Question

How do you use the rational roots theorem to find all possible zeros of `f(x) = 3x^3 + 39x^2 + 39x + 27`?

Answer 1

See explanation...

Explanation:

Before applying the rational roots theorem, note that all of the coefficients are divisible by `3`, so separate that out as a factor first:

`f(x) = 3x^3+39x^2+39x+27 = 3(x^3+13x^2+13x+9)`

Then applying the rational roots theorem to the remaining cubic factor, we can deduce that any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p`, `q` where `p` is a divisor of the constant term `9` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-3`, `+-9`

In addition note that all of the coefficients are positive, so there are no zeros for positive values of `x`. So the only possible rational zeros are:

`-1`, `-3`, `-9`

None of these is a zero, so `f(x)` has no rational zeros.

That is as much as we can learn from the rational roots theorem.

In fact `f(x)` has one Real irrational zero near `-12` and a pair of Complex zeros.