Factorization Factorization is one of the most important identical deformations in middle school mathematics. It is widely used in elementary mathematics and is a powerful tool for us to solve many mathematical problems. The factorization method is flexible and skillful. Learning these methods and skills is not only necessary to master the content of factorization, but also necessary to cultivate students' problem-solving skills and develop their thinking ability. They all have very unique functions. In junior high school mathematics textbooks, methods such as extracting common factors, using formulas, grouping decomposition and cross multiplication are mainly introduced. In the competition, there are also item addition method, undetermined coefficient method, binary multiplication method, rotational symmetry method and so on. (1) improve the common factor method (1) common factor: the factor of the common factor contained in each item is called ~. (2) Improve the common factor method: general. This common factor can be put outside parentheses, and the polynomial can be written in the form of factor product. This method of factorization is called the method of improving the common factor. AM+BM+CM = M (A+B+C) ③ Specific methods: When all the coefficients are integers, the coefficients of the common factor should take the greatest common divisor of all the coefficients; The letter takes the same letter for each item, and the index of each letter takes the lowest degree. If the first term of a polynomial is negative, a "-"sign is usually put forward. Make the coefficient of the first item in brackets positive. (2) Formula method ① Variance formula:. A 2-b 2 = (a+b) (a-b) ② Complete square formula: a 2 2ab+b 2 = (a b) 2 ※ Polynomials that can use the complete square formula to decompose factors must be trinomial. The other term is twice the product of these two numbers (or formulas). ③ Cubic sum formula: A 3+B 3 = (A+B) (A 2-AB+B 2). Cubic difference formula: a3-B3 = (a-b) (a2+ab+) [a (n-1)+a (n-2) b+...+b (n-2) a+b (n-1)] am+. The grouping decomposition method must have a clear purpose, that is, the common factors can be extracted directly after grouping, or formulas can be used. (4) Decomposition and supplement method: one term of a polynomial is decomposed or filled with two (or more) mutually opposite terms, so that the original formula is suitable for factorization by extracting common factors, and the formula method or group decomposition method is adopted; It should be noted that the deformation must be carried out under the principle of equality with the original polynomial. (5) Factorization of1× 2+(PQ) x+PQ type cross multiplication formula. The characteristics of this quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomials with the coefficient of1:x2+(p q) x+pq = (x+p) (x+q) 2kx2+MX+n. If it can be decomposed into K =, then kx2+MX+n = (AXB) (CX d) a \-/b. (3) If the above methods cannot be decomposed, you can try to decompose by grouping, splitting and adding items; ④ Factorization must be carried out until each polynomial factor can no longer be decomposed. (6) Applying factorial theorem: If f(a)=0, then f(x) must contain the factor (x-a). If f (x) = x 2+5x+6 and f(-2)=0, it can be determined that (x+2) is a factor of x 2+5x+6. Classic example: 1. Decomposition factor (1+y) 2-2x2 (1+y 2)+x 4 (1-y) 2 solution: original formula = (1+) x 2 (1-). ^2-2( 1+y)x^2( 1-y)-2x^2( 1+y^2)=[( 1+y)+x^2( 1-y)]^2-(2x)^2 =[( 1+y)+x^2( 1-y)+2x][( 1+y) +x^2( 1-y)-2x]=(x^2-x^2y+2x+y+ 1)(x^2-x^2y-2x+y+ 1)=[(x+ 1)^2-y(x^2- 1)][(x- 1)2-y(x ^ 2- 1)]=(x+ / Kloc-0/) (x+1-xy+) = (x+3y) (x4-5x2y2+4y4) = (x+3y) (x2-4y2) = (x+3y) (x+y). This deformation is called factorization of this polynomial. There are many methods of factorization, which can be summarized as follows: 1. If all terms of a polynomial contain a common factor, then this common factor can be put forward, so that the polynomial can be transformed into the product of two factors. Example 1, factorization factor x-2x-x (the senior high school entrance examination in Huai 'an in 2003) x -2x -x=x(x -2x- 1) 2. Applying formula method Because factorization factor has reciprocal relationship with algebraic expression multiplication, if the multiplication formula is reversed, it can be used to decompose some polynomials. Example 2, factorization factor a +4ab+4b (2003 Nantong senior high school entrance examination in 2003) solution: a +4ab+4b =(a+2b) 3. The polynomial am+an+bm+bn is decomposed by grouping factorization method. The first two terms can be divided into one group, and the common factor A is put forward, and then the last two terms can be divided into one group. Put forward the common factor B. Thus, we get the solution of (a+b)(m+n) Example 3 and the decomposition factor m +5n-mn-5m: m+5n-Mn-5m = m-5m-Mn+5n = (m-5m)+(-Mn+5n) = m (m-5). Example 5, factorization factor x +3x-40 = X+3x+()-()-40 = (X+)-() = (X++) (X+-) = (X+8) (X-5) 6, decomposition and addition. Example 6, decomposition factor bc(b+c)+ca(c-a)-ab(a+b) solution: BC (b+c)+ca (c-a)-ab (a+b) = BC (c-a+a+b)+ca (c-a). (c-a)(a+b) 7。 The substitution method can sometimes select the same part of a polynomial, replace it with another unknown, then decompose it and finally convert it back. Example 7, factorization factor 2x -x -6x -x+2 solution: 2x-x-6x-x+2 = 2 (x+1)-x (x+1)-6x = x [2 (x+)-(x+)-. = x(y+2)(2y-5) The root method makes the polynomial f(x)=0, and its roots are x, x, x, ... x, then the polynomial can be decomposed into f (x) = (x-x) (x-x)...(x-x) Example 8. Factorization factor 2x +7x -2x-13x+6 solution: let f (x) = 2x+7x-2x-6556. 1, then 2x+7x-2x-13x+6 = (2x-1) (x+3) (x-1) 9, image command y=f(x), function y=f(x). Factorizing x +2x -5x-6 solution: Let y= x +2x -5x-6 as its image, as shown in the right figure. Then x+2x-5x-6 = (x+1) (x+3) (x-2)10. The principal component method first selects a letter as the principal component, then arranges the items from high to low according to this letter number, and then factorizes them. Example 10, decomposition factor a (b-c)+b (c-a)+c (a-b) Analysis: A can be selected as the main element in this question. The solutions are arranged in descending order: a (b-c)+b (c-a)+c (a-b) = a (b-c)-a (b-c)+(b-c) = (b-c) [a-a (b+c)+BC]. The number p is decomposed into prime factors, and the prime factors are properly combined. The combined factors are written as the sum and difference of 2 or 10, and 2 or 10 is simplified to X, that is, the factorization formula. Example 1 1, factorization factor x +9x +23x+ 15 solution: let x=2, then x+9x+23x+15 = 8+36+46+15. That is, 105=3×5×7. Note that the coefficient of the highest term in the polynomial is 1, while 3, 5 and 7 are x+ 1, x+3 and x+5, respectively. When x=2, the value is x+9x+23x+ 15 =. Example 12. Factorization Factor x -x -5x -6x-4 Analysis: It is easy to know that this polynomial has no principal factor, so it can only be decomposed into two quadratic factors. Solution: Let X-X-5x-6x-4 = (X+AX+B) (X+CX+D) = X+(A+C) X+(AC+B+D) X+(AD+BC) X+BD, then the solution is X-X-5x-6x-BD.

/question/3623 16 1 1 . html？ ansup 1