History of Polynomial Equations

1
History of Polynomial Equations

• Quadratic, Cubic, and Quartic

2
Contents

• History
• Early Polynomials
• Cubic
• Quartic
• Quadratic
• Completing the Square
• Reduced Polynomial
• Cubic
• Quartic
• Radical Extensions

3
Early Polynomials

• First seen in Mesopotamia, during Babylonian
  period
• Solved verbally
• Systems of linear equations used
• Egypt
• Ahmes Papyrus
• Recorded by scribe Ahmes around 1650 BC
• Contains 87 problems and solutions
• Practical problems
• Math exercises
• Asia
• Nine Chapters on Mathematical Art

4
Early Polynomials

• Greece
• Only concerned with positive solutions
• The Elements, Euclid
• Foundation of geometry, brief mention of
  geometric solution to quadratic equation
• Arithmetica, Diophantus
• Based on theory of numbers
• Could solve

5
Early Polynomials

• Arab and Indian Mathematicians
• Dominated during period between Greek mathematics
  and rebirth of math in Europe
• Muhammad ibn Musa al-Khwarizmi
• Kitab al-jabr wa al-mugabalah origin of the term
  algebra
• First occurrence of general quadratic equation
• Europe, 12th century
• Hibbur ha-meshihah ve-ha-tishboret (Treatise on
  Measurement and Calculation), Abraham bar Hiyya
  Ha-nasi

6
Cubic

• No solution found until 16th century
• Scipione del Ferro, found solution to x3mx n
  but did not publish
• At his death passed to student, Antonio Maria
  Fior
• Fior challenges Niccolo Fontana Tartaglia to
  contest
• Each proposed 30 problems
• Tartaglia finishes within 2 hours, Fior completed
  zero problems
• Tartaglia had discovered the solution only days
  earlier

7
Cubic

• Giralamo Cardano and Tartaglia
• Cardano contacted Tartaglia for the solution to
  put in his book
• Tartaglia refused initially, but eventually gives
  Cardano the solution
• With the solution to x3mx n , Cardano is able
  to find proof and begin working on more general
  solution

8
Cubic and Quartic Solutions

• Cardano and Ludovico Ferrari
• Ferrari was student of Cardano
• Discovered solution to quartic at the time
  Cardano found cubic
• Results published in Ars Magna (1545)
• First time solutions are seen to be negative,
  irrational, and involve square roots of negative
  numbers

9
Completing the Square
Adding and subtracting the area of the smallest
square gives,
Assuming z is a root of the equation,
10
Reduced polynomial

• Definition (1) A polynomial f(x) of degree n is
  reduced if it has no terms
• Lemma (1) The substitution of
  changes
  where or into a
  reduced polynomial . If
  u is a root of , then is a
  root of .

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Quadratic (Reduced polynomial method)

• Let with
  .
• Then, and the
  roots are
• and the roots for
  are

(1)
12
Corollary/Theorem

• Corollary (2) Given numbers c and d, there
  exists numbers a and ß with a ß c amd aß
  d.
• Theorem (3) If f(x) of kx were k is a
  field then a k is a root if and only if x-a
  divides f(x) in kx.

13
Cubic

• Cardanos trick was to write the root u as aß
  u. Then substituting into the equation,
  using Corollary (2) , which
  forces the u term to drop out and we arrive at,
  We can rewrite this as,

(2)
(3)
(4)
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Cubic (continued)

• Since Eq (4) is a quadratic in a3 it and can be
  solved using the quadratic formula.Taking the
  cube roots of a and ß, we find are able to find
  one root of the equation by substitution.

(4)
(5)
15
Cubic (continued)

• Then using Theorem (3), dividing f(x) with (x-u)
  we arrive at a quadratic g(x) which can be solved
  with the quadratic formula to find the remaining
  two roots.

(6)
16
Quartic

• Similar to quadratic method using a reduced
  polynomial Substitue and
  then factor into
  quadratics,

(7)
(8)
17
Quartic (continued)

• To find j,m, and l combine like terms and set the
  coefficients equal,you arrive at,which is a
  cubic in j2. Then use the cubic equation already
  found to solve for m and l. Subsitute j, m and l
  to find the four roots using the quadratic
  equation.

(9)
(10)
18
Pure Extensions

• Definition (4) A pure extension of type m is an
  extension k(u)/k where um k for some m1. An
  extension K/k is a radical extension if there is
  a tower of fields, where each Ki1/Ki is a
  pure extension of type mi.

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The Quadratic as a Radical Extension

• For the equation
  , let where . Then
  by definition K1 is a radical extension of
  since u2 .
• So, by Definition (4), f(x) is solvable by
  radicals.

20
Conclusion

• The Equation That Couldnt Be Solved by Mario
  Livio