Question 189028: Given P(x)=6x^3+x^2+2x+3, how many POSSIBLE roots of P(x) are there?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
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 =6x^3+x^2+2x+3}) , how many POSSIBLE rational 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 p. So 
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 q. So 
The list of possible rational roots of your polynomial function can then be derived by substituting all of the possible values for p and q into:
So we have:
 (but you already have this in the list)
 (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 not in this list is not a root of this function.
John
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