September 18th, 2011, 07:38 AM  #1 
Senior Member Joined: Mar 2011 Posts: 105 Thanks: 0  Precalculus
When we using Descarte's Rules of Sign if the functions like: f(x) = 4x^3  5x + 8 how do we know how many variation sign that it has?? 
September 18th, 2011, 08:55 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,716 Thanks: 1532 
The coefficients are positive, negative and positive (in that order), so there are two changes of sign. Hence f(x) has either two positive zeros or no positive zeros. In fact, it has no positive zeros. The coefficients of f(x) are 4, 5, 8, so f(x) has one change of sign, and that tells you that f(x) has exactly one negative zero. 

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