Answer
Positive real zeros: 3 or 1
Negative real zeros: 2 or none
Work Step by Step
*Refer to page 384 for a complete description of Descarte's Rule of Signs*
Descarte's Rule of Signs applies for two scenarios:
1) Positive real zeros: we identify how many sign changes there are within the polynomial from left to right:
$$f(x) = +2x^{5} -3x^{3} - 5x^{2} + 3x - 1$$
where we see 3 changes: from positive $2x^{5}$ to negative $3x^{3}$, from negative $5x^{2}$ to positive $3x$ and from positive 3x to negative 1. This means that, according to the Rule, there are either $3$ positive real zeros for this polynomial or $3-2 = 1$ positive real zero.
2) Negative real zeros. we substitute $-x$ into the original function:
$$f(-x) = +2(-x)^{5} -3(-x)^{3} - 5(-x)^{2} + 3(-x) - 1$$
which simplifies to:
$$f(-x) = +2(-1)x^{5} -3(-1)x^{3} - 5(1)x^{2} + 3(-1)x - 1$$
$$f(-x) = -2x^{5} + 3x^{3} - 5x^{2} - 3x - 1$$
where we see 2 changes: from negative $2x^{5}$ to positive $3x^{3}$ and from positive $3x^{3}$ to negative $5x^{2}$. This means that, according to the Rule, there are either $2$ negative real zeros for this polynomial or $2-2=$ no negative real zeros.
