因式分解：(x^2+3x+2)(4x^2+8x+3)-90用换元法
问题描述：
因式分解：(x^2+3x+2)(4x^2+8x+3)-90用换元法
答
原式=（x+1）(x+2)(2x+3)(2x+1)-90
=[(x+1)(2x+3)][(x+2)(2x+1)]-90
=(2x^2+5x+3)(2x^2+5x+2)-90
=(2x^2+5x)^2+5(2x^2+5x)-84
=(2x^2+5x-7)(2x^2+5x+12)
=(x-1)(2x-7)(2x^2+5x+12)
答
(x^2+3x+2)(4x^2+8x+3)-90
=(x+2)(x+1)(2x+1)(2x+3)-90
=(x+2)(2x+1)(x+1)(2x+3)-90
=(2x^2+5x+2)(2x^2+5x+3)-90
令t=2x^2+5x
原式=(t+2)(t+3)-90
=(t+2)(t+2+1)-90
=(t+2)^2+(t+2)-90
=(t+2+10)(t+2-9)
=(t+12)(t-7)
=(2x^2+5x+12)(2x^2+5x-7)
=(2x^2+5x+12)(2x+7)(x-1)