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find all the zeros of this equation f(x) = 3x^3-7x^2+6x-8 please show work.
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best bet is to factor by grouping
(3x^3-7x^2)+(6x-8), it is a start
oh okay. I guess that's all I really needed to know.
Finding zeros is a pain. First find the possible rational zeros, then test them to see if one will turn up a zero
this is real and non real though. The idea is to find 3 zeros so that we understand that the degree usually determines the # of zeros
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For this equation: \[3x ^{3}-7x ^{2}+6x-8\] p= \[\pm1,\pm2,\pm4,\pm8\]q=\[\pm1,\pm2\] these are the possible zeros then do p/q, make a chart and plug the p/q values in.
thank you :)
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