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Introduction Quadratic Equations Cubic Equations Quartic Equations

Solving Polynomial Equations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0

(Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)

2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0

(This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)

3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0

(This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)

4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0

(This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)

5. For higher order there can be no formula. (Separate set ofpresentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula.

(Separate set ofpresentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition.

Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.

A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.

Notation: r = n√

a or r = a1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation:

r = n√

a or r = a1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a

or r = a1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n .

Second roots will be called squareroots and they will be denoted

√a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n . Second roots will be called square

roots

and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n . Second roots will be called square

roots and they will be denoted√

a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quadratic formula.

Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0.

Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F.

Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula).

Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c

(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).

(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca

(x+

b2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).

ax21 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b

√b2−4ac+b2−4ac

4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof

(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?)

For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ

, the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

r

ei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n )

, wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step I: Canonical form.

Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0.

Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff

y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0,

where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution).

x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d

= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d

= ay3 +3aαy2 +3aα2y+aα

3 +by2 +2bαy+bα2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3

+by2 +2bαy+bα2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2

+ cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα

+d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3

+(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2

+(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (substitution). x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof.

With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a

=3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2

− 2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2

+ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2

+ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a

=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3

+b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3

− bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2

+da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3

− bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2

+da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step 2: Cardano’s formula.

Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0.

Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27

, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof.

Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution.

Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q

= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q

= α3 +3α

2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q

= α3 +β

3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof.

Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof.

u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27.

Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β .

α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3.

It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3.

Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa.

Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.

WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1.

But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.).

Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3

= α1,2,3−p

3α1,2,3, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1

(or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).

The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term.

Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F.

Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff

y = x+a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0

where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b

, q =a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof.

Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise.

(I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem.

The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic.

Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C.

Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff

x solves the equation(x2 +

p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0

, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0

and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r

x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2

−2x2(p

2+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2

+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2

−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).

(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations

logo1

Introduction Quadratic Equations Cubic Equations Quartic Equations

Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Solving Polynomial Equations