Exercise 41 Page 203

The number of positive real zeros of a polynomial function $f$ with real coefficients is either equal to the number of sign changes between consecutive coefficients of $f (x)$ or is less than that by an even number.
The number of negative real zeros of a polynomial function $f$ with real coefficients is either equal to the number of sign changes between consecutive coefficients of $f (- x)$ or is less than that by an even number.

Let's check the sign changes in

f (x)

and

f (- x) .

f (x) f (- x) = x^{5} - (1) 4 x^{3} + (2) 6 x^{2} + 2 x - (3) 6 = (- x)^{5} - 4 (- x)^{3} + 6 (- x)^{2} + 2 (- x) - 6 = - x^{5} + (1) 4 x^{3} + 6 x^{2} - (2) 2 x - 6

There are $three$ sign changes in $f (x),$ so there are three or one positive zeros.
There are $two$ sign changes in $f (- x),$ so there are two or no negative zeros.

Option C mentions four negative real zeros. According to Descartes Rule of Signs this is not possible, so the correct answer is C.

Extra

Which option is the true classification?

We used Descartes' Rule of Signs to eliminate C as a possible classification of zeros. This only means that the other options do not contradict this rule. We can check which is the actual classification by drawing a graph with a graphing calculator. We begin by pushing the $Y =$ button and typing the equation in the first row.

To see the graph, you will need to adjust the window. Push $WINDOW,$ change the settings, and push $GRAPH .$

There are two negative and one positive real zeros of the function. Since the degree is $5,$ this means that there are also two imaginary zeros.