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babydoll332

  • 2 years ago

Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the function f). k = 2; f(x) = 5x4 + 4x3 - 2x2 + 2x + 4; Upper bound?

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  1. babydoll332
    • 2 years ago
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    yes correct?

  2. babydoll332
    • 2 years ago
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    @ZeHanz

  3. ZeHanz
    • 2 years ago
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    If I get your question right, you do need to divide f(x) by (x-2). This goes as follows: 2 [ 5 4 -2 2 4 ] 10 28 52 108 ---------------------+ 5 14 26 54 112 <-- remainder

  4. babydoll332
    • 2 years ago
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    so is it lower or upper level?

  5. babydoll332
    • 2 years ago
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    **bound

  6. babydoll332
    • 2 years ago
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    @zordoloom

  7. ZeHanz
    • 2 years ago
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    What do you mean: an upper bound or lower bound for zeros?

  8. babydoll332
    • 2 years ago
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    i believe so

  9. zordoloom
    • 2 years ago
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    It seems like it's lower bound. Take a look at this description of bounds: Upper and Lower Bounds (http://people.richland.edu/james/lecture/m116/polynomials/zeros.html) If you have a polynomial with real coefficients and a positive leading coefficient, then ... Upper Bound If synthetic division is performed by dividing by x-k, where k>0, and all the signs in the bottom row of the synthetic division are non-negative, then x=k is an upper bound (nothing is larger) for the zeros of the polynomial. Lower Bound If synthetic division is performed by dividing by x-k, where k<0, and the signs in the bottom row of the synthetic division alternate (between non-negative and non-positive), then x=k is a lower bound (nothing is smaller) for the zeros of the polynomial. The zero in the bottom row may be considered positive or negative as needed.

  10. ZeHanz
    • 2 years ago
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    f(2) = 112, there are no real zeros, see attached graph

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  11. babydoll332
    • 2 years ago
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    ohh ok thank u

  12. ZeHanz
    • 2 years ago
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    Thanks, zordoloom, I didn't know that!

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