What are the possible number of negative zeros of f(x) = 2x7 – 2x6 + 7x5 – 7x4 – 4x3 + 4x2 ?

Question

anonymous · Answer

descartes rule of sign for this one

$$f(x) = 2x^7 – 2x^6 + 7x^5 – 7x^4 – 4x^3 + 4x^2$$
your first job is to compute 
$$f(-x)$$ and then count the number of change in sign of the coefficients

anonymous · Answer

$$f(-x)=2(-x)^7-2(-x)^6+7(-x)^5-7(-x)^4-4(-x)^3+4(-x)^2$$
$$f(-x)=-2x^7-2x^6-7x^5-7x^4+4x^3+4x^2$$

anonymous · Answer

there is only one change of sign, from the coefficient -7 to the coefficient 4

anonymous · Answer

so one negative?

anonymous · Answer

exactly

anonymous · Answer

if there were three changes in sign, it could be 3 or 1 you have to count down by twos. but since there is only one, you cannot count down by twos, and therefore there must be one a good clear readable explanation and method is here http://www.purplemath.com/modules/drofsign.htm