Formula: find the right common factor and extract it all at once; The whole family moved out and left 1 to look after the house; The negative sign should be changed, and the deformation depends on parity. For example:-am+BM+cm =-(a-b-c) ma (x-y)+b (y-x) = a (x-y)-b (x-y) = (a-b) (x-y). Note: Replacing 2a+ 1/2 with 2(a+ 1/4) is not a common factor.

Formula method

If the multiplication formula is reversed, some polynomials can be factorized. This method is called formula method. Square difference formula: (A+B) (A-B) = A 2-B 2, followed by A 2-B 2 = (A+B) (A-B) Complete square formula: (A+B) 2 = A 2+2AB+B 2, followed by a. 2 Note. Two formulas: ax 2+bx+c = a (x-(-b+√ (b 2-4ac))/2a) cube and formula: a 3+b 3 = (a+). Cubic difference formula: a 3-b 3 = (a-b) (a 2+ab+b 2) Complete cubic formula: a 3 3a 2b+3ab 2 b 3 = (a b) 3. Formula: a 3+b 3.

Factorization technique

1. Master factorization skills: ① The left side of the equation must be a polynomial; ② The result of factorization must be expressed in the form of a product; ③ Each factorization must be an algebraic expression, and the number of factorizations must be lower than that of the original polynomial; ④ Factorization factors must be decomposed until each polynomial factor can no longer be decomposed. Note: Find the common factor before decomposing the factor, and consider the coefficient and factor before determining the common factor. 2. The basic steps of extracting the common factor method: (1) Find the common factor (2) Extract the common factor and determine another factor: ① The first step of finding the common factor is to determine the coefficient first and then the letters; ② The second step is to extract the common factor and determine another factor. Pay attention to determine another factor. You can divide the original polynomial by the common factor, and the quotient is the one left after extracting the common factor.

Edit the methods used in this competition.

Group multiplication

Group decomposition is a simple method to solve equations. Let's learn this knowledge. There are four or more terms in an equation that can be grouped, and there are two forms of general grouping decomposition: dichotomy and trisection. For example: AX+AY+bx+BY = A (X+Y)+B (X+Y) = (A+B) (X+Y) We divide AX and AY into a group, BX and BY into a group, and use the law of multiplication and distribution to match each other, which solves the difficulty at once. Similarly, this problem can be done. Ax+ay+bx+by = x (a+b)+y (a+b) = (a+b) (x+y) Examples: 1.5ax+5bx+3ay+3by Solution: = 5x (a+b)+3y (a).

=(5x+3y)(a+b) Description: Different coefficients can be decomposed into groups. As mentioned above, 5ax and 5bx are regarded as a whole, and 3ay and 3by are regarded as a whole, which can be easily solved by using the multiplication and distribution law. 2.x 3-x2+x- 1 solution: = (x3-x2)+(x-1) = x2 (x-1)+(x-1) = (. 3.X 2-X-Y 2-Y correlation formula

Solution: = (x 2-y 2)-(x+y) = (x+y) (x-y)-(x+y) = (x+y) (x-y-1) Use dichotomy, and then use the formula A 2-b.

Cross multiplication

There are two situations in this method. Factorization of (1)x2+(P+Q)X+PQ formula The characteristics of this kind of quadratic trinomial 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 trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q). For example: x2-2x-8 = (x-4) (x+2) ② kx. Then kx 2+MX+n = (ax+c) (bx+d). The diagram is as follows: a ╲╱ c b ╲ d For example, in (7x+2)(x-3), a = 1 b = 7 c =

Method of splitting and adding items

This method refers to disassembling one term of a polynomial or filling two (or more) terms that are opposite to each other, so that the original formula is suitable for decomposition by improving the common factor method, using the formula method or grouping decomposition method. It should be noted that the deformation must be carried out under the principle of equality with the original polynomial. For example: BC (B+C)+CA (C-A)-AB (A+B) = BC (C-A+A+B)+CA (C-A)-AB (A+B) = BC (C-A)+BC (A+B)+CA (.

Method of completing a square

For some polynomials that cannot be formulated, they can be fitted in a completely flat way, and then factorized by the square difference formula. This method is called matching method. It belongs to the special case of the method of splitting items and supplementing items. It should also be noted that the deformation must be carried out under the principle of equality with the original polynomial. For example: x2+3x-40 = x2+3x+2.25-42.25 = (x+1.5) 2-(6.5) 2 = (x+8) (x-5).

Using factorial theorem

For the polynomial f(x)=0, if f(a)=0, then f(x) must contain the factor x-a. For example, 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. (Actually, x 2+5x+6 = (x+2) (x+3). ) Note: 1. For a polynomial whose coefficients are all integers, if x = q/p (when p and q are coprime integers) and the polynomial value is zero, then q is a divisor of constant terms, and the coefficient of the highest term of p is about 2.

Alternative method

Sometimes in factorization, you can choose the same part of the polynomial, replace it with another unknown, then factorize it and finally convert it back. This method is called substitution method. Note: don't forget to return the RMB after exchange. For example, when decomposing (x2+x+1) (x2+x+2)-12, y = x 2+x, then the original formula = (y+1) (y+2)-12 = y.

Root-seeking method

Let the polynomial f(x)=0 and find its roots as x 1, x2, x3, ... xn, then the polynomial can be decomposed into f (x) = (x-x1) (x-x2) (x-x3) ... (x-xn). For example, in the decomposition of 2x. Then by comprehensive division, we can know that the roots of this equation are 0.5, -3, -2, 1. So 2x4+7x3-2x2-13x+6 = (2x-1) (x+3) (x+2).

method of images

Let y=f(x) be the image of function y=f(x), and find the intersection of function image and x axis, X 1, X2, X3...Xn, then the polynomial can be factorized into f (x) = f (x) = (x-x1) (x For example, when x 3+2x 2-5x-6 is decomposed, you can make y = x 3; +2x 2-5x-6。 Make an image, the intersection with the X axis is -3,-1, and 2 is x3+2x2-5x-6 = (x+1) (x+3) (x-2).

Principal component method

First, choose a letter as the main element, then arrange the items from high to low according to the number of letters, and then factorize them.

Special value method

Substitute 2 or 10 into x, find the number p, decompose the number p into prime factors, properly combine the prime factors, write the combined factors as the sum and difference of 2 or 10, and simplify 2 or 10 into x, thus obtaining factorization. For example, in the decomposition of x 3+9x 2+23x+15, let x=2, then x 3+9x 2+23x+15 = 8+36+46+15 =/kloc. When x=2, x 3+9x 2+23x+ 15 may be equal to (x+ 1) (x+3) (x+).

method of undetermined coefficients

Firstly, the form of factorization factor is judged, then the letter coefficient of the corresponding algebraic expression is set, and the letter coefficient is calculated, thus decomposing polynomial factor. For example, when x 4-x 3-5x 2-6x-4 is decomposed, the analysis shows that this polynomial has no primary factor, so it can only be decomposed into two quadratic factors. So let x4-x3-5x2-6x-4 = (x2+ax+b) (x2+CX+d) related formula.

= x 4+(a+c) x 3+(AC+b+d) x 2+(ad+BC) x+BD, so that a+c=- 1, ac+b+d=-5, ad+bc=-6, BD =-6

Double cross multiplication

Binary multiplication is a kind of factorization, similar to cross multiplication. Double cross multiplication is binary quadratic sixth power. The initial formula is as follows: ax 2+bxy+cy 2+dx+ey+FX, where y is unknown and the rest are constants. And illustrate how to use it with examples. Example: factorization: x2+5xy+6y2+8x+18y+12. Analysis: This is a quadratic sextuple, so we can consider factorization by binary multiplication. Solution: As shown in the figure below, cross connect all the numbers to get the original formula = (x+2y+2) (x+3y+6). The steps of the binary multiplication are as follows: ① First, the quadratic term is decomposed by the binary multiplication, such as x2+5xy+6y 2 = (x+22) in the graph of the binary multiplication. For example, 6y in Figure ② of cross multiplication? +18y+12 = (2y+2) (3y+6) ③ Check according to the first coefficient of another letter (such as X), as shown in the cross multiplication diagram ③. This step cannot be omitted, otherwise it is easy to make mistakes. Decomposition of quadratic polynomial by the relationship between roots and coefficients: for quadratic polynomial AX 2+BX+C (A ≠ 0) AX 2+BX+C = A [X 2+(B/A) X+(C/A) X]. When △ = B 2-4ac.

General steps for editing this polynomial factorization.

(1) If the polynomial term has a common factor, then the common factor should be raised first; (2) If there is no common factor, try to decompose it by formula and cross multiplication; (3) If the above methods can't be decomposed, you can try to decompose (4) factorization factors by grouping, splitting and supplementing, which must be carried out until every polynomial factor can't be decomposed any more. It can also be summarized in one sentence: "First, look at whether there is a common factor, and then look at whether there is a formula. Try cross multiplication, and group decomposition should be appropriate. " A few examples 1. Decomposition factor (1+y) 2-2x2 (1+y 2)+x 4 (1-y) 2. Solution: The original formula = (1+) x2 (1-y)-2x2 (1+y) (complement) = [(1+y)+x2 (1-y)]. Kloc-0/+y)+x 2 (1-y)-2x] = [(x+1) 2-y (x2-1)] [(x-1) 2-2 solution. (x 4-5x 2y 2+4y 4) = (x+3y) (x 2-y 2) = (x+3y) When y is not equal to 0, x+3y, x-y, x+2y and x-2y are mutual. Prove that this triangle is an isosceles triangle. Analysis: This question is essentially factorizing the polynomial on the left side of the relation equal sign. Prove: ∫-C2+a2+2ab-2bc = 0, ∴ (a+c) (a-c)+2b (a-c) = 0. ∴ (a-c) (a+2b+c) = 0。 0.∴ A-C = 0, that is, a=c and △ABC is an isosceles triangle. 4. Factorization-12x2n× y n+18x (n+2) y (n+1)-6xn× y (n-1). Solution:-12x2n× y n+18x (n+2) y (n+1)-6xn× y (n-1) =-6xn× y (.