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ax3+bx2+c=(x2+px+c)(kx+m) where kx+m is a linear factor
expandinding the above and comparing coefficients of same powers of x we get
p=-a/c,ap+c=b,put p=-a/c in ap+c=b
we get -a2/c+c=b
-a2+c2=bc
so a2-c2=-bc
hence proved
suppose (ex+f) is the other root. Therefore,
(x^2+px+1)(ex + f) = ax^3 + bx^2 + c
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