NCTM presentation

1. 12 Annua l Meeting
20
rvice & In -Service
Prese
Gallery W orkshop
Merging History and Technology:
Immersive Strategies for Solving Cubic Polynomials
Frederick W. Sakon, Angelina Kuleshova, &
Christopher G. Thompson

2. Workshop Overview
• Build historical timeline regarding the resolution of cubic
polynomials.
• Explore multiple representations and technological advances
that help students conceptualize and visualize cubic
equations.
• Utilize historical methods and technology to solve cubic
polynomials.
• Discuss relationship between historical development of
mathematics content and pedagogical content knowledge.

3. Workshop Applications
Mathematics Teacher Educators
Preservice Teacher Training
In-service Teacher Professional Development
Mathematics Teachers
Classroom Instruction

4. The Problem Begins
x + ax + bx + c = 0
3 2
400 B.C.
Interest in solving cubic equations arose
from geometric problems considered by
Greek mathematicians.
Particularly, the question of how to
β
trisect a given angle amounted to solving β
a cubic equation. β

5. Some Progression
Omar Khayyam
1048 - 1131
Khayyam still approached the problem from the geometric
perspective, using intersecting conic sections to construct a
line segment with a length satisfying the equation.
Arabic algebra was expressed entirely in words, from which
Khayyam constructed and worked with 14 different types of
cubic equations (e.g. “a cube and roots are equal to a
number” equivalent to the form x3 + bx = c).

6. The Italian Renaissance
Scipione del Ferro
1465 - 1526
Niccolò Fontana
1500 - 1557
• Del Ferro and Fontana, known as Tartaglia, both
discovered methods to solve particular cubics but kept them
secret.
• Italian scholars of the time had to prove their academic
competence in public competition against another scholar.
• Del Ferro passed on his method to solve cubics of the form
x3 + bx = c to his student, Antonio Maria Fiore.

7. The Italian Renaissance
1535
1539
In 1535, Tartaglia boasts about his ability to solve cubic
equations, prompting a challenge from Fiore.
News of Tartaglia’s victory against Fiore spread and
inspired Girolamo Cardano (1501 – 1576), to contact
Tartaglia in 1539.
With promises of secrecy, Cardano requested that Tartaglia
share his methods, to which Tartaglia eventually agreed.

8. The Italian Renaissance
1545
1539
Cardano had obtained methods for solving cases of the
cubic in the form x3 + ax2 = c and x3 + bx = c.
After six years of work, Cardano and his assistant, Lodovico
Ferrari (1522 – 1565) managed to solve the general
equation completely.
Cardano published the Ars Magna (The Great Art),
documenting a complete account of how to solve any type
of cubic equation.

9. Cardano’s Explanation:
Solving a cubic equation
A cube and things equal to a number
Cube the third part of the number of things, to
which you add the square of half the number and
take the square root of the whole which you will
use, in the one case adding the half of the number
which you just multiplied by itself, in the other
case subtracting the same half, and you will have a
“binomial” and “apotome” respectively; then
subtract the cube root of the apotome from the
cube root of the binomial. This is the value of the
thing.

10. Translating Cardano’s Explanation:
“A cube and things x3+cx = d
equal to a number”
3
c
“cube the third part of  
the number of things” 3
2
d 
“add the square of half + 
the number” 2

11. Translating Cardano’s Explanation:
“take the square root of 3 2
c d 
the whole”   + 
3  2 
“in the one case adding
the half of the number 3
c d 
2
d
which you just   +  +
3  2  2
multiplied by itself”

12. Translating Cardano’s Explanation:
“in the other case subtracting
3 2
the same half” c d  d
  +  −
3  2  2
“then subtract the cube root
of the apotome from the
cube root of the binomial”
3
3
c d  d 3
2
c   d 2 d
3   +  + −   +  −
3  2  2 3  2 2
 

13. Activity 1
Derive solutions to each of the given cubic
equations by applying Cardano’s method:
3
3
c d 
2
d 3 c  d  d
2
3   +  + −   +  −
3  2  2 3  2 2
 

14. Reflection
There are some limitations or problems we might run
into given only Cardano’s method to solve cubic
equations.
Let’s consider 3 cases

15. Case 1: x3 + 6x = 20
3
3
 6   20 
2
20  6   20  20
2
x= 3
  +  + − 3
3  + 2  − 2
 3  2  2    
x= 3
108 + 10 − 3
108 − 10
When we apply the method, we cannot simplify to the solution,
x = 2, without utilizing a calculator or computer to resolve the
non-perfect square and cube roots.

16. Case 2: x3 = 30x + 36
3
 −30   36  36
3 2
 −30   36  36 2
x= 3
  +  + − 3
 3  + 2  − 2
 3   2 2    
x= 3
−676 + 18 − 3
−676 − 18
Here, we cannot simplify to the solution, x = 6, without complex
number theory. After Cardano’s publication of Ars Magna, Rafael
Bombelli (1526 – 1572) developed methods for resolving such cases.

17. Case 3: x3 + 6x2 + 11x + 6 = 0
• A key element to Cardano’s discovery is that he actually
proved that his method worked.
• However, Cardano sought the solution of a particular type of
cubic: x3 + cx = d. Obviously, not all cubic equations resemble
this particular form.
• For cases such as this, he proposed a means of reducing, or as
he called it depressing, the original cubic to the form which fit
his method.
• How might cubic equations be reduced so that they fit
Cardano’s proven method?

18. Reducing Cubic Equations
Given a cubic equation, x + ax + bx + n = 0,
3 2
a
we can apply the substitution, x = y − ,
3
to obtain an equivalent equation, x 3 + cx + d = 0.

19. A Reduction Example
Given a cubic equation, x + 6x + 11x + 6 = 0,
3 2
6
we can apply the substitution, x = y − , or x = y − 2.
3
(y − 2)3 + 6(y − 2)2 + 11(y − 2) + 6 = 0
y 3 − 4y 2 + 4y − 2y 2 + 8y − 8 + 6y 2 − 24y + 24 + 11y − 22 + 6 = 0
y3 − y = 0
Now, Cardano’s method could be applied, where c = – 1 and d = 0.
Note, the solution to the original cubic equation requires
substitution of the intermediary solution for y into x = y – 2.

20. Activity 2
Convert each cubic equation to the form x3 + cx = d,
for which Cardano’s method may be applied.
a
x = y−
3

21. Reflection
Solving cubic equations was a mathematical process that took
many years to develop. Each additional development granted
us greater flexibility in our problem solving capacity.
Let’s consider some of the tools we have developed and how they
can expand our perspective on solving cubic equations

27. Notes from prior course
material

28. A “simple” cubic example, continued
The equation, x3 + 60x2 + 1200x = 4000, can be
thought of as being very close to a perfect cube.
With this in mind, can you “complete the cube”?
Try this now, thinking of what you would need to
do to complete the cube on x3 + 60x2 + 1200x.
(Recall that: (a+b)3=a3+3a2b+3ab2+b3)

29. Example, continued
If they considered the
x + bx + cx = d
“general” cubic as: 3 2
Then the “cubic formula”
3
became:
c c
x=3   +d −
However, not all cubics were of
b b
this form!!!
(And similar to other
equations solved, we only find
one solution with this
“formula.”)

30. Various forms of cubic equations…and ways to
solve them
Example 1 (from the Italian
Renaissance):
Solve
x3 + 60x2 + 1200x = 4000
The solution is:
x = 3 (1200 / 60) 3 + 4000 − (1200 ) = 3 12000 − 20
60