Question

一道初二因式分解题

（x²-1)(x+3)(x+5)+12。最好要有过程，谢谢

Answer 1

（x²-1)(x+3)(x+5)+12。
=(X+1)(X-1)(x+3)(x+5)+12。
=(x²+4X+3)(X²+4X-5)+12
=(x²+4X+3)²-8(x²+4X+3)+12
=(x²+4X+3-2)(x²+4X+3-6）
=(x²+4X+1)(x²+4X-3)

Answer 2

(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).

Answer 3

(x²-1)(x+3)(x+5)+12
=(X+1)(X-1)(x+3)(x+5)+12
=(x²+4X+3)(X²+4X-5)+12
设t=x²+4X+4
原式=(t-1)(t-9)+12
=t²-10t+9+12
=t²-10t+21
=(t-3)(t-7)
=(x²+4X+1)(x²+4X-3)