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 Lower bound of polynomial zeros

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moon1130 Posted - 08/10/2013 : 14:47:48

I have the polynomial function F(x)= x+11x+31x+21, which factors to (x+1)*(x+3)*(x+7). From the factorization, it is clear that x=-8 is a lower bound of the real zeros. Yet, when I divide the function by -8 using synthetic division, the last row of the synthetic division is 1 3 7 -35 rather than 1 -3 7 -35, which it should be if -8 were a lower bound according to the Upper and Lower Bounds Theorem.

I have checked my division several times and I come up with the same result. Why is the lower bound theorem failing here?

Thank you for any help you can give me.
4   L A T E S T    R E P L I E S    (Newest First)
moon1130 Posted - 09/10/2013 : 17:23:58
Thank you Ultraglide and Royhaas for your responses.

According to my text, the lower bound of a polynomial function is any number to the left of the smallest negative zero. Since -7 is the smallest negative zero, -8, -9, etc. are each lower bounds for this function. This being the case, when I divide the polynomial by -8 using synthetic division, one would expect that the quotient would have the alternating sign form of + - + - or - + = +.....
rather than + + + -. I could not figure out why this non-alternating form.

Thanks for your help,


Ultraglide Posted - 08/13/2013 : 09:45:23
Just a note: since you have factored the polynomial (I curious as to how the word 'factorize' has crept into use in some areas), you will note that it the zeros occur at -1, -3, and -7 but not at -8. Maybe you need to try using one of the first three numbers in your synthetic division.
royhaas Posted - 08/13/2013 : 08:03:41
The section "Polynomials with real roots" in http://en.wikipedia.org/wiki/Properties_of_polynomial_roots
should give some help.
Ultraglide Posted - 08/12/2013 : 22:03:46
This being a cubic function, it has no bounds, horizonally or vertically.

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