因式分解(x^2+2x+2)^2+4(3x^2+7x+3)(2x^2+5x+1)

1个回答

  • 设(x+1)^2=y,那么原式整理后

    =[(x+1)^2+1]^2+4[3(x+1)^2+x][2(x+1)^2+(x-1)]

    =(y+1)^2+4(3y+x)[2y+(x-1)]

    =y^2+2y+1+4[6y^2+(5x-3)y+x(x-1)]

    =25y^2+(20x-10)y+4x^2-4x+1

    =(5y)^2+10(2x-1)y+(2x-1)^2

    =[5y+(2x-1)]^2

    =[5(x+1)^2+(2x-1)]^2

    =(5x^2+12x+4)^2

    =[(5x+2)(x+2)]^2

    解法2

    原式=(x^2+2x+2)^2+4[3(x^2+2x+2)+(x-3)][2(x^2+2x+2)+(x-3)]

    =25(x^2+2x+2)^2+20(x-3)(x^2+2x+2)+4(x-3)^2

    =[5(x^2+2x+2)+2(x-3)]^2

    =(5x^2+12x+4)^2

    =[(5x+2)(x+2)]^2