SOLUTION: How many zeros does P(x) have and list all the possible suspects for finding the real zeros for P(x). P(x)=x^4+6x^3+6x^2+2x+8

Question 171683: How many zeros does P(x) have and list all the possible suspects for finding the real zeros for P(x).
P(x)=x^4+6x^3+6x^2+2x+8
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many zeros does P(x) have and list all the possible suspects for finding the real zeros for P(x).
P(x)=x^4+6x^3+6x^2+2x+8
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How many?: four because the degree of the polynomial is four.
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Real Number Candidates:
If p/q is a zero, p must be a divisor of 8 and q must be a divisor of 1
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+-1,+-2,+-4,+-8 are the possibilities
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Cheers,
Stan H.
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