1 / 15

And the zeros are x = - 3, x = -1 , and x = 2

And the zeros are x = - 3, x = -1 , and x = 2. Quotient. Dividend. Divisor. Since the remainder is –64, we know that x + 3 is not a factor. -(. ). ). -(. 2. –4. 1. 6. 2. 0. 50. –8. –14. –16. –28. –56. –4. –7. –8. –14. –28. –6.

lynton
Download Presentation

And the zeros are x = - 3, x = -1 , and x = 2

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. And the zeros are x = -3, x = -1, and x = 2

  2. Quotient Dividend Divisor Since the remainder is –64, we know that x + 3 is not a factor.

  3. -( ) ) -(

  4. 2 –4 1 6 2 0 50 –8 –14 –16 –28 –56 –4 –7 –8 –14 –28 –6 Use synthetic division to find the quotient and remainder. The quotient is – 4x4 – 7x3 – 8x2 – 14x – 28 and the remainder is –6, also f (2) = -6. Note: We must write a 0 for the missing term.

  5. List all possible rational zeros. • Find all rational zeros and factor P(x).

  6. 2 1 3 Example 6 Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. P(x) = +x4 – 6x3 + 8x2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(x). P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1 = x4 + 6x3 + 8x2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root.

  7. Example 7Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1 satisfy the following conditions. (a) No real zero is greater than 3. (b) No real zero is less than –1. Solution a) c > 0 b) c < 0 All are nonegative. No real zero greater than 3. The numbers alternate in sign. No zero less than 1.

More Related