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 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?
 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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ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0If I get your question right, you do need to divide f(x) by (x2). This goes as follows: 2 [ 5 4 2 2 4 ] 10 28 52 108 + 5 14 26 54 112 < remainder

babydoll332
 2 years ago
Best ResponseYou've already chosen the best response.0so is it lower or upper level?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0What do you mean: an upper bound or lower bound for zeros?

zordoloom
 2 years ago
Best ResponseYou've already chosen the best response.2It 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 xk, where k>0, and all the signs in the bottom row of the synthetic division are nonnegative, 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 xk, where k<0, and the signs in the bottom row of the synthetic division alternate (between nonnegative and nonpositive), 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.

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0f(2) = 112, there are no real zeros, see attached graph

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks, zordoloom, I didn't know that!
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