text: Dascartes rule of signs says that the number of positive roots of a polynomial (with real coefficients and ordered in descending powers) is equal to the number of sign changes between consecutive coefficients or is less than that by an exact multiple of 2 (since double roots are counted twice)  Negative roots are counted by changingthe signs of the coefficients of all odd powers of the variable and counting the sign changes the same as for positive roots.\n\nIn your first problem there are 3 sign changes so there are 3 positive roots (one of which may be a repeated root).  Changing the signs of the x^3 and x terms leads to no sign changes (all of the coefficients are negative) so there are no negative roots and the total is 3 roots, all positive.\n\nIn the 2'nd example, there are 6 sign changes so there are 6 positive roots.  Changing the sign of the x^5, x^3, and x terms leaves no sign changes so there are no negative roots.  Since a polynomial of order n has n roots, there are 8-6 = 2 imaginary or complex roots in the polynomial and 6 real roots (which may contain repeated roots)\n\n\nDoug