A few math factoring in the second year of junior high school Scoring is not a problem Finding a mas

m--m-a's m-1=a's m-1 power (a-1)4-a to the nth power - a to the n + 1st power + a to the n + 2nd power to a = -a to the nth power (1 + a-a).

7.(2x+y)²-9(x+2y)²=4x²+4xy+y²-9x²-36xy-36y²=-5x²-32xy-35y²=-(x+5y)(5x+7y)

8.n+1-3a to the nth power of a + n-1 power of 3a = n-1 power of a (n-3 to the nth power of a *n-1 power of a+3).

1. .m(a-b)+2(b-a)=(m-2)(a-b)2..

6x y -5x y = x y (6x-5y) to the m power --a m-1 = a m-1 * (a-1) 4.-a to the nth power - a to the n+1st power + a to the n+2nd power = a to the nth power * (a -a-1).

m+n+p)(1-x)

4q-4qp+2)(1-p)²

7.（2x+y)²-9(x+2y)²=(5x+7y)(-x-5y)

1.(m-2)(a-b)

M-1 to the power (A-1).

to the nth power (-1-a+a).

5.（m+n+p）(1-x)

6.（p-1)²(4qx(1-p)+2)

7.Is it wrong?

n-1 power (a-3a+3).

Check again, I may have made a mistake in the exam.

1.（a-b)(m-2)

5.(1-x)(m+n+p)

6.(1-p)*2(4q+2)

7.-(x+5y)(5x+7y)

to the n-1 power (a-3 to the n-power * a+3 to the n-1 power).

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A: 1Original.

2.Original. =[4(x+y)]^2-[5(x-y)]^2=(4x+4y+5x-5y)(4x+4y-5x+5y)=(9x-y)(9y-x)

3.Is 3(a+b) 2 missing a c on the left side of the question? It's 3C(A+B) 2, right?

Original. =3c[(a+b)^2-9c^2]

3c(a+b+3c)(a+b-3c)

4.Original. =(x+2)(x+4)+(x+2)(x-2)=(x+2)(x+4+x-2)

2(x+2)(x+1)

1 (x minus root number 7) multiplied by (x plus root number 7).

2 (x * root number 2 minus root number 3) multiplied by (x * root number 2 plus root number 3).

3 (a+b) to the nth power multiplied by (a+b-1) squared.

1, (x-y-2) (x-y a square b (x square + 4) (x-2) (x+2).

3. (4th power of x + 4th power of y) (square of x + square of y) (x+y) (x-y).

If the original formula can take a+b as x and 2b as y, then the original formula =x 2-2xy+y 2=(x-y) 2=(a+b-2b) 2=(a-b) 2

a+b)²-4b（a+b)+4b²

a²+2ab+b²-4ab-4b²+4b²a²-2ab+b²

a-b) Factorization of similar types of questions is also done in the same way First, all, ,