几道因式分解的题y(y+1)(x^2+1)+x(2y^2+2y+1)6x^2-5xy-6y^2-2xz-23yz-20z

1个回答

  • 1.

    y(y+1)(x^2+1)+x(2y^2+2y+1)

    =y(y+1)x^2+y(y+1)+x(2y^2+2y+1)

    =y(y+1)x^2+(2y^2+2y+1)x+y(y+1)——十字相乘法

    =[yx+y+1][(y+1)x+y]

    =[yx+y+1][yx+x+y]

    2.

    6x^2-5xy-6y^2-2xz-23yz-20z^2

    =(2x-3y)(3x+2y)-2xz-23yz-20z^2 (把前三项用十字相乘法因式分解)

    所以设:原式=(2x-3y+az)(3x+2y+bz)

    展开得:=(2x-3y)(3x+2y)+(2x-3y)bz+az(3x+2y)+abz^2

    =(2x-3y)(3x+2y)+(2b+3a)xz-(3b-2a)yz+abz^2

    左右对应相等,则有:

    2b+3a=-2

    3b-2a=23

    ab=--20

    解得:a=5,b=-4

    所以原式=(2x-3y+5z)(3x+2y-4z)

    3.

    2x^2-7xy+6y^2+2x-y-12

    =(x-2y)(2x-3y)+2x-y-12

    =(x-2y)(2x-3y)+6x-4x-9y+8y-12

    =(x-2y)(2x-3y)+6x-9y-4x+8y-12

    =(x-2y)(2x-3y)+3(2x-3y)-4(x-2y+3)

    =(x-2y+3)(2x-3y)-4(x-2y+3)

    =(x-2y+3)(2x-3y-4)