Factorization, 50 questions, several factorization problems

1.-2b

2.Question 2: No.

3.=1/11（a-b)(121-a^2-ab-b^2)

4.=-2/3a(x^2-6x+9)

2/3(x-3)^2

5.==125x(

6.=x^2-2x+1+4x^2-4+4x^2+8x+4

9x^2+6x+1

3x+1)^2

When x=1, the original value is 16

7.=-(x-y)^2+1

1-x+y)(1+x-y)

1）-4ab

2) The second did not understand.

3) Substitute (b-a) cube = (b-a) * (a square + a * b + b squared) to extract the common factor.

4) First, propose: -2 3a, and the rest in parentheses is (x squared - 6*x + 9) = (x-3) squared.

5) (6) All the original formula = 9x2 + 6x + 1 = (3x + 1) squared, when x = 1 is equal to 16

7)1+2xy-x2-y2=1-(x2-2xy+y2)=1-(x-y)2=(1+x-y)*(1-x+y)

（1）-4ab

2) The second did not understand.

3) Substitute (b-a) cube = (b-a) * (a square + a * b + b squared) to extract the common factor.

4) First, propose: -2 3a, and the rest in parentheses is (x squared - 6*x + 9) = (x-3) squared.

5) (6) All the original formula = 9x2 + 6x + 1 = (3x + 1) squared, when x = 1 is equal to 16

7)1+2xy-x2-y2=1-(x2-2xy+y2)=1-(x-y)2=(1+x-y)*(1-x+y)

1.Let t=x 2-3x

Therefore, the original formula =t*(t+2)+1=(t+1) 2 then substitutes t=x 2-3x back to get :

Original = (x 2-3x+1) 2

2.Let t=x 2+5x

Original formula = (t+6)(t+4)-120=t 2+10t-96=(t-6)(t+16).

Then substitute t=x 2+5x back to get:

Original = (x 2+5x-6)(x 2+5x+16) = (x-1)(x+6)(x 2+5x+16).

Criss-cross)