Ca8e Ppt 5 6

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Ca8e Ppt 5 6

  1. 1. Section 5.6 Complex Zeros; Fundamental Theorem of Algebra
  2. 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. 3. By the conjugate pairs theorem, if x = -3i and x = 2+4i then x = 3i and x = 2-4i
  4. 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. 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. 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. 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. 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. 9. - 2i 1 - 4 + 2i - 8i - 2i 8i 1 -4 0 4 1 -4 4 1 0 Zeros: 2i, -2i, 4
  10. 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. 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. 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. 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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