Definition: transforming a polynomial into the product of several simplest algebraic expressions is called factorization of this polynomial (also called factorization). Significance: It 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. Factorization is flexible and ingenious. Learning these methods and skills is not only necessary to master the content of factorization, but also plays a very unique role in cultivating students' problem-solving skills and developing their thinking ability. Learning it can not only review the four operations of algebraic expressions, but also lay a good foundation for learning scores; Learning it well can not only cultivate students' observation, thinking development and calculation ability, but also improve students' comprehensive analysis and problem-solving ability. Factorization and algebraic expression multiplication are opposite deformations.
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There is no universal method for factorization, and junior high school mathematics textbooks mainly introduce common factor method and formula method. There are division and addition and subtraction, grouping decomposition and cross multiplication, undetermined coefficient method, double cross multiplication, symmetric polynomial method, rotational symmetric polynomial method, remainder theorem method, radical method, method of substitution, long division, short division and division. (In fact, a classic example: 1. Decomposition factor (1+y) 2-2x2 (1+y 2)+x4 (1-y) 2 solution: original formula = (65438+) x 2 (1-y)-2x. kloc-0/+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( / kloc-0/-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+3y) (x 4-5x 2y 2+4y 4) = (x+3y) (x 2-4y 2) (x 2-y 2) = (x+3y) (x+y) (x+2y). ) Pay attention to the thorough decomposition of the three principles 1. The final result is only brackets. 3 The first term coefficient of the polynomial in the final result is positive (for example,-3x2+x = x (-3x2+1)). Induction: 1, the common factor method in the textbook of Shanghai Science 7th Edition. 2. Formula method. 3. Grouping decomposition method. 4. Number method. [x ^ 2+(a+b)x+ab =(x+a)(x+b)]5。 Combined decomposition method. 6. Cross multiplication. 7. Double cross multiplication. 8. Matching method. 9. Demolition method. 10, substitution method. 1 1, long division. 12, addition and subtraction. 13, root method. 14, mirror method. 15, principal component method. 16, undetermined coefficient method. 17, special value method. 18, factorial theorem method.
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Methods of improving common factor
The common factor of each term is called the common factor of each term of this polynomial. If every term of a polynomial has a common factor, we can put forward this common factor, so that the polynomial can be transformed into the product of two factors. This method of decomposing factors is called the improved common factor method. Specific methods: when all the coefficients are integers, the coefficients of the common factor formula should take the greatest common divisor of all the coefficients; The letter takes the same letter of each item, and the index of each letter takes the smallest number; Take the same polynomial with the lowest degree. If the first term of a polynomial is negative, a "-"sign is usually put forward to make the coefficient of the first term in brackets become positive. When the "-"sign is put forward, the terms of the polynomial should be changed. Formula: find the right common factor and clean it up 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) m; A (x-y)+b (y-x) = a (x-y)-b (x-y) = (a-b) (x-y). Note: replace 2a+ 1/2 with 2(a+ 1/4).
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: ax2+bx+c = a (x-(-b+√ (B2-4ac))/2a) cube and formula: a 3+b 3 = (a+b) (a2-). Cubic difference formula: A3-B3 = (a-b) (A2+AB+B2); Complete cubic formula: A3 32B+3B2 B3 = (A B) 3. Formula: A3+B3+C3-3abc = (A+B+C) (A2+B2+C2-AB-BC-CA) For example: A 2+4A B+. (3) Factorization skills 1. Factorization and algebraic expression multiplication are reciprocal deformation. 2. 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 product; ③ Each factor must be an algebraic expression, and the degree of each factor 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. 3. Basic steps of common factor method: (1) Find common factor; (2) Put forward the common factor and determine another factor: ① First, determine the coefficient, and then determine the letters according to the method of determining the common factor to find the common factor; (2) The second step is to put forward 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 remainder after improving the common factor. You can also use the common factor to remove each term of the original polynomial and find the remaining factors. (3) After extracting the common factor, the number of terms of another factor is the same as that of the original polynomial.
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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). 2.x 3-x2+x- 1 solution: = (x3-x2)+(x-1) = x2 (x-1)+(x-1) = (. 3.X 2-X-Y 2-Y solution: = (X2-Y2)-(X+Y) = (X+Y) = (X+Y) (X-Y-1.
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 factorize some quadratic trinomial terms with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q). ② if k=ac and n=bd, then kx 2+MX+n = (AX+B) (CX+D). The diagram is as follows: a b ╳ c d For example, because 1 -3 ╳ 7 2 -3×7=-2 1, 1× 2 =
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 (. +(BC-ab)(a+b)= c(c-a)(b+a)+b(a+b)(c-a)=(c+b)(c-a)(a+b)。
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)=x2+5x+6 and f(-2)=0, it can be determined that x+2 is a factor of x2+5x+6. (In fact, x2+5x+6 = (x+2) (x+3). Note: 1, for polynomials with integer coefficients, if x = q/p (when p and q are coprime integers) and the polynomial value is zero, then q is the divisor of constant term and p is the divisor of the highest degree; 2. For the polynomial f (a) = 0, where b is the coefficient of the highest degree and c is a constant term, then a is the divisor of C/B..
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. Correlation formula
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 = y2.
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. Let 2x 4+7x 3-2x 2- 13x+6 = 0, then we can know that the roots of this equation are 0.5, -3, -2, 1, so 2x 4+7x 3-2x 2- 13x.
method of images
Let y=f(x), make the image of function y=f(x), and find the intersection of function image and x axis, x 1, x2, x3, ... Xn, ... xn, then the polynomial can be factorized into f (x) = f (x) = (x-x/kloc-0. 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-0. 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 3+9x 2+23x+.
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. (2) according to the first coefficient of a letter (such as y) to score constant items. For example, 6y in Figure ② of cross multiplication? + 18y+ 12 =(2y+2)(3y+6)； (3) according to the first coefficient of another letter (such as x), such as cross plot (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 cannot be decomposed, you can try to decompose by grouping, splitting and adding items; (4) Factorization must be carried out until every polynomial factorization can no longer be decomposed. 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)+2x] [(1+y)+x 2 (1-y)-2x] = [(x+1) 2-y (x2-1)]] = (x+3y) (x 4-5x2y2+4y4) = (x+3y) (x 2-4y2) When y=0, the original formula = x 5 is not equal to 33; When y is not equal to 0, x+3y, x+y, x-y, x+2y and x-2y are different from each other, and 33 cannot be divided into products of more than four different factors, so the original proposition holds. 3. The three sides A, B and C of 3.△ ABC have the following relationship: -C 2+A 2+2AB-2BC = 0. 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。 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 (.
Four points for attention in editing this paragraph
The four points in factorization can be summarized as follows in four sentences: the first term is negative and often negative, each term is "male" and the first term is "male", a certain term is 1, and the brackets are divided into "bottom". The following examples can be used for reference: Example 1 Decomposition -A2-B2+2ab+4. Solution:-A2-B2+2AB+4 =-(A2-2AB+B2-4) =-(A-B+2) (A-B-2) The "negative" here means "negative sign". If the first term of a polynomial is negative, it is generally necessary to put forward a negative sign to make the coefficient of the first term in brackets positive. Prevent students from having two examples--9x2+4y2 = (-3x) 2-(2y) 2 = (-3x+2y) = (3x+2y)-12x2nyn+ 18xn. Solution:-12x2nyn+18xn+2yn+1-6xnyn-1=-6xnyn-1(2xny-3xy22+1) where If each term of a polynomial contains a common factor, first extract this common factor, and then further decompose this factor; "1" here means that when the whole term of the polynomial is a common factor, put forward this common factor first, and don't miss the 1 in brackets. Factorization must be carried out until each polynomial factor can no longer be decomposed. That is, break it down to the end, not give up halfway. The common factor contained in it should be "clean" at one time, leaving no "tail", and the polynomial in each bracket can not be decomposed again. Prevent students from making mistakes such as 4x4Y2-5x2Y2-9Y2 = Y2 (4x4-5x2-9) = Y2 (x2+1) (4x2-9). Attention should be paid to the exam: when there is no explanation for real numbers, it is generally enough to explain only rational numbers. If you have an explanation of real numbers, you usually turn to integers! From this point of view, the four attentions in factorization run through the four basic methods of factorization, which are in the same strain as the four steps of factorization or the four sentences of general thinking order: "First, see if there is a common factor, then see if a formula can be established, try cross multiplication, and group decomposition should be appropriate."
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1, applied to polynomial division. 2. It is applied to finding the roots of higher-order equations. 3. By the way, Mei Sen composite decomposition has made some insignificant progress: 1, p=4r+3. If 8r+7 is also a prime number, then: (8r+7) | (2 p- 1). That is, (2p+1) | (2p-1); For example: 23 | (211-1); ; 1 1=4×2+3; 47|(2^23- 1); ; 23=4×5+3; 167|(2^83- 1); ,,,.83=4×20+3; . . . . 2, p = 2n× 3 2+1,then (6p+ 1) | (2 p- 1), for example: 223 | (2 37-1); ; 37=2×2×3×3+ 1; 439|(2^73- 1); 73=2×2×2×3×3+ 1; 3463|(2^577- 1); ; 577=2×2×2×2×2×2×3×3+ 1; ,,,。 3, p = 2 n× 3 m× 5 s- 1, then (8p+1) | (2p-1); . For example; 233|(2^29- 1); 29=2×3×5- 1; ; 1433|(2^ 179- 1); 179=2×2×3×3×5- 1; 19 13|(2^239- 1); 239=2×2×2×2×3×5- 1。