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anonymous
 5 years ago
factor into 2 factors with integral coefficients:
x^8+98x^4+1
anonymous
 5 years ago
factor into 2 factors with integral coefficients: x^8+98x^4+1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wolframalpha shows: (1 + 4 x + 8 x^2  4 x^3 + x^4) (1  4 x + 8 x^2 + 4 x^3 + x^4) but how to get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(x ^{4}+ax ^{3}+bx ^{2}cx +1)(x ^{4}ax ^{3}+bx ^{2}+cx+1)\] \[= x ^{8} +(2ba ^{2})x ^{6}+(b ^{2}+2ac+2)x ^{4} +(2bc ^{2})x ^{2}+1\] \[2ba ^{2} =0\] \[b^{2}+2ac+2=98\] \[2bc^{2}=0\] solve system for a,b,c a=c=4 b=8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well dumbcow..u knew the answer before solving...:)LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol true, but if you in general know the form (alternating signs so odd exponents cancel) you could solve for the coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wouldnt know if both the factors had the same set of integer coefficients:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no but that would be the assumption since terms cancel

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think tats too much assuming...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so here goes a medal...!:) lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean the signs alternate?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get signs to alternate?
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