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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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ZeHanzBest ResponseYou've already chosen the best response.0
If 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
 one year ago

babydoll332Best ResponseYou've already chosen the best response.0
so is it lower or upper level?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
What do you mean: an upper bound or lower bound for zeros?
 one year ago

zordoloomBest ResponseYou've already chosen the best response.1
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 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.
 one year ago

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

ZeHanzBest ResponseYou've already chosen the best response.0
Thanks, zordoloom, I didn't know that!
 one year ago
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