Merging History and Technology:

Immersive Strategies for Solving

Cubic Polynomials

Frederick W. Sakon, Angelina Kuleshova, &Christopher G. Thompson

2012 Annual Meeting

Preservice & In-Service

Gallery Workshop

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.

Workshop Applications

Mathematics Teacher EducatorsPreservice Teacher Training

In-service Teacher Professional Development

Mathematics TeachersClassroom Instruction

The Problem Begins

Interest in solving cubic equations arose from geometric problems considered by Greek mathematicians.

Particularly, the question of how to trisect a given angle amounted to solving a cubic equation.

x3 ax2 bx c 0

ββ

β

400 B.C.

Some Progression

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).

1048 - 1131

Omar Khayyam

The Italian Renaissance

1500 - 1557

Niccolò Fontana

1465 - 1526

Scipione del Ferro

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.

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.

1535

1539

The Italian Renaissance

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.

1545

1539

The Italian Renaissance

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.

Translating Cardano’s Explanation:

“A cube and things equal to a number”

“cube the third part of the number of things”

“add the square of half the number”

3

3

c

2

2

d

x3+cx = d

Translating Cardano’s Explanation:

“take the square root of the whole”

“in the one case adding the half of the number which you just multiplied by itself”

23

23

dc

223

23ddc

Translating Cardano’s Explanation:

“in the other case subtracting the same half”

“then subtract the cube root of the apotome from the cube root of the binomial”

223

23ddc

3

23

3

23

223223

ddcddc

Activity 1

Derive solutions to each of the given cubic equations by applying Cardano’s

method:

3

23

3

23

223223

ddcddc

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

Case 1: x3 + 6x = 20

x 6

3

3

20

2

2

20

23

6

3

3

20

2

2

20

23

33 1010810108 x

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.

Case 2: x3 = 30x + 36

x 30

3

3

36

2

2

36

23

30

3

3

36

2

2

36

23

x 676 183 676 183

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.

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?

Reducing Cubic Equations

Given a cubic equation,

we can apply the substitution,

to obtain an equivalent equation,

x3 ax2 bx n 0,

,3

ayx

x3 cx d 0.

A Reduction Example

Given a cubic equation,

we can apply the substitution, or

x3 6x2 11x 6 0,

x y 6

3, x y 2.

(y 2)3 6(y 2)2 11(y 2) 6 0

y3 4y2 4y 2y2 8y 8 6y2 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.

Activity 2

Convert each cubic equation to the form x3 + cx = d, for which Cardano’s method

may be applied.

x y a

3

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

Notes from prior course material

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)

Example, continued

If they considered the “general” cubic as:

Then the “cubic formula” became:

However, not all cubics were of this form!!!

(And similar to other equations solved, we only find one solution with this “formula.”)

dcxbxx 23

b

cd

b

cx

3

3

Various forms of cubic equations…and ways to solve them

Example 1 (from the Italian Renaissance):

Solve

x3 + 60x2 + 1200x = 4000

The solution is:

2012000)601200(4000)60/1200( 33 3 x