Question

Let a,b,c and be strictly positive real numbers. Can the polynomial x^8+ax^7+bx^6-cx^5+x^2-1 have exactly six positive zeroes? Can it have exactly three positive zeroes? Explain.

My question: Does it make sense to value a,b and c =1, plug them and use Descartes? Because when I do, I get only 5 alternating signs which indicates non of the possible choices unless it is a trick question

Answer 1

Not a trick question...

Answer 2

Okay, so where did I go wrong?

Answer 3

How do you gather information about how many solutions will be positive in this type of polynomial?

Answer 4

I use Descartes Rule of sign?

Answer 5

@ranga