1 / 14

Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial

Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial. Pamela Leutwyler. (2x – 5)(x + 3)(7x – 2) =. (2x – 5)(x + 3)(7x – 2) = 14x 3 + 3x 2 – 107x + 30 = 0. The roots are:. -3. (2x – 5)(x + 3)(7x – 2) = 14x 3 + 3x 2 – 107x + 30 = 0. The roots are:.

lumina
Download Presentation

Review SYNTHETIC DIVISION to find roots of third degree characteristic polynomial

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


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

  2. (2x – 5)(x + 3)(7x – 2) =

  3. (2x – 5)(x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0

  4. The roots are: -3 (2x – 5)(x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0

  5. The roots are: -3 (2x – 5)(1x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0

  6. The roots are: -3 (2x – 5)(1x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0

  7. The roots are: -3 (2x – 5)(1x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0 If is a root of the polynomial equation

  8. The roots are: -3 2 7 1 (2x – 5)(1x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0 If is a root of the polynomial equation Then qis a factor of 14

  9. 5 2 The roots are: -3 -3 2 7 1 (2x – 5)(1x + 3)(7x – 2) = 14x3 + 3x2 – 107x + 30 = 0 If is a root of the polynomial equation Then qis a factor of 14 andpis a factor of 30

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

  11. 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

  12. The remainder is NOT ZERO. +1 is not a root. 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 -3 -22 1 -36 -22 1 -3

  13. The remainder is ZERO. +7 is a root. 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 21 14 7 0 2 1 3

  14. The remainder is ZERO. +7 is a root. 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 21 14 7 0 2 1 3 factor this or use quadratic formula or continue with synthetic division to get the other roots.

More Related