Question 189028
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This is a 3rd degree polynomial function, hence the Fundamental Theorem of Algebra tells us that there are exactly 3 factors of the polynomial and therefore 3 roots -- no more and no less.  But I don't think that is the question you meant to ask.  The 3 roots are at least one real number which may be irrational and the other two may be either a conjugate pair of complex numbers or they may be real also but, again, no guarantee that they are rational.


The question you meant to ask, I think, is:


Given *[tex \Large P(x)=6x^3+x^2+2x+3}], how many POSSIBLE <i><b>rational</b></i> roots of P(x) are there?


To answer that, we need the Rational Root Theorem.


First, find all of the integer factors of the constant term, namely 3 in your problem.  3 is prime so the only integer factors are 1 and 3.  Call this <i>p</i>.  So *[tex \Large p\,\in\,\{1,3}]


Next, find all of the integer factors of the lead coefficient, namely 6 in your problem.  The factors are 1, 2, 3, and 6.  Call this <i>q</i>.  So *[tex \Large q\,\in\,\{1,2,3,6}]


The list of possible rational roots of your polynomial function can then be derived by substituting all of the possible values for <i>p</i> and <i>q</i> into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{p}{q}]


So we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{1} = \pm 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{6}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{3}{1} = \pm 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{3}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{3}{3} = +\pm 1] (but you already have this in the list)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \frac{3}{6} = \pm \frac{1}{2}] (you already have this in the list, too)


Remember, this is a list of possibilities.  No more than three but possibly none of these may actually be roots.  The only guarantee is that any rational number <i><b>not</b></i> in this list is <i><b>not</b></i> a root of this function.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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