# Factoring a cubic function

• Jul 29th 2009, 06:37 PM
jordangiscool
Factoring a cubic function
who would you factor this?
k^3+6k^2+11k+6
• Jul 29th 2009, 06:42 PM
VonNemo19
By the rational zero theorem
• Jul 29th 2009, 07:10 PM
Stroodle
• Jul 29th 2009, 07:49 PM
VonNemo19
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 $a_nx^n+a{n-1}x^{n-1}+...+a_1x+a_0$ has rational zeros, then they must be given by the quotient of the divisors of $a_0$ divided by the divisors of $a_n$.

So, begin by listing all of the divisors of the constant term:

$\pm{1},\pm2,\pm3,\pm6$

Then list the divisors of the leading coefficient:

$\pm1$

So, therefore, the possible zeros of your are $\pm1,\pm2,\pm3,\pm6$.

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 $(x-c)$ is a factor of f(x).
• Jul 30th 2009, 05:00 AM
DeMath
Quote:

Originally Posted by jordangiscool
who would you factor this?
k^3+6k^2+11k+6

Hint

$\begin{gathered}{k^3} + 6{k^2} + 11k + 6 = \hfill \\= {k^3} + 1 + 6{k^2} - 6 + 11k + 11 = \hfill \\= \underbrace {{k^3} + 1}_{{a^3} + {b^3}} + 6\underbrace {\left( {{k^2} - 1} \right)}_{{a^2} - {b^2}} + 11\left( {k + 1} \right). \hfill \\ \end{gathered}$