It's simple. Listening carefully in class is mainly about junior high school mathematics, and "factorization" is the key chapter. This chapter plays a vital role in future mathematics learning. Four methods are mainly explained in the textbook, among which common factor method, formula method and cross multiplication are all introduced in detail, so I won't study them here. The following are mainly summarized as follows.

First, several common methods of grouping factorization.

1. Decomposition by common factor

Example 1 factorization factor 7x2-3y+xy+21x.

Analysis: items 1 and 4 contain common factor 7x, items 2 and 3 contain common factor y, and there is common factor (x-3) after grouping.

Solution: The original formula = (7x2-21x)+(xy-3y) = 7x (x-3)+y (x-3) = (x-3) (7x+y).

2. Decomposition by coefficient

Example 2 Factorization factor x3+3x2+3x+9.

Analysis: the coefficient ratio of item 1, item 2, item 3 and item 4 is 1: 3, and they are grouped by coefficients.

Solutions; The original formula = (x3+3x2)+(3x+9) = x2 (x+3)+3 (x+3) = (x+3) (x2+3).

Group by number

Example 3 factorization factor m2+2m n-3m-3n+N2.

Analysis: 1, 2 and 5 are quadratic terms, and 3 and 4 are linear terms. After grouping by time, the common factor can be extracted by formula.

Solution: The original formula = (m2+2mn+N2)+(-3m-3n) = (m+n) 2-3 (m+n) = (m+n) (m+n-3).

4. Group by multiplication formula

Analysis: The combination of item 1, item 3 and item 4 happens to be a complete square formula, and the square difference formula of the second item is used after grouping.

5. Expand and then group

Example 5 factory AB (C2+D2)+CD (A2+B2).

Analysis: Expand parentheses before regrouping.

Solution: The original formula = ABC 2+Abd 2+CD A2 = (ABC 2+CD A2)+(CD B2+A BD2) = AC (BC+AD)+BD (BC+AD) = (BC+AD) (AC+BD).

6. Break down the items and group them again.

Example 6 X2 factory -Y2+4x+2y+3.

Analysis: group constants and use multiplication formula.

Solution: The original formula = x2-y2+4x+2y+4-1= (x2+4x+4)+(-y2+2y-1) = (x+2) 2-(y-1) 2 = (.

7. Add items before grouping

Example 7 Factorization factor x4+4.

Analysis: the number of items in the above formula is small and it is difficult to decompose. You can add items before grouping.

Solution: The original formula = x4+4x2-4x2+4 = (x2+2) 2-(2x) 2 = (x2+2x+2) (x2-2x+2).

Second, factorization with method of substitution.

Factorization with the idea of adding auxiliary elements means that the original formula is complex and difficult to decompose directly, and it can be completed step by step by simplifying the complex.

Example 8 Factorization factor (x2+3x-2) (x2+3x+4)- 16.

Analysis: If y=x2+3x, the original formula is transformed into (y-2)(y+4)- 16, and then it is easy to decompose.

Solution: Let y=x2+3x, then

The original formula = (y-2) (y+4)-16 = y2+2y-24 = (y+6) (y-4).

So the original formula = (x2+3x+6) (x2+3x-4) = (x-1) (x+4) (x2+3x+6).

Third, factorization by root method.

Example 9 Decomposition factor X2+7x+2.

Analysis: x2+7x+2 cannot be completed by the above methods, but it can still be decomposed. You can first find the root of the equation corresponding to the polynomial, and then decompose it.

Fourthly, the factor is decomposed by undetermined coefficient method.

Example 10 factorization factor x2+6x- 16.

Analysis: If it can be decomposed, it should be decomposed into the product form of two linear terms, namely (x+b 1)(x+b2), and expanded into.

X2+(b 1+b2)x X b 1 b2 contrast x2+6x- 16

You can get B 1+b2=6, b 1 b2 =- 16, B 1, and you can decompose b2.

Solution: Let x2+6x-16 = (x+b1) (x+b2).

Then x2+6x-16 = x2+(b1+B2) x+b1B2.

∴x2+6x- 16=(x-2)(x+8).