因式分解:(x^2+3x+2)(4x^2+8x+3)-90用换元法
2个回答
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(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)
=(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)
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