m m2 2m3 factorization

m(1+m-2m²)

m(2m²-m-1)

m(m-1)(2m+1)

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1。x 3-xy2 +2 xy-= (x2-y2 2 -1) = [2 - y 2-2y +1)] = plate (xy +1) (times + y-1).

2。2-bx-2 + ab = 2-2-bx + ab = (+) xa) -b option (round) = (xa) (+ab).

3。 x ^ 2-4y ^ 2-3x +6?==（x +2 y）（x-2y）-3（x-2y）=（x-2y）（x +2 y-3）

4.- 3-2x 2-x +4 starting xy2 =- (2 +2 deporting +1-4 y2) = (+1-2 y) (x +1 2y).

5。9a, 2-4b, 2-6a+1 = 9a, 2-6a+1-4b2 = (Figure 3a-1), 2-4b, 2=(Figure 3a-1+2b) (Figure 3a-1-2b), >6. x 2-ax-y 2 + ay = x 2-y 2-ax + ay = (x + y) (xy) (x,y) = (x,y) (x + sub).

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

8。4a2-b2-4a1 = 4a2-4 +1-b2 allocation = (Figure 2a-1), 2-b2 allocation = (2a-1 + b) (Figure 2a-1-b).

m(1+m-2m^2)

m(2m^2-m-1)

m(m-1)(2m+1)

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Jelly m as -as m) as -as min nai m as +4m-z (m as -as m) as -as (m as -as m) -z(m as -as m-z) as -as m-as m-m-as m-as m-1 (m-z) as

This requires a little skill, first of all, this is a problem involving high school math, which can be solved quickly with high school knowledge!

Since you haven't learned the Leaky Oak Game yet, I can simplify it!

First, let m 3-3m-2=0

Substituting -3 -2 -1 0 1 2 3 are these six numbers, see which one can be true! (These are the seven, of course, special but plus or minus 4, often used as plus or minus 1).

It can be calculated that -1 can be , so the above equation must have the factor m+1, and start to make up:

m 3-3m-2=m 3-3m-3+1=m 3-3(m+1)+1=m 3+1-3(m+1)=(m+1) (e.g., m 2-m-2).

Let's get back!

The original Qingbo large formula =-m[m 2-2m+1+n 2]=-m[(m-1) +n ] After that, there is no way to decompose it, and I suspect that the question is wrong, it should be +mn 2