Question

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

Answer 1

@mjoxprm360

Answer 2

there are if i count them correctly 5 changes in sign, so it can have at most 5 positive zeros (and therefore not 6)

Answer 3

Descartes rule of sign. I dig it. Thanks.

Answer 4

If you get a MAXIMUM of 6, you cannot end up with 3. They drop out in pairs.