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Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
Ca8e Ppt 5 6
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Ca8e Ppt 5 6

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  • 1. Section 5.6 Complex Zeros; Fundamental Theorem of Algebra
  • 2. Example 1 A polynomial of degree 5 whose coefficients are real numbers has the zeros -2, -3i, and 2+4i. Find the remaining two zeros.
  • 3. By the conjugate pairs theorem, if x = -3i and x = 2+4i then x = 3i and x = 2-4i
  • 4. Example 2: Find the remaining zeros of f. Degree 5; zeros: 1, i, 2i The degree We have three indicates the zeros, so we are number of zeros a missing two!!! polynomial has. By the Conjugate Pairs Theorem Remaining zeros: -i, -2i
  • 5. Example 3 Find a polynomial f of degree 4 whose coefficients are real numbers and that has the zeros 1, 1, 4+i.
  • 6. By the Conjugate Pairs Theorem x = 1, x = 1, x = 4 + i, x = 4 − i P ( x) = ( x − 1)( x − 1)( x − 4 − i )( x − 4 + i ) P( x) = ( x − x − x + 1)( x − 4 x + xi − 4 x + 2 2 16 − 4i − xi + 4i − i ) 2 P( x) = ( x − 2 x + 1)( x − 8 x + 16 − (−1)) 2 2 P( x) = ( x − 2 x + 1)( x − 8 x + 17) 2 2
  • 7. P( x) = x − 8 x + 17 x − 2 x + 16 x − 4 3 2 3 2 34 x + x − 8 x + 17 2 P( x) = x − 10 x + 34 x − 42 x + 17 4 3 2
  • 8. Example 4: Use the given zero to find the remaining zeros of each function. f ( x) = x − 4 x + 4 x − 16; zero : 2i 3 2 2i 1 -4 4 - 16 2i - 4 − 8i 16 1 - 4 + 2i - 8i 0 Note: 2i (-4 + 2i ) = -8i + 4i = -8i + 4(-1) = -4 − 8i 2 2i (-8i ) = -16i = -16(-1) = 16 2
  • 9. - 2i 1 - 4 + 2i - 8i - 2i 8i 1 -4 0 4 1 -4 4 1 0 Zeros: 2i, -2i, 4
  • 10. Example 5 Find the complex zeros of the polynomial function x 4 + 2 x3 + x 2 − 8 x − 20 ±1, ± 2, ± 4, ± 5, ± 10, ± 20
  • 11. 2 1 2 1 -8 - 20 2 8 18 20 1 4 9 10 0 -2 1 4 9 10 -2 -4 - 10 1 2 5 0 x + 2x + 5 = 0 2 +5 Note: The resulting quadratic equation can not be factored since there is no number that multiplied gives you five and at the same time added gives you two.
  • 12. We have to use completing the square or the quadratic formula 2 x + 2x + 5 = 0 a =1 b=2 c=5 Quadratic Formula − b ± b − 4ac 2 x= 2a Substitute − (2) ± (2) − 4(1)(5) 2 x= 2(1)
  • 13. Simplify and solve! − 2 ± 4 − 20 x= 2 − 2 ± − 16 x= 2 − 2 ± 4i x= = −1 ± 2i 2 Complex zeros: x = −1± 2i Real zeros: x = ±2 Remember the problem only asks for the complex zeros!

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