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

Question

anonymous · Answer

any ideas?

anonymous · Answer

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 · Answer

jeez

anonymous · Answer

mad problem, :)

anonymous · Answer

anyone?

dumbcow · Answer

$$(x ^{4}+ax ^{3}+bx ^{2}-cx +1)(x ^{4}-ax ^{3}+bx ^{2}+cx+1)$$
$$= x ^{8} +(2b-a ^{2})x ^{6}+(b ^{2}+2ac+2)x ^{4} +(2b-c ^{2})x ^{2}+1$$
$$2b-a ^{2} =0$$
$$b^{2}+2ac+2=98$$
$$2b-c^{2}=0$$
solve system for a,b,c
a=c=4
b=8

anonymous · Answer

well dumbcow..u knew the answer before solving...:)LOL

dumbcow · Answer

lol true, but if you in general know the form (alternating signs so odd exponents cancel) you could solve for the coefficients

anonymous · Answer

i wouldnt know if both the factors had the same set of integer coefficients:)

dumbcow · Answer

no but that would be the assumption since terms cancel

anonymous · Answer

i think tats too much assuming...

dumbcow · Answer

haha ok well i tried

anonymous · Answer

so here goes a medal...!:) lol

anonymous · Answer

what do you mean the signs alternate?

anonymous · Answer

how did you get signs to alternate?