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Math Help - Factoring a cubic function

  1. #1
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    Factoring a cubic function

    who would you factor this?
    k^3+6k^2+11k+6
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  2. #2
    No one in Particular VonNemo19's Avatar
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    By the rational zero theorem
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  3. #3
    Senior Member Stroodle's Avatar
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    This should help you: The Factor Theorem
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  4. #4
    No one in Particular VonNemo19's Avatar
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    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).
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by jordangiscool View Post
    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}
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