who would you factor this?
k^3+6k^2+11k+6
This should help you: The Factor Theorem
You're not responding so I'll take that to mean that you need an extra push.
The rational zero theorem basically states that a polynomial of the form $\displaystyle a_nx^n+a{n-1}x^{n-1}+...+a_1x+a_0$ has rational zeros, then they must be given by the quotient of the divisors of $\displaystyle a_0$ divided by the divisors of $\displaystyle a_n$.
So, begin by listing all of the divisors of the constant term:
$\displaystyle \pm{1},\pm2,\pm3,\pm6$
Then list the divisors of the leading coefficient:
$\displaystyle \pm1$
So, therefore, the possible zeros of your are $\displaystyle \pm1,\pm2,\pm3,\pm6$.
So, now you can test these values by substition, or synthetic division. Now understand that (by the factor theorem as stroodle posted) if x=c is a zero of f(x), then $\displaystyle (x-c)$ is a factor of f(x).
Hint
$\displaystyle \begin{gathered}{k^3} + 6{k^2} + 11k + 6 = \hfill \\= {k^3} + 1 + 6{k^2} - 6 + 11k + 11 = \hfill \\= \underbrace {{k^3} + 1}_{{a^3} + {b^3}} + 6\underbrace {\left( {{k^2} - 1} \right)}_{{a^2} - {b^2}} + 11\left( {k + 1} \right). \hfill \\ \end{gathered}$