Question 319714: given:
P(x)=x^4-x^3-5x^2-x+6
Find all real and non- real zeroes of P(x).
Thanks for taking the time to look at this problem, the concept of this type of problem is a bit confusing to me at the moment and I am not really sure if I am working them right. Appreciate any input that can be given.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! given:
P(x)=x^4-x^3-5x^2-x+6
Find all real and non- real zeroes of P(x).
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Note: Since the coefficients add up to zero, x = 1 is a zero.
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Use synthetic division to find other zeroes:
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1)....1....-1....-5....-1....6
.......1.....0.....-5....-6...|..0
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Quotient: x^3 -5x -6
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This quotient has a zero around x = 2.6891,
but that is the only Real zero my calculator
will pick up.
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It appears there are two complex zeroes, but
I cannot pin them down.
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Cheers,
Stan H.
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