因式分解(x²-1)(x+3)(x+5)+12要过程
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(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)
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)
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(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+4x+1)(x^2+4x-3)
=(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+4x+1)(x^2+4x-3)
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