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

Mar 25, 2016

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:

$f \left(x\right) = 3 {x}^{3} + 39 {x}^{2} + 39 x + 27 = 3 \left({x}^{3} + 13 {x}^{2} + 13 x + 9\right)$

Then applying the rational roots theorem to the remaining cubic factor, we can deduce that any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ 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:

$\pm 1$, $\pm 3$, $\pm 9$

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 $f \left(x\right)$ has no rational zeros.

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

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