Use Descartes' rule of signs to determine the possible number of:
a) positive real roots of
b) negative real roots of
My answer to a) (zero positive real roots), was correct.
Zero positive real roots because there are zero sign changes..
and I had thought there would be zero negative roots as well. But it came back wrong.
Can someone explain to me why? Thanks.
Using Descarte's method, to determine the number of negative roots you replace the x in the polynomial with -x, and see how many sign changes there are in the resulting equation:
Note that there are 3 changes of sign as you read left to right. Thus there may be 3 negative roots, or possibly just 1 (since it's possible that 2 of the roots may actually be complex).
If you plot this function you'll see that there is one real negtive root. The two other roots are therefore complex.
There's another "trick" that you could have used. Remember that a polynomial of order N must have N roots. So since you already knew that there were zero positive roots, you know that there must be either (a) 3 negative roots, or (b) 1 negative root and 2 complex roots.