Question 142440
For f(x)=-2x^5+3x^4-6x^2+1 what does Descartes rule of signs tell you about
The number of positive real zeros?
<pre><font size =4 color = "indigo"><b>
f(x) = -2x<sup>5</sup> + 3x<sup>4</sup> - 6x<sup>2</sup> + 1

The signs go:

       -    +     -     +         

That makes 3 sign changes.  So there are either

3 positive real zeros or
2 less than 3 positive real zeros

Answer: There are either 3 or 1 positive real zeros.

The number of negative real zeros?

Now find f(-x)

f(x) = -2x<sup>5</sup> + 3x<sup>4</sup> - 6x<sup>2</sup> + 1
f(-x) = -2(-x)<sup>5</sup> + 3(-x)<sup>4</sup> - 6(-x)<sup>2</sup> + 1
f(-x) = -2(-x<sup>5</sup>) + 3x<sup>4</sup> - 6x<sup>2</sup> + 1

f(-x) = +2x<sup>5</sup> + 3x<sup>4</sup> -6x<sup>2</sup> + 1

Erasing all but the signs:

        +    +     -    +

There are two sign changes, so 

2 negative real zeros or
2 less than 2 negative real zeros

Answer: There are either 2 or 0 negative real zeros.

Edwin</pre>