# finding rational roots of polynomials

Could someone please explain me how to apply the Descartes's Criterion? For example , how do I find the rational roots of $x^3 -x +1$.

I've been looking at some examples, but I get confused.

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Here $a_0=1$ and $a_n=1$

Factors of $a_0=1$ only and for $a_n$ too, it is $1$ only.

Thus, possible rational root is $\pm1/1=\pm1$

When we substitute $\pm 1$ back to the equation, they don't satisfy it $\implies \pm 1$(the only possible rational candidates) are not roots of the given equation $\implies$there is no rational root of $x^3-x+1=0$

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No, the factors are $\pm 1$, so the possibilities are $\pm 1$. Try it for $x^2+2x+1$. In the case of $x^3-x+1,\ -1$ doesn't work either. –  Ross Millikan Oct 1 '12 at 2:54

For a better example, consider $6x^2-35$. The factors of $6$ are $\pm(1,2,3,6)$ and the factors of $35$ are $1,5,7,35$ (you only need the sign on one set). The possible rational roots are then $\pm(1,5,7,35,\frac 12, \frac 52, \frac 72 ,\frac {35}2,\frac 13, \frac 53, \frac 73 ,\frac {35}3,\frac 16, \frac 56, \frac 76 ,\frac {35}6)$. In this case, none works as the factorization is $(\sqrt 6x + \sqrt{35})(\sqrt 6x - \sqrt{35})$

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@Maesumi: done. Missed the x's –  Ross Millikan Oct 1 '12 at 3:29