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anonymous

  • 5 years ago

Determine the zeros of f(x) = x3 – 12x2 + 28x – 9 help please:)

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  1. amistre64
    • 5 years ago
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    1 -12 28 -9 0 <-- add these ------------- x ) 1 <- multiply these your "x" will be whatever you wanna test for a zero

  2. amistre64
    • 5 years ago
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    1,3,9 loos to be our options

  3. amistre64
    • 5 years ago
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    1 -12 28 -9 0 3 -27 3 ------------- 3 ) 1 -9 1 -6: -6 is NOT a zero so "3" is out

  4. amistre64
    • 5 years ago
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    1 -12 28 -9 0 -3 45 ------------- -3 ) 1 -15 ...... this aint gonna zero out either

  5. amistre64
    • 5 years ago
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    1 -12 28 -9 0 9 -27 9 ------------- 9 ) 1 -3 1 0 : x=9 works as a zero

  6. amistre64
    • 5 years ago
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    x^2 -3x +1 is a quadratic form now that can be done with the quad formula if need be

  7. anonymous
    • 5 years ago
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    thank you!

  8. anonymous
    • 5 years ago
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    If guess that one of the roots is rational, the possible roots from the rational root theorem are ±1 , ±3 , ±9. If I try x = 9, I find that f(9) = 0. x^3 - 12x^2 + 28x - 9 = (x-9)(x^2 - 3x + 1) Using the quadratic formula, the other 2 solutions are (3 ± sqrt(9-4))/2 = (3 ± sqrt(5))/2 Solutions to f(x) = 0 : 9 , (3 + sqrt(5))/2 , (3 - sqrt(5))/2

  9. amistre64
    • 5 years ago
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    youre welcome :)

  10. anonymous
    • 5 years ago
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    understand?

  11. anonymous
    • 5 years ago
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    yes thank you:)

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