I have a low fever and go home for isolation, but I don't want to delay my class. Can you help me explain the factorization?

1 decompose thoroughly.

The final result is only parentheses.

3 In the final result, the first term coefficient of the polynomial is positive (for example, -3x 2+x =-x (3x- 1)).

[Edit this paragraph] Basic method

(1) common factor method

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 =-m (a-b-c);

a(x-y)+b(y-x)= a(x-y)-b(x-y)=(x-y)(a-b).

Note: Changing 2A 2+ 1/2 to 2 (A 2+ 1/4) is not a common factor.

⑵ Formula method

If the multiplication formula is reversed, some polynomials can be factorized. This method is called formula method.

Square difference formula: A2-B2 = (a+b) (a-b);

Complete square formula: a22ab+b 2 = (ab) 2;

Note: Polynomials that can be decomposed by the complete square formula must be trinomial, two of which can be written as the sum of squares of two numbers (or formulas), and the other is twice the product of these two numbers (or formulas).

Cubic sum formula: A3+B3 = (a+b) (A2-AB+B2);

Cubic difference formula: A3-B3 = (a-b) (A2+AB+B2);

Complete cubic formula: a 3 3a 2b+3ab 2 b 3 = (a b) 3.

Formula: A 3+B 3+C 3+3 ABC = (A+B+C) (A 2+B 2+C 2-AB-BC-CA)

For example: a 2+4ab+4b 2 = (a+2b) 2.

(3) Factorization skills

1. Factorization factor and algebraic expression multiplication are reciprocal deformation.

2. 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 the common factor;

(2) Take the common factor and determine another factor:

(1) The first step to find the common factor can be determined according to the method of determining 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.

[Edit this paragraph] The method used in the competition

⑶ Grouping decomposition method

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 put ax and ay in a group, bx and by in a group, and matched each other by multiplication and division and distribution, which immediately solved the difficulty.

Similarly, this problem can be done.

ax+ay+bx+by

=x(a+b)+y(a+b)

=(a+b)(x+y)

A few examples:

1.5ax+5bx+3ay+3by

Solution: =5x(a+b)+3y(a+b)

=(5x+3y)(a+b)

Note: 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-x^2+x- 1

Solution: = (x 3-x 2)+(x-1)

=x^2(x- 1)+ (x- 1)

=(x- 1)(x2+ 1)

Using dichotomy, the common factor method is used to propose x2, and then it is easy to solve it.

3.x2-x-y2-y

Solution: =(x2-y2)-(x+y)

=(x+y)(x-y)-(x+y)

=(x+y)(x-y- 1)

Use dichotomy, then use the formula a2-b2=(a+b)(a-b), and then skillfully solve it.

(4) Cross multiplication

There are two situations in this method.

①x？ Factorization of +(p+q)x+pq Formula

The characteristics of this kind of 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 trinomial factors with coefficients of 1: x? +(p+q)x+pq=(x+p)(x+q)。

②kx？ Factorization of +mx+n Formula

If there is k=ac, n=bd, and there is ad+bc=m, then kx? +mx+n=(ax+b)(cx+d)。

The chart is as follows:

×

c d

For example, because

1 -3

×

7 2

-3× 7 =-2 1, 1× 2 = 2, and 2-2 1=- 19,

So 7x? - 19x-6=(7x+2)(x-3)。

Formula of cross multiplication: head-tail decomposition, cross multiplication and summation.

5] 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)+ca(c-a)+BC(a+b)-ab(a+b)

=c(c-a)(b+a)+b(a+b)(c-a)

=(c+b)(c-a)(a+b)。

[6] Matching method

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: x? +3x-40

=x？ +3 times +2.25-42.25

=(x+ 1.5)？ -(6.5)?

=(x+8)(x-5)。

Application of factorial theorem.

For the polynomial f(x)=0, if f(a)=0, then f(x) must contain the factor x-a. 。

For example: f(x)=x? +5x+6, f(-2)=0, then it can be determined that x+2 is X? The coefficient of +5x+6. (actually, x? +5x+6=(x+2)(x+3)。 )

Note: 1. For polynomials whose coefficients are all integers, if x = q/p(p (when p and q are coprime integers) and the polynomial value is zero, then q is the divisor of constant terms 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..

Substitution 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, in the decomposition (x? +x+ 1)(x？ +x+2)- 12, which can make y=x? Then +x

The original formula =(y+ 1)(y+2)- 12.

=y？ +3y+2- 12=y？ +3y- 10

=(y+5)(y-2)

=(x？ +x+5)(x？ +x-2)

=(x？ +x+5)(x+2)(x- 1)。

You can also see the picture on the right.

(9) 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, when 2x 4+7x 3-2x 2- 13x+6 is decomposed, let 2x 4+7x 3-2x 2- 13x+6 = 0.

By comprehensive division, the roots of the equation are 0.5, -3, -2, 1.

So 2x4+7x3-2x2-13x+6 = (2x-1) (x+3) (x+2) (x-1).

⑽ image method

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.

Compared with ⑼ method, it can avoid the complexity of solving equations, but it is not accurate enough.

For example, when x 3+2x 2-5x-6 is decomposed, you can make y = x 3; +2x^2 5x 6。

Make an image, and the intersection with the X axis is -3,-1, 2.

Then x3+2x2-5x-6 = (x+1) (x+3) (x-2).

⑾ 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.

⑿ 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, when x 3+9x 2+23x+ 15 is decomposed, let x=2, then

x^3 +9x^2+23x+ 15=8+36+46+ 15= 105，

105 is decomposed into the product of three prime factors, namely 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,

Then x 3+9x 2+23x+ 15 may be equal to (x+ 1)(x+3)(x+5), which is true after verification.

[13] undetermined coefficient method

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).

=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd

Therefore, a+c=- 1,

ac+b+d=-5，

ad+bc=-6，

bd=-4。

The solutions are a= 1, b= 1, c=-2 and d =-4.

Then x4-x3-5x2-6x-4 = (x2+x+1) (x2-2x-4).

You can also see the picture on the right.

[14] Double cross multiplication.

Binary multiplication is a kind of factorization, similar to cross multiplication.

Double cross multiplication is a binary quadratic sextuple, and the initial formula is as follows:

ax^2+bxy+cy^2+dx+ey+f

X and y are unknowns and the rest are constants.

Use an example to illustrate how to use.

Example: the decomposition factor: x2+5xy+6y2+8x+18y+12.

Analysis: This is a quadratic six-term formula, and you can consider factorization by binary multiplication.

Solution: As shown below, just cross-connect all the numbers.

x 2y 2

① ② ③

x 3y 6

∴ Original formula = (x+2y+2) (x+3y+6).

Double cross multiplication includes the following steps:

(1) first decompose the quadratic term by cross multiplication, such as x2+5xy+6y 2 = (x+2y) (x+3y) in the cross multiplication diagram (1);

(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.

[Edit this paragraph] General steps of 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. "

Several examples

1. Decomposition factor (1+y) 2-2x2 (1+y 2)+x 4 (1-y) 2.

Solution: The original formula = (1+y) 2+2 (1+y) x2 (1-y)+x4 (1-y) 2-2 (1+y).

=[( 1+y)+x2( 1-y)]2-2( 1+y)x2( 1-y)-2x 2( 1+y)

=[( 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+ 1)(x+ 1-xy+y)(x- 1)(x- 1-xy-y)。

2. Verification: For any real number x, y, the value of the following formula will not be 33:

x^5+3x^4y-5x^3y^2- 15x^2y^3+4xy^4+ 12y^5.

Solution: The original formula = (x 5+3x 4y)-(5x 3y 2+15x 2y 3)+(4xy 4+12y 5).

=x^4(x+3y)-5x^2y^2(x+3y)+4y^4(x+3y)

=(x+3y)(x^4-5x^2y^2+4y^4)

=(x+3y)(x^2-4y^2)(x^2-y^2)

=(x+3y)(x+y)(x-y)(x+2y)(x-2y)。

(The factorization process can also be seen in the picture on the right. )

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.

∵a, B and C are three sides of △ABC,

∴a+2b+c>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).

=-6x^n×y^(n- 1)(2x^n×y-3x^2y^2+ 1).

[Edit this paragraph] Factorization four notes:

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". Here are some examples for your reference.

Example 1 factorization -A2-B2+2AB+4.

Solution:-A2-B2+2AB+4 =-(A2-2AB+B2-4) =-(A-B+2) (A-B-2)

The "negative" here means "minus 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 making mistakes such as-9x2+4y2 = (-3x) 2-(2y) 2 = (-3x+2y) (-3x-2y) = (3x-2y).

Example 2 Factorization-12x2nn+18xn+2yn+1-6xnyn-1. Solution:-12x2nyn+18xn+2yn+1-6xnyn-1=-6xnyn-1(2xny3xy2+1).

"Gong" here means "common factor". 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 during inspection:

When there is no explanation for real numbers, it is generally enough to explain only rational numbers.

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."

[Edit this paragraph] The application of factorization

1, applied to polynomial division.

2. It is suitable for finding the roots of higher-order equations.

3, applied to the operation of fractions