Can anyone help me write ten factorizations, speed, 50 points

1.（x+2)^2+6(x+2)+8

x+2)+2][(x+2)+4]

x+4)(x+6)

2.-x 4y+x 2y-xy(Note: x4 refers to the fourth power of x) xy(-x +xy-1).

x²-4y²)+x-2y)

x-2y)(x+2y)+(x-2y)

x-2y)(x+2y+1)

x+y)²-x+y)-6

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

i.e. a quarter).

a²+a²+a+

6.(x^2+3x-3)(x^2+3x+4)-8(x²+3x)²+x²+3x)-20

x²+3x+5)(x²+3x-4)

x²+3x+5)(x+4)(x-1)

x²+y²)-x²y²

x²+y²+xy)(x²+y²-xy)

a(a-1)]²32a(a-1)+60

a(a-1)-2][a(a-1)-30](a-2)(a+1)(a-6)(a+5)

2x-1)(2x+7)

10.Knowing a-2b=5,3a+4b=6, find the value of polynomial 3a2-2ab-8b2.

3a²-2ab-8b²

3a+4b)(a-2b)

(1) should be x 3

x^4+2x^3y-3x^2y^2-4xy^3-y^4

x^4+2x^2y+x^2y^2-4x^2y^2-4xy^3-y^4

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

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

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

If you defactor the factor within the range of real numbers, continue, otherwise it will end there.

(x+3/2*y)^2-5/4*y^2)((x-y/2)^2-3y^2/4)

x+3y/2-√5y/2)(x+3y/2+√5y/2)(x-y/2-√3y/2)(x-y/2+√3y/2)

2) 2x^4-15x^3+38x^2-39x+14

2(x^4-2x^3+x^2)-11x^3+36x^2-39x+14

2x^2(x-1)^2-11x(x^2-2x+1)+14(x^2-2x+1)

2x^2-11x+14)(x-1)^2

2x-7)(x-2)(x-1)^2

3) a^3b-ab^3+a^2+b^2+1

a^3b-a^2b^2+ab+a^2b^2-ab^3+b^2+a^2-ab+1

ab(a^2-ab+1)+b^2(a^2-ab+1)+1(a^2-ab+1)

ab+b^2+1)(a^2-ab+1)

4) a^4+2a^3+3a^2+2a+1

This is the perfect square formula.

a^2+a+1)^2

5) 5x^3+2xy^2+7y^3

It should be y 3, so that there is a solution of x=-y.

Original formula = 5x(x-y)(x+y)+7y2(x+y).

x+y)(5x^2-5xy+7y^2)

It is no longer possible to divide within the range of real numbers.

6) x^8+x^4+1

x^8+2x^4+1-x^4

x^4+1)^2-x^4

x^4+x^2+1)(x^4-x^2+1)

Same recipe. (x^4+2x^2+1-x^2)(x^4+2x^2+1-3x^2)

x^2-x+1)(x^2+x+1)(x^2-√3x+1)(x^2+√x+1)

7) x^6-y^6+x^4+x^2y^2+y^4

x^4-y^4)(x^4+x^2y^2+y^4)+(x^4+x^2y^2+y^4)

x^4-y^4+1)(x^4+x^2y^2+y^4)

x^4-y^4+1)((x^2+y^2)^2-x^2y^2)

x^4-y^4+1)(x^2+y^2+xy)(x^2+y^2-xy)

=（2x^4-15x^3+28x^2）+（10x^2-39x+14）

2x-7）（x-4）x^2+（2x-7）（5x-2）

2x-7）2x-7）（x^3-4x^2+5x-2）

a^3b-a^2b^2+a^2b^2+ab-ab-ab^3+a^2+b^2+1

a^3b-a^2b^2+ab+a^2b^2-ab^3+b^2+a^2-ab+1

ab(a^2-ab+1)+b^2(a^2-ab+1)+1(a^2-ab+1)

ab+b^2+1)(a^2-ab-1)

It seems that the question is wrong 5x 3+2xy 2+7y 3

5x^3-5xy^2）+（7xy^2+7y^3）

5x（x^2-y^2)+7y^2(x+y)

5x(x+y)(x-y)+7y^2(x+y)

x+y)(5x^2-5xy+7y^2)

6 solution: x 8+x 4+1

x^8+2x^4+1-x^4

x^4+1)^2-x^4

x^4+1+x^2)(x^4+1-x^2)

x^4+2x^2+1-x^2)(x^4-x^2+1)

(x^2+1)^2-x^2](x^4-x^2+1)

x^2+x+1)(x^2-x+1)(x^4-x^2+1)

7 formula: x 3-y 3=(x-y)(x 2+xy+y 2).

x^6-y^6+x^4+x^2y^2+y^4

x^4-y^4）（x^4+x^2y^2+y^4）+（x^4+x^2y^2+y^4）

x^4-y^4+1）（x^4+x^2y^2+y^4）

I'll think about it again.

Plagiarism is prohibited!

The first question is wrong, right?

..Tell you how...

The first step is to extract the common factor.

Part 2 You'll find that you can use perfect squares or squares.

Basically, these two cloths.,All the factorization in junior high school can be made.。。

Thank you..

1. (4x+3y)2=16x2+9y2 (

2.The square of (a-b) is equal to the square of (b-a).

Radio 4If (2a+3b)2=(2a-3b)2+( is true, then the formula in parentheses is [

5.If (x-y)2=0, the equation that follows is [

6.The following equation holds for [

a.(a-b)2=a2-ab+b2 b.(a+3b)2=a2+9b2

c.(a+b)(a-b)=(b+a)(-b+a) d. (x-9)(x+9)=x2-9

A1 ×

4. b5. a

6.c judgment 1 (4x+3y)2=16x2+9y2 (

2.The square of (a-b) is equal to the square of (b-a).

Radio 4If (2a+3b)2=(2a-3b)2+( is true, then the formula in parentheses is [

5.If (x-y)2=0, the equation that follows is [

6.The following equation holds for [

a.(a-b)2=a2-ab+b2 b.(a+3b)2=a2+9b2

c.(a+b)(a-b)=(b+a)(-b+a) d. (x-9)(x+9)=x2-9

A1 ×

4. b5. a

6. c