Let 8x^3+3x^2-6x-2=0 be a polynomial equation

Question

Use Descartes' Rule of signs to help list the possibilities for the nature of the solutions for the polynomial equation. Do not solve the equation.

 

List the possible number(s) of negative real solutions:

 

List the possible number(s) of non-real complex solutions

Answer 1

one change in sign, so maximum one positive real solution

substitute -x for x and count the changes in sign

two changes in sign for f(-x) so max 2 negative real solutions

non-real complex solutions come in conjugate pairs, so 2 is the maximum possible.

either 0 or 2

f(x) = 8x^3 + 3x^2 -6x -2 = 0

x f(x)

0 -2

1 3

1/2 1+3/4-3-2 = -3 1/4

3/4 27/16 + 27/8 -36/8 - 32/16 = -25/16 real solution between 3/4 and 1

-1 -1

-2 -64+12+12-2= --42 no sign changes in f(x) for negative x

odds are there's 1 real with 3/4<x<1 and two imaginary solutions or two negative real solutions

turns out it's 3 real solutions 1 positive 2 negative

just graph the cubic and find the x intercepts = the real solutions