因式分解(x²-1)(x+3)(x+5)+12要过程

(x²-1)(x+3)(x+5)+12
解:原式=(x+1)(x-1)(x+3)(x+5)+12
=[(x-1)(x+5)][(x+3)(x+1)]+12
=(x²+4x-5)(²+4x+3)+12
=[(x²+4x-1)-4][(x²+4x-1)+4]+12
=(x²+4x-1)²-4²+12
=(x²+4x-1)²-2²
=(x²+4x-1+2)(x²+4x-1-2)
=(x²+4x+1)(x²+4x-3)

(x^2-1)(x+3)(x+5)+12 =(x+1)(x-1)(x+3)(x+5)+12 =[(x-1)(x+5)][(x+3)(x+1)]+12 =(x^2+4x-5)(x^2+4x+3)+12 =[(x^2+4x-1)-4][(x^2+4x-1)+4]+12 =(x^2+4x-1)^2-16+12 =(x^2+4x-1)^2-4 =(x^2+4x-1+2)(x^2+4x-1-2) =(x^2...