T O P I C R E V I E W 
moon1130 
Posted  08/10/2013 : 14:47:48 Hi,
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 nonalternating form.
Thanks for your help,
moon1130

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. 

