use the rational zeros theorem to write a list of all possible rational zeros of the function. f(x)=3x^3+39x^2+39x+27
all fractions \(\frac{p}{q}\) where the numerator \(p\) divides \(27\) and the denomnator \(q\) divides \(3\)
there are a bunch the numerator could be either 1, 3, 9 or 27 the denominator could be either 1 or 3 and don't forget the \(\pm\)
can you show me the working please. im having problems with this specific problem
ok lets make a list \[\pm1,\pm3,\pm9,\pm27\] is a start there i took the numerator to be divisors of 27 and the denominator to be 1
now we can take the denominator to be 3, but we only get two new ones \[\pm\frac{1}{3}\] because if you divide any of the other by 3 you get a number already in the list
that is the complete list of possible rational zeros
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