review synthetic division to find roots of third degree characteristic polynomial pamela leutwyler

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Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

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Page 1: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

Review SYNTHETIC DIVISION

to find roots of third degree characteristic polynomial

Pamela Leutwyler

Page 2: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(x + 3)(7x – 2) =

Page 3: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

Page 4: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

Page 5: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(1x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

Page 6: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(1x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

Page 7: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(1x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

q

pIf is a root of the polynomial equation

Page 8: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(1x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

q

pIf is a root of the polynomial equation

Then q is a factor of 14

2 1 7

Page 9: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

(2x – 5)(1x + 3)(7x – 2) =

14x3 + 3x2 – 107x + 30 = 0

The roots are:

2

5

7

2-3

q

pIf is a root of the polynomial equation

Then q is a factor of 14 and p is a factor of 30

2 1 7

5-3

2

Page 10: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

A characteristic polynomial will always have lead coefficient = 1.

Rational eigenvalues will be integral factors of the constant coefficient of the characteristic polynomial .

example: find the eigenvalues for the matrix

124

322

331

014194

124

322

331

det 23

polynomialsticcharacteri

potential rational roots are factors of 14. +1, -1, +2, -2, +7, -7, +14, -14

Page 11: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

014194 23 polynomialsticcharacteri

potential rational roots are factors of 14. +1, -1, +2, -2, +7, -7, +14, -14

Test the potrats using synthetic division:

1 -4 -19 -14

Page 12: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

014194 23 polynomialsticcharacteri

potential rational roots are factors of 14. +1, -1, +2, -2, +7, -7, +14, -14

Test the potrats using synthetic division:

+1 1 -4 -19 -14

1

1

-3

-3

-22

-22

-36

The remainder is NOT ZERO.+1 is not a root.

Page 13: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

014194 23 polynomialsticcharacteri

potential rational roots are factors of 14. +1, -1, +2, -2, +7, -7, +14, -14

Test the potrats using synthetic division:

+7 1 -4 -19 -14

1

7

3

21

2

14

0

The remainder is ZERO.+7 is a root.

Page 14: Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial Pamela Leutwyler

014194 23 polynomialsticcharacteri

potential rational roots are factors of 14. +1, -1, +2, -2, +7, -7, +14, -14

Test the potrats using synthetic division:

+7 1 -4 -19 -14

1

7

3

21

2

14

0

)23)(7(

14194

2

23

polynomialsticcharacteri

The remainder is ZERO.+7 is a root. factor this or use quadratic formula or continue with

synthetic division to get the other roots.