Question 634387
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The <b><i>possible</i></b> rational zeros of 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ a_0x^n\ +\ a_1x^{n-1}\ +\ \cdots\ +\ a_{n-1}x\ +\ a_n]


are elements of the set *[tex \LARGE \left{\frac{p}{q}\,|\,p,\,q\,\in\,\mathbb{Z},\,a_n\,\text{mod}\,p\,=\,0,\,a_0\,\text{mod}\,q\,=\,0\right}]


So, for


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho_1(x)\ =\ x^3\ -\ x^2\ -\ 8x\ -\ 2]


*[tex \LARGE a_0\ =\ 1], *[tex \LARGE a_n\ =\ 2].  The only integer divisors of 1 are *[tex \LARGE \pm1], and the only integer divisors of 2 are *[tex \LARGE \pm1] and *[tex \LARGE \pm2] so the possible rational roots are *[tex \LARGE \pm1] and *[tex \LARGE \pm2]


Use synthetic division to test each of the possibilities, since the synthetic division remainder for an *[tex \LARGE x_1] divisor is equal to *[tex \LARGE \rho(x_1)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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