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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.

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

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  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.

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