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Math Help - Binomial function

  1. #1
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    Binomial function

    How would I factorize

    x^3+2x^2-9x-18

    and then solve

    f(x)=0?

    Thanks a lot.
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  2. #2
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    Quote Originally Posted by Yppolitia View Post
    How would I factorize

    x^3+2x^2-9x-18

    and then solve

    f(x)=0?

    Thanks a lot.
    Factor by grouping: (Typically if you are asked to factor a cubic it will be a variation of this method.)
    (x^3 + 2x^2) - (9x + 18)

    x^2(x + 2) - 9(x + 2)

    (x^2 - 9)(x + 2)

    Now note that x^2 - 9 factors:

    So we get:
    (x + 3)(x - 3)(x + 2)

    So if we have f(x) = x^3+2x^2-9x-18 = 0

    Then
    (x + 3)(x - 3)(x + 2) = 0

    So x = -3, -2, or 3.

    -Dan
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  3. #3
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    also if you do not know how to factor that well you could try and find one root, with the rational root theorem.

    lets say that u have f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 now since this has a root lets call it r we then have

    f(x)=(x-r)Q(x)
    where Q(x)=y_nx^{n-1}+y{n-1}x^{n-1}+\dots+y_2x+y_1. as you notice the degree of Q is one less than that of f.
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