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Find Descartes rule x^3+3x^2+3x+1

Solution:
Your problem `->` Descartes rule x^3+3x^2+3x+1


Here `f(x)=x^3+3x^2+3x+1`

`f(x)=x^3color{blue}{+}3x^2color{blue}{+}3xcolor{blue}{+}1`

look first at `f(x)`: (positive case)

`f(x)=``color{blue}{+}``x^3``color{blue}{+}``3x^2``color{blue}{+}``3x``color{blue}{+}``1`
Sign change count`color{blue}{+}` to `color{blue}{+}`
 
`color{blue}{+}` to `color{blue}{+}`
 
`color{blue}{+}` to `color{blue}{+}`
 


There are no sign changes, so there are no positive roots.


Now look at `f(–x)`: (negative case)

`f(-x)=(-x)^3color{blue}{+}3(-x)^2color{blue}{+}3(-x)color{blue}{+}1`

`f(-x)=color{red}{-}x^3color{blue}{+}3x^2color{red}{-}3xcolor{blue}{+}1`

`f(-x)=``color{red}{-}``x^3``color{blue}{+}``3x^2``color{red}{-}``3x``color{blue}{+}``1`
Sign change count`color{red}{-}` to `color{blue}{+}`
1
`color{blue}{+}` to `color{red}{-}`
2
`color{red}{-}` to `color{blue}{+}`
3


There are 3 sign changes, so there are 3 or counting down in pairs, 1 negative roots.





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