Question 63890
f(x)= -6x^5+x^4+5x^3+x+1
# of changes in signs = 1
Therefore one positive real zero.
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f(-x)=6x^+x^4-5x^3-x+1
# of changes in signs = 2
Therefore 2 or 0 negative real zeroes.
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Cheers,
Stan H.