Factorization. No, it won t. The second year of junior high school. To step

Solution: Original formula = x 2 + 4x-12

x+6) (x-2) (cross method).

Original formula = [3(2x+3)+2(2x-5)][3(2x+3)-2(2x-5)] a 2-b 2=(a+b)(a-b))).

10x-1)(2x+19)

Original = x 2-2y 2-7x 2-7xy-xy-6x 2-2y 2-8xy

2(3x^2+4xy+y^2)

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

The last item in this question should be "-xy", not "+xy.""；If it's "+xy."", this problem cannot be factored.

Factorization: The transformation of a polynomial into the product of several simplest formulas is called factoring.

1. Extract the common factor.

2. Square difference formula, perfect square formula.

3. Cross multiplication.

4. Group decomposition method.

5. Matching method.

I'll post it to you, 27 questions are a bit troublesome, and I reasoned it out in reverse.

Let's make two courses for you, both are similar, mention the common factor if you can, and disassemble it if you can, and there will be a result.

23，（19x－31）（13x－17）－（13x－17）（11x－23）

13x－17）[（19x－31）－（11x－23）]

13x 17) (8x 8), 11x 23) and (ax b) (8x c) equivalents, so a = 11, b = 23, c = 8

then, a + b + c = 11 + 23 + 8 = 42

24，a（a－1）－（a²－b）=2

a+b=2，（a－b）²=4，a²+b²－2ab=4，a²+b²）/2－ab=2。

26, a, b, c are the three sides of abc, and a +c = 2ab + 2bc 2b , a +b 2ab = 2bc b c , a b) = (b c), the above equation only holds when a = b=c, abc is equilateral.

28,2(x 1)(x 9)=2x 20+18, one item error;

2 (x 2) (x 4) = 2x 12+16, the constant term is wrong, so the original formula is 2x 12+18

Decomposition factor: 2x 12+18=2(x 3).

27. If a rational number a is equal to the square of another rational number b, then this rational number a is called a perfectly squared number.

Four consecutive natural numbers: (n 2), (n 1), n, (n + 1), then n (n + 1) (n 1) (n 2) + 1

n²－1）（n²－2n）+1

n^4－2x^3－n²+2n+1

n²－n －1） ²

So, the product of four consecutive natural numbers plus 1 is a perfectly square number.

The same conclusion would be drawn if (n 1), n, (n+1), (n+2) were used to denote four consecutive natural numbers. Details.

Solution: Since Xiaohong only misreads the coefficients of the terms once, the quadratic and constant terms of the decomposed polynomial are the same as those of the original polynomial, which are 2x*2 and 18

Since Xiaoliang only misreads the coefficients of the constant terms, the quadratic and primary terms of the decomposed polynomial are the same as the original polynomials of 2x*2 and -12x

Therefore, the original polynomial of 2x*2+(-12x)+ 18 can be decomposed into 2(x-3)(x-3) = 2(x-3)*2

2(x-1)(x-9).

2(x^2-10x+9)

2x^2-20x+18

The constant is 182 (x-2) (x-4).

2（x^2-6x+8）

2x^2-12x+16

The coefficient of the primary term is -12

Original = 2x 2-12x+18

2(x^2-6x+9)

2(x-3)^2

Restore the formulas of the two people back to each other, and take the constant term of the coefficient of the first term wrongly. If you make a mistake in taking a constant term, you can get the original formula by combining it here, and you can learn to solve it yourself.

It can be made by mentioning the common factor method and the formula method (square difference, perfect square, cross multiplication), and the process must be careful.

（1）（2x+y)²-7（2x+y）-18=[(2x+y)-9][(2x+y)+2]=(2x+y-9)(2x+y+2)

2) 4x to the fourth power - 13x to the power y + 9y to the fourth power = (4x -9y) (x -y).

2x+3y)(2x-3y)(x+y)(x-y)（3）（x²-4x）（x²-4x-2）-15=（x²-4x)²-2(x²-4x)-15=（x²-4x-5)(x²-4x+3)

x-5) (x+1) (x-3) (x-1) (4) ab to the third power + b + b-a + 1

a(b³-1)+(b²+b+1)

a(b-1)(b²+b+1)+(b²+b+1)=(b²+b+1)(ab-a+1)