Which Descartes rule of signs would apply to the following 
P(x)=x^2+5x+6

First term, x^2 is positive, +
Second term +5x is positive, +
Third term 6 is positivs +

The signs of the terms left to right go +++, 
So there are no sign changes, so
there are no positive zeros (or roots).

Substitute (-x) for x and simplify

P(-x)=(-x)^2+5(-x)+6
P(-x)=x^2-5x+6

The signs of the terms going left to right is +-+

That's two sign changes, so there are either 2 negative
zeros (roots) or 2 less that 2 negative or 0 negative
zeros (roots)

Conclusion: There are no positive zeros (roots), and 
either 2 or 0 negative zeros (roots).

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P(x)=8x^3+2x^2-14x+5

The signs of the terms go ++-+

That's 2 sign changes, so there are either 2 or 0 positive
zeros (roots).

Substitute -x for x and simplify:

P(-x)=8(-x)^3+2(-x)^2-14(-x)+5
P(-x)=-8x^3+2x^2+14x+5

The signs of terms go -+++
That's 1 sign change, so there is exactly 1 negative zero.

Conclusion:  There is 1 negative and either 2 or 0 positive
zeros (roots).

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P(x)=9x^3-4x^2+10

The signs of the terms go +-+
That's 2 sign changes going left to right.
So there are either 2 or 0 positive zeros (roots).

Substitute -x for x and simplify:

P(-x)=9(-x)^3-4(-x)^2+10
P(-x)=-9x^3-4x^2+10

The signs of the terms go --+
That's 1 sign change going left to right.
So there is 1 negative zero (root).

Conclusion: There is 1 negative zero (root) and
either 2 or 0 positive zeros (roots).

Edwin