who would you factor this?

k^3+6k^2+11k+6

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- Jul 29th 2009, 06:37 PMjordangiscoolFactoring a cubic function
who would you factor this?

k^3+6k^2+11k+6 - Jul 29th 2009, 06:42 PMVonNemo19
By the rational zero theorem

- Jul 29th 2009, 07:10 PMStroodle
This should help you: The Factor Theorem

- Jul 29th 2009, 07:49 PMVonNemo19
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). - Jul 30th 2009, 05:00 AMDeMath
Hint

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