Factorization from one to two will not do the solution of emergency 30

The transformation of a polynomial into the product of several simplest expressions is called factoring the polynomial (also known as factoring). For example: m -n = (m+n)(m-n), so you can do this for your question.

Let m+n=a, x+y=b, then.

Original = (a-b) 2+4ab

Collation yields =(a-b) 2

If you change it back to the setting, then =(m+n-x-y) 2=(m+n-x-y)(m+n-x-y).

In fact, factorization is easy to open and ruin, I have already understood the factorization in the second year of junior high school in the sixth grade, the key lies in a rule: for example.

Perfect Flat, Cubic Method, Cross Multiplication, Extracting Common Factors, and more. Take cross-multiplication, for example, x 2+6x+5; x^2+22x+40;

2x 2+7x+5, where x 2+6x+5 can be analyzed like this: 6x=5x+1x=4x+2x=3x+3x, the idea is that the coefficients in front of 5x+1x are multiplied exactly 5, and the clever staring is equal to the constant term 5, so there is (x+1)(x+5); The third formula 2x 2+7x+5 can be analyzed in this way: since the coefficient of the quadratic term is not one, the sub-equation of this type has (ax + constant 1) after the confession of filial piety

bx + constant 2

a b 2, where a and b can only be 1 and 2, constant 1 constant 2 5, where constants 1 and 2 can only be 1 and 5, substituting and removing the parentheses to get (2x+5) (x+1); Once you're proficient, you'll know it at a glance.

5x^3＋x^2－5x－1=(x+1)(x-1)(5x+1)

The minus sign means to turn b-a into a-b, which is b-a=-(-b)-=-=-(-b+a)=-(a-b).

6a(b-a)^5-2(a-b)^3=6a^5-2(a-b)^3=6a-2(a-b)^3

6a-2(a-b)^3=-6a(a-b)^5-2(a-b)^3=-2(a-b)^3*

This is the common factor].

2(a-b)^3*(3a²+3b²-6a²b+1)

6a(b-a)

5-2(a-b)31

Because a-b-(b-a).

So (a-b)3=

(b-a)]3

And because of (-a).

3=-a3, so (a-b)3=

b-a)3 substitutes the above formula into the original formula.

6a(b-a)

5-2(a-b)

3=6a(b-a)

5-2[-(b-a)3]

2=6a(b-a)

5+2(b-a)3=

2(b-a)

3[3a(b-a)

2+1] (omitted hereinafter).

Among the steps. 1 is the application of the opposite number, step.

2 is the application of the negative sign.

The main test points of this type of question are A-B

(b-a) and (-a).

3=-a3 application, as long as these two steps are done correctly, the remaining problems are not big. When encountering such a problem, first look at whether the inside of the parentheses is opposite to each other, if it is a single power or a double power outside the parentheses, if it is a double power, then according to.

a-b）2n=

b-a)2n, substitute it into the original formula, that is, replace b-a in parentheses with a-b.

The symbol outside the parentheses remains unchanged.

If it is a single power, it is based on.

a-b）2n+1

(b-a)2n+1 substitutes it into the original formula, replacing b-a in parentheses with a-b.

The symbol outside the parentheses is changed.

4m squared minus 9n squared minus 4m plus 1

4m(m-1)-(3n+1)(3n-1)2a squared plus 2b squared + 2c squared minus 2ab minus 2bc minus 2ac = (a-b) 2+(b-c) 2+(a-c) 2

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.

=2012²-2002×2012+2000×2002-2000×2012

2012 (2012-2002) + 2000 (2002-2012) = (2012-2002) (2012-2000) The factorization ends here.

1、(x-1/2)^2

2、（x+y-4z)(x+y-8z)

3. A n-1 (1 2a+3) (1 2a-3) 4.

5、-（x^n-2)(x+3）^2

6. (2x-1 2+3y)(2x-1 2-3y)7、-[1+(a+b) 2][-1+(a-b) 2]8, (a+1 8b 3n)(a-1 8b 3n) If there is a specific methodological problem, I really don't know it.

I didn't know how to go to high school, and I forgot all about what I didn't study in high school.

Remove parentheses Merge similar terms Use the formula method.

<> picture, Jian You, this stop and fight.