polynomial
equation that possesses 3 real roots.
To solve the equation, we will need to accomplish only 6
steps, as seen in the algorithm’s organigram.
START
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
4
5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #1 – Pick a 3rd degree polynomial function
SELECT
A
3RD
DEGREE POLYNOMIAL FUNCTION IN THE FORM OF
.
Let’s apply our algorithm again to solve the equation
As we will see, this function only possess 3 real roots.
START
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #2 – Get rid of the 2nd degree term
DEFINE
BY SUBSTITUTING
FOR
.
DIVIDING BY
, we obtain:
Since
START
AND THEN
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #3 – Find the number of real roots
REWRITE
IN THE FORM OF
WHERE
.
AND
, we find that
Since
and
START
.
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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5
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Solving a third degree polynomial equation
with 3 real roots
Step #3 – Ascertain the number of real roots
,
THERE IS SINGLE REAL SOLUTION.
THERE ARE 3 REAL SOLUTIONS.
IF
and
Since
,
OTHERWISE,
.
We conclude that our equation has 3 real solutions.
START
2
3
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5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #4 – Apply a substitution
DEFINE
BY
FOR
, WHERE
IN
.
, we find:
Since
START
SUBSTITUTING
2
3
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Solving a third degree polynomial equation
with 3 real roots
Step #5 – Solve the equation
COMPUTE
THE
3
:
ROOTS OF
,
.
AND
And so we find:
START
2
3
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5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #6 – Compute the roots of the original function
USING
THE
3
ROOTS OF THE FUNCTION
ROOTS OF THE FUNCTION
Since
:
START
,
COMPUTE THE
3
.
, we compute the 3 roots of
2
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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5
6
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Solving a third degree polynomial equation
with 3 real roots
Step #6 – Compute the roots of the original function
USING
THE
3
COMPUTE THE
, we compute the 3 roots of
Since
4
4
4
2
3
.
ROOTS OF THE FUNCTION
START
,
ROOTS OF THE FUNCTION
3
2
· sin
4
3
2
· sin
4
3
2
· sin
4
3
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
4
5
:
3
3
3
6
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Implementation of a new TI-Nspire function
This new function has been implemented by Michel
Beaudin and can be downloaded. Go to
https://cours.etsmtl.ca/seg/mbeaudin/
And save the TI-Nspire CX CAS library “Kit_ETS_MB” into
MyLib. The name of the function is “compact_cubic”.
Live examples: on the TI-Nspire CX CAS file.
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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Conclusion
Teaching mathematics using CAS is, as far as we are
concerned, an excellent opportunity for mathematics
teachers to stay enthusiastic even though the curriculum
has been quite the same for the several past years!
TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd
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