2x^3-17x^2-11x-1=0

Question

Solve the given polynomial equation. Use the rational zero theorem and descartes's rule of signs.

Answer 1

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.