设x^2+x=y
原式=(y-3)(y-5)-3
=y^2-8y+15-3
=y^2-8y+12
=(y-2)(y-6)
因为x^2+x=y
所以原式=(x^2+x-2)(x^2+x-6)
=(x+2)(x-1)(x+3)(x-2)
x^2-x-3
=x^2-x+1/4-1/4-3
=(x-1/2)^2-13/4
=(x-1/2-√13/2)(x-1/2+√13/2)
x^2-x-3=(x-1/2)^2-13/4=[x-(1+√13)/2][x-(1-√13)/2]
x^2-x-3 =x^2-2(1/2)x+(1/2)^2-(1/2)^2-3
= (x-(1/2))^2-((1/4)+3)
= (x-(1/2))^2-(13/4)^(1/2 *2)
=((x-(1/2)+(13/4)^(1/2))((x-(1/2)-(13/4)^(1/2))
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