356 CHAPTER 3 Polynomial and Rational Functions 5. A polynomial function having degree 6 and only real coefficients may have no real zeros. 6. The polynomial function ƒ1x2 = 2x5 + 3x4 - 8x3 - 5x + 6 has three variations in sign. 7. If z = 7 - 6i, then z = -7 + 6i. 8. The product of a complex number and its conjugate is always a real number. Use the factor theorem and synthetic division to determine whether the second polynomial is a factor of the first. See Example 1. 9. x3 - 5x2 + 3x + 1; x - 1 10. x3 + 6x2 - 2x - 7; x + 1 11. -2x4 - 5x3 + 8x2 -3x -13; x +1 12. -3x4 + x3 - 5x2 + 2x + 4; x - 1 13. -x3 + 3x - 2; x + 2 14. -2x3 + x2 - 63; x + 3 15. 4x2 + 2x + 54; x - 4 16. 5x2 - 14x + 10; x + 2 17. x3 + 2x2 + 3; x - 1 18. 2x3 + x + 2; x + 1 19. 2x4 + 5x3 - 2x2 + 5x + 6; x + 3 20. 5x4 + 16x3 - 15x2 + 8x + 16; x + 4 For each polynomial function, one zero is given. Find all other zeros. See Examples 2 and 6. 33. ƒ1x2 = x3 - x2 - 4x - 6; 3 34. ƒ1x2 = x3 + 4x2 - 5; 1 35. ƒ1x2 = x3 - 7x2 + 17x - 15; 2 - i 36. ƒ1x2 = 4x3 + 6x2 - 2x - 1; 1 2 37. ƒ1x2 = -x4 - 5x2 - 4; -i 38. ƒ1x2 = -x4 - 26x2 - 25; i For each polynomial function, (a) list all possible rational zeros, (b) find all rational zeros, and (c) factor ƒ1x2 into linear factors. See Example 3. 39. ƒ1x2 = x3 - 2x2 - 13x - 10 40. ƒ1x2 = x3 + 5x2 + 2x - 8 41. ƒ1x2 = x3 + 6x2 - x - 30 42. ƒ1x2 = x3 - x2 - 10x - 8 Factor ƒ1x2 into linear factors given that k is a zero. See Example 2. 21. ƒ1x2 = 2x3 - 3x2 - 17x + 30; k = 2 22. ƒ1x2 = 2x3 - 3x2 - 5x + 6; k = 1 23. ƒ1x2 = 6x3 + 13x2 - 14x + 3; k = -3 24. ƒ1x2 = 6x3 + 17x2 - 63x + 10; k = -5 25. ƒ1x2 = -6x3 - 25x2 - 3x + 4; k = -4 26. ƒ1x2 = -8x3 - 50x2 - 47x + 15; k = -5 27. ƒ1x2 = x3 + 17 - 3i x2 + 112 - 21i2x - 36i; k = 3i 28. ƒ1x2 = 2x3 + 13 + 2i2x2 + 11 + 3i2x + i; k = -i 29. ƒ1x2 = 2x3 + 13 - 2i2x2 + 1-8 - 5i2x + 13 + 3i2; k = 1 + i 30. ƒ1x2 = 6x3 + 119 - 6i2x2 + 116 - 7i2x + 14 - 2i2; k = -2 + i 31. ƒ1x2 = x4 + 2x3 - 7x2 - 20x - 12; k = -2 (multiplicity 2) 32. ƒ1x2 = 2x4 + x3 - 9x2 - 13x - 5; k = -1 (multiplicity 3)
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