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.

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.

anonymous · Answer

@Loser66

loser66 · Answer

I read Decartes rule for it but it doesn't say that you can put the value of a, b, c, d, then because they don't affect the answer. http://www.purplemath.com/modules/drofsign.htm

loser66 · Answer

@ranga

anonymous · Answer

I picked those values as part of the explanation to prove that if they were positive numbers, bla bla bla.