Formula method, complete square with formula, square difference

Cross multiplication, (x+p) (x+q) = the square of x+(p+q) x+pq.

Methods of improving common factor

The grouping decomposition method is to separate several items so that the square difference is completely squared.

abstract

Definition: Convert a polynomial into the product of several algebraic expressions. This deformation is called factorization of this polynomial, and it is also 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 ability, attention and calculation ability, but also improve students' comprehensive analysis and problem-solving ability.

Factorization and algebraic expression multiplication are inverse deformations.

Factorization method

There is no universal method for factorization, and junior high school mathematics textbooks mainly introduce common factor method and formula method. There are split addition method, grouping decomposition method, cross multiplication method, undetermined coefficient method, double cross multiplication method, rotational symmetry method, residue theorem method and so on.

basic approach

(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.

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).

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.

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

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.x3-x2+x- 1

Solution: =(x3-x2)+(x- 1)

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

=(x- 1)(x2+ 1)

Using dichotomy, 2x is proposed by common factor method, and then it is easy to solve.

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) Methods of splitting and supplementing projects

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)。

5] 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 2+3x-40

=x^2+3x+2.25-42.25

=(x+ 1.5)^2-(6.5)^2

=(x+8)(x-5)。

[6] Cross multiplication.

There are two situations in this method.

① factorization of x2+(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 the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).

② Factorization of KX2+MX+N formula

If k=ac, n=bd, ad+bc=m, then kx 2+MX+n = (ax+b) (CX+d).

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

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)+x^2( 1-y)]^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+ 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)。

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).

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, 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, it is x 2+5x+6 = (x+2) (x+3). )

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, when decomposing (x 2+x+1) (x 2+x+2)-12, you can make y = x 2+x, then

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

=y^2+3y+2- 12=y^2+3y- 10

=(y+5)(y-2)

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x- 1).

(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, if you decompose x 3+2x 2-5x-6, 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).

[14] Double cross multiplication.

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

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, 6y2+18y+12 = (2y+2) (3y+6) in the cross multiplication diagram ②;

(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.

That's all I know. O(∩_∩)O laughed ~ ~ ~ ~ ~.