anonymous
  • anonymous
factor into 2 factors with integral coefficients: x^8+98x^4+1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
any ideas?
anonymous
  • anonymous
wolframalpha 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
  • anonymous
jeez

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anonymous
  • anonymous
mad problem, :)
anonymous
  • anonymous
anyone?
dumbcow
  • dumbcow
\[(x ^{4}+ax ^{3}+bx ^{2}-cx +1)(x ^{4}-ax ^{3}+bx ^{2}+cx+1)\] \[= x ^{8} +(2b-a ^{2})x ^{6}+(b ^{2}+2ac+2)x ^{4} +(2b-c ^{2})x ^{2}+1\] \[2b-a ^{2} =0\] \[b^{2}+2ac+2=98\] \[2b-c^{2}=0\] solve system for a,b,c a=c=4 b=8
anonymous
  • anonymous
well dumbcow..u knew the answer before solving...:)LOL
dumbcow
  • dumbcow
lol true, but if you in general know the form (alternating signs so odd exponents cancel) you could solve for the coefficients
anonymous
  • anonymous
i wouldnt know if both the factors had the same set of integer coefficients:)
dumbcow
  • dumbcow
no but that would be the assumption since terms cancel
anonymous
  • anonymous
i think tats too much assuming...
dumbcow
  • dumbcow
haha ok well i tried
anonymous
  • anonymous
so here goes a medal...!:) lol
anonymous
  • anonymous
what do you mean the signs alternate?
anonymous
  • anonymous
how did you get signs to alternate?

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