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anonymous

  • one year ago

what are the factors of x^3-2x^2-29x+30 Please explain the steps too?

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  1. anonymous
    • one year ago
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    I know the answers are (x-6) (x-1) (x+5) but how do I explain that?

  2. zzr0ck3r
    • one year ago
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    Well you can foil those out or you can notice that \(1\) is a solution, and then divide the whole thing by \((x-1)\) then factor that.

  3. zzr0ck3r
    • one year ago
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    Do you know how to do that?

  4. anonymous
    • one year ago
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    yes thank you

  5. UsukiDoll
    • one year ago
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    1. use the rational root theorem - find all factors of the last term in the polynomial so in that case it's 30 . We need the positive and negative versions 1, 2, 3, 5, 6, 10, 15, 30 -1, -2, -3, -5, -6, -10, -15, -30 2. Plug in one of these numbers and see if our result is 0. If our result is 0, then we have found one of the roots. If not, then we continue until we find a number that can give us 0 For example, if we let x = 1 \[1^3-2(1^2)-29(1)+30 \rightarrow 1-2-29+30 =-30+30 = 0 \] 3.since we obtained one of the roots. Use long division or synthetic division to find the remaining roots.

  6. UsukiDoll
    • one year ago
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    I'm gonna use synthetic division because long division will take some time |dw:1439178108445:dw|

  7. UsukiDoll
    • one year ago
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    now we factor \[x^2-x-30 \] the middle term has to be -1 and the last term is -30 suppose we use this combination 5 x -6 = -30 then 5-6 = -1 so we need a positive 5 and a negative 6 (x+5)(x-6)

  8. UsukiDoll
    • one year ago
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    so now we got the remaining roots using synthetic division (you can use long division too... takes time but the results are the same) we have (x-1) from the rational root theorem and (x+5)(x-6) with synthetic division therefore (x-1)(x+5)(x-6)

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