Respuesta :

apologiabiology
so first factor out the big equation on top
very hard
we can cancel out the 0x terms since 0x=0

so we notice that 3 is a factor of both equations
(6x+3)=(3)(2x+1)

and
(6x^5+21x^4+33x^3+18x+12)=(3)(2x+1)(x^4+3x^3+4x^2-2x+4
so we notice that there is a (3)(2x+1) in bot h so divide and cross them out

the answer is (x^4+3x^3+4x^2-2x+4)

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