how to get the roots of u^3-3u^2-4
to get the possible zeroes: a = get +&- factors of -4: \[\pm1,\pm2,\pm4\] b = then get the +&- factors of coefficient of the highest degree, which is u^3, with coefficient of 1: \[\pm1\] then get all possible quotients of a/b: \[\pm1,\pm2,\pm4\] then use trial and error synthetic division to get at least one factor. let's use x+1: 1 -3 0 -4 ||-1 -1 -4 4 1 -4 -4 0<----- you got a zero, meaning x+1 is a root! to get the other root: 1x^2-4x-4=0 (from the 1 -4 -4 0 that we got) (x-2^2)=0 roots:(x+1), (x+2) of multiplicity 2
nice one, bro :))
(x-2)^2=0, i mean, not (x-2^2)=0
don't understand... did you do guess work to see who the equation equals to zero?
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