Pre-Calculus

Is it possible to factor x^3 - 9x^2 + 24x - 20 and solve for x by hand?

I tried using the grouping method and got x^2(x-9) + 4(6x-5) but how can I continue solving for x?

If I use a calculator I get x= 2 and 5 but is it possible to find this by hand? thanks

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  1. You use rational root theorem.
    Let p = constant term
    Let q = numerical coefficient of highest degree of x
    x = +/- p/q

    In here, p = -20 and q = 1. Thus,
    x = (+/-) 1 , 2 , 4 , 5 , 10, 20

    You have to substitute each factor (for instance x = -1) to the expression, and if you get a zero answer, that value of x is a factor. Well, it's kind of like trial and error, but at least you know which ones are to be substituted.

    Hope this helps :3

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  2. *I mean you get all the factors of x = +/- p/q. In the problem, x = +/- 20, and its factors are written there above. The factors are the ones you have to check by substituting them to the expression. :)

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  3. After showing that both f(1) and f(-1) ≠ 0
    try f(2)
    = 8 -36+48-20 = 0
    so (x-2) is a factor
    Now use synthetic division and
    x^3 - 9x^2 + 24x - 20
    = (x-2)(x^2 - 7x + 10)
    = (x-2)(x-2)(x-5)

    so x^3 - 9x^2 + 24x - 20
    = (x-5)(x-2)^2

    For a cubic, once you find one root, you can reduce your cubic to a quadratic, by either long or synthetic division.
    Of course you could just feel lucky and keep going to find other roots from the list that Jai gave you, and you would hit x = 5

    And of course, as soon as you found 2 factors of a cubic you can "reason" out what the third factor would be.

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