The Descartes rule of signs starts with counting the number of changes of sign between terms.
2x3-17x2-11x-1=0. There is only one change of sign, that is, between +2x3 and -17x2. So there is one real solution or 3 real solutions (because 3-1=even number). There can be no more than 3 because we have a cubic, which has a degree of 3.
Rational zeroes are obtained by looking at the factors of the highest degree and lowest degree (that is, the constant term). So 2 only has the factors 2 and 1, and 1 has only 1 as a factor.
So one factor could be x±1 or 2x±1. To test if x=±1 we simply write down all the coefficients as a sum and using plus-minus where we see an odd power of x:
±2-17∓11x-1 gives us either 2-17-11-1=-27, or -2-17+11-1=-9. Neither of these sum to zero so neither x-1 nor x+1 is a factor.
Next we try 2x±1, that is, x=±½ and substitute for x in the cubic:
2/8-17/4-11/2-1=-21/2 or -2/8-17/4+11/2-1=(-1-17+22-4)/4=0, so we have the factor 2x+1, that is, the zero x=-½.
We can use algebraic division to find the quadratic:
x2 -9x -1
2x+1 ) 2x3-17x2-11x-1
2x3 +x2
-18x2-11x
-18x2 -9x
-2x -1
-2x -1
0
To solve the quadratic x2-9x-1 we can use the formula: x=(9±√(81+4))/2=(9±√85)/2=9.1098 or -0.1098.
So x=-½ (-0.5), 9.1098, or -0.1098.