Question

(X^2+3X+2)(4X^2+8X+3)

(X^2+3X+2)(4X^2+8X+3)-90分解因式过程要用“换元法”做！！急需完整过程，那位高手帮一下谢谢

Answer 1

(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)

Answer 2

(x+1)(x+2)(2x+3)(2x+1)-90=[(x+1)(2x+3)][(x+2)(2x+1)]-90=(2x^2+4x+3)(2x^2+4x+2)-90
令2x^2+4x=t
(t+3)(t+2)-90=t^2+5t-84=(t+12)(t-7)=(2x^2+4x+12)(2x^2+4x-7)