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anonymous

  • 5 years ago

factor into 2 factors with integral coefficients: x^8+98x^4+1

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  1. anonymous
    • 5 years ago
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    any ideas?

  2. anonymous
    • 5 years ago
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    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?

  3. anonymous
    • 5 years ago
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    jeez

  4. anonymous
    • 5 years ago
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    mad problem, :)

  5. anonymous
    • 5 years ago
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    anyone?

  6. dumbcow
    • 5 years ago
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    \[(x ^{4}+ax ^{3}+bx ^{2}-cx +1)(x ^{4}-ax ^{3}+bx ^{2}+cx+1)\] \[= x ^{8} +(2b-a ^{2})x ^{6}+(b ^{2}+2ac+2)x ^{4} +(2b-c ^{2})x ^{2}+1\] \[2b-a ^{2} =0\] \[b^{2}+2ac+2=98\] \[2b-c^{2}=0\] solve system for a,b,c a=c=4 b=8

  7. anonymous
    • 5 years ago
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    well dumbcow..u knew the answer before solving...:)LOL

  8. dumbcow
    • 5 years ago
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    lol true, but if you in general know the form (alternating signs so odd exponents cancel) you could solve for the coefficients

  9. anonymous
    • 5 years ago
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    i wouldnt know if both the factors had the same set of integer coefficients:)

  10. dumbcow
    • 5 years ago
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    no but that would be the assumption since terms cancel

  11. anonymous
    • 5 years ago
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    i think tats too much assuming...

  12. dumbcow
    • 5 years ago
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    haha ok well i tried

  13. anonymous
    • 5 years ago
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    so here goes a medal...!:) lol

  14. anonymous
    • 5 years ago
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    what do you mean the signs alternate?

  15. anonymous
    • 5 years ago
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    how did you get signs to alternate?

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