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anonymous
 one year ago
what are the factors of x^32x^229x+30
Please explain the steps too?
anonymous
 one year ago
what are the factors of x^32x^229x+30 Please explain the steps too?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know the answers are (x6) (x1) (x+5) but how do I explain that?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Well you can foil those out or you can notice that \(1\) is a solution, and then divide the whole thing by \((x1)\) then factor that.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Do you know how to do that?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.11. 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 \[1^32(1^2)29(1)+30 \rightarrow 1229+30 =30+30 = 0 \] 3.since we obtained one of the roots. Use long division or synthetic division to find the remaining roots.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I'm gonna use synthetic division because long division will take some time dw:1439178108445:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1now we factor \[x^2x30 \] the middle term has to be 1 and the last term is 30 suppose we use this combination 5 x 6 = 30 then 56 = 1 so we need a positive 5 and a negative 6 (x+5)(x6)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so 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 (x1) from the rational root theorem and (x+5)(x6) with synthetic division therefore (x1)(x+5)(x6)
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