Answer
2 real zeros
Work Step by Step
For Descartes' Rule of Signs, the number of sign changes (+ then - or - then +) for P(x) will be the number of possible positive real roots. The number of sign changes for P(-x) will be the number of possible negative real roots.
Given $P(x) = 2x^6 + 5x^4 - x^3 - 5x - 1$
For P(x), there is one sign change. Thus, there is 1 possible positive real root.
For $P(-x) = 2(-x)^6 + 5(-x)^4 - (-x)^3 - 5(-x) - 1$
$P(-x) = 2x^6 + 5x^4 + x^3 + 5x - 1$
There is 1 sign change. Thus there is 1 negative real root.
Thus the answer is 2 real zeros
