To factor the following formulas, there is a specific process

Updated on educate 2024-06-07
3 answers
  1. Anonymous users2024-02-11

    = x-y)^2*(x-y)

    x-y)^3

    a-b)^2*(18a-18b-12b)

    6(a-b)^2*(3a-5b)

    3.(2a+b)(2a-3b)-3a(2a+b) =2a+b)(2a-3b-3a)

    2a+b)(a+3b)

    x(x^2+3x-3y-xy-x^2-2xy-y^2)

    x(3x-3y-3xy-y^2),7,x(x-y)^2-y(y-x)^2 =x(x-y)^2-y(x-y)^2=(x-y)(x-y)^2=(x-y)^3

    18(a-b)^3-12b(b-a)^2=18(a-b)^3-12b(a-b)^2=[18(a-b)-12b](a-b)^2

    18a-30b)(a-b)^2=6(3a-5b)(a-b)^2

    2a+b)(2a-3b)..2. Divide the following types of defeat into solution formulas, and have specific friends who have quietly passed a good slag process!

    x(x-y)^2-y(y-x)^2

    18(a-b)^3-12b(b-a)^2

    2a+b)(2a-3b)-3a(2a+b)

    x(x+3)(x-y)-x(x+y)^2

  2. Anonymous users2024-02-10

    squared - t-12 = (t+3)(t-4).

    The square of Hu + 11m-12 = (m + 12) (m-1) of the square of -2xy + y of the square of -c = (x-y) -c = (x-y + c) (x-y-c).

    4 mx-my-ny+nx=m(x-y)+n(x-y)=(m+n)(x-y)

    +6a-3b-b.

    4a²-b²+6a-3b

    2a+b)(2a-b)+3(2a-b)

    2a+b+3)(2a-b),6,Defactor the following types of rent promotions: (To complete the process)!

    squared - t+12

    +11m-12 squared

    The square of -2xy + the square of y-the square of c.

    4 mx-my-ny+n-x

    +6a-3a-b squared.

  3. Anonymous users2024-02-09

    (1) Original formula = 6x n (3x-4).

    2) Original formula = 3 (a-b) [5 x (a-b) -y] = 3 (a-b) (5ax-5bx-y).

    3) Original formula = -5a (4+3x).

    4) Original formula = (m+n)(p-q)[1-(p-q)]= (m+n)(p-q)(1-p+q).

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