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find all the zeros of f(x) = 3x^3-7x^2+6x-8
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x = 1/6 (1-i sqrt(47)) x = 1/6 (1+i sqrt(47)) x = 2 Would you like an explanation?
RIght, real root is 2.
yes please
Factor (x-2) (3 x^2-x+4) x = 2 Then use the quadratic formula.
for 3x^2-x+4 use quadratic?
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Yes
I see I see :D thank you.
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