Ask a factoring 4, ask a factoring

x 2+6x+8)(x 2+14x+48)+12(x+4)(x+2)(x+6)(x+8)+12(x+4)(x+6)(x+2)(x+8)+12(x 2+10x+24)(x 2+10x+16)+12(x 2+10x) 2+40(x 2+10x)+16*24+12(x 2+10x) 2+40(x 2+10x)+396(x 2+10x+22)*(x 2+10x+18) if this continues.

x^2+10x+25-3)*(x^2+10x+25-7)((x+5)^2-3)*(x+5)^2-7)(x+5+sqrt(3))*x+5-sqrt(3))*x+5+sqrt(7))*x+5-sqrt(7))

sqrt(y) represents the y root number.

18x²-3x-1

18x²+3x-6x-1

3x(6x+1)-(6x+1)

3x-1) (6x+1), you can also use cross multiplication 9a -6ab+b -6a + 2b+1

3a)²-2(a)(b)+(b)²-6a+2b+1=(3a-b)²-2(3a-b)+1

(3a-b)-1]²

3a-b-1)²

The addition and subtraction method is to add (and subtract) one of the missing terms of one of the polynomials to achieve the factoring factor, which is canceled out at the time.

a^4-11a^4+1

1-10a^4

1²-(10a²)²

1-√10a²)(1+√10a²)

1+√10a²)[1-10^(1/4)a][1+10^(1/4)a]

If it is a 4-11a +1

a^4-2a²+1-9a²

a²)²2a²+1-9a²

a²-1)²-3a)²

a²-1-3a)(a²-1+3a)

x³+3x²-4

x³-x²+4x²-4x+4x-4

x²(x-1)+4x(x-1)+4(x-1)=(x-1)(x²+4x+4)

x-1)(x+2)²

With the exception of Article 2, the other three articles used the method of addition and subtraction.

1, original formula = (x+1)2-y2

x+1-y)2

2, ax2-4a=a(x2-4)=a(x-2)(x+2)x2-4x+4=(x-2)2, so the common factor is (x-2)3,16+8xy-16x2-y2=-(16x2+8xy+y2)+16=16-(4x+y)=(4-4x-y)(4+4x+y)

4. Select C: 3x-6=3(x-2), x2-2x=x(x-2).

The first one (x2+2x+1)—y2=(x+1)2-y2=(x+1+y)(x+1-y).

The second ax2-4a=a(x2-4)=a(x+2)(x-2) x2-4x+4=(x-2)(x-2), so the common factor is (x-2).

The third 16+8xy-16x2-y2=16-(16x2-8xy+y2)=4*4-(4x-y)(4x-y)=(4+4x-y)(4-4x+y).

The fourth option is c 3x-6=3(x-2) x2-2x=x(x-2) Okay, that's it.

1) Undefeated in this meeting.

2）25(a+b)^2-9(a-b)^2

5(a+b)]^2-[3(a-b)]^2[5(a+b)+3(a-b)][5(a+b)-3(a-b)][5a+5b+3a-3b][5a+5b-3a+3b][8a+2b][2a+8b]

2(4a+b)*2(a+4b)

4(4a+b)(a+4b)

3）-x^2-4y^2+4xy

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

x-2y)^2

4）(a-b)^2+2(a-b)+1

Slow withering a-b) disturbance (2 + a-b) + 1

2) [5(a b) 3(a b)][5(a b)+3(a laughing potato b)].

3）-（x＾2＋4y＾2－4xy）＝－x－2y）＾24）〔（a－b）＋1〕＾2

1) (Hand Wisdom Touches Eyes (

After the formula of the first question, (a-2) 2+(b-3) 2=0 is solved to obtain a=2, b=3

a+b=5 Question 3: Original formula = x4 + 2 x 2 + 1 + x3 + x = (x 2 + 1) 2 + x (x 2 + 1) = (x 2 + x + 1).

Question 4: Original formula = (n 2-4) 2+36 = (n-2) 2(n+2) 2+36

n=3 is a prime number.

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

Original = (4x+5y)(4x-5y).

Original = (a +4) (a -4).

Original = (x+y) -xy).

x+y)+xy][(x+y)-xy]

I fought for a long time.