factoring

Factorization refers to the decomposition of a polynomial into the product of several algebraic expressions. 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. 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, the methods of extracting common factors, using formulas, grouping decomposition and cross multiplication are mainly introduced. There are also sub-item addition, undetermined coefficient method, double cross multiplication, rotational symmetry method and so on.

(1) common factor method

(1) common factor: the common factor of each term is called ~ of this polynomial term.

② Extraction method of common factor: Generally speaking, if every term of a polynomial has a common factor, you can put this common factor outside brackets and write the polynomial as a factor product. This method of decomposing factors is called extracting common factors.

am+bm+cm=m(a+b+c)

③ 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 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 to make the coefficient of the first term in brackets positive.

⑵ Use the 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 be decomposed by the complete square formula must be trinomials, 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). ※ 。

③ Cubic sum formula: A 3+B 3 = (A+B) (A 2-AB+B 2).

Cubic difference formula: a 3-b 3 = (a-b) (a 2+ab+b 2).

④ Complete cubic formula: a 3 3a 2b+3ab 2 b 3 = (a b) 3.

⑤a^n-b^n=(a-b)[a^(n- 1)+a^(n-2)b+……+b^(n-2)a+b^(n- 1)]

A m+b m = (a+b) [a (m-1)-a (m-2) b+...-b (m-2) a+b (m-1)] (m is an odd number).

⑶ Grouping decomposition method

Grouping decomposition: a method of grouping polynomials and then decomposing factors.

The grouping decomposition method must have a clear purpose, that is, the common factor can be directly extracted or the formula can be used after grouping.

(4) Methods of splitting and supplementing projects

Decomposition and supplement method: one term of polynomial is decomposed or filled with two terms (or several terms) which are opposite to each other, so that the original formula is applicable to common factor method, formula method or group decomposition method; It should be noted that the deformation must be carried out under the principle of equality with the original polynomial.

5] Cross multiplication.

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 of 1: x 2+(p q) x+PQ = (x+p) (x+q).

② Factorization of KX2+MX+N formula

If it can be decomposed into k = AC, n = BD and AD+BC = M, then

kx^2+mx+n=(ax b)(cx d)

a \ - /b ac=k bd=n

c / - \d ad+bc=m

General steps of polynomial decomposition. ※:

(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.

(6) Applying factorial theorem: If f(a)=0, then f(x) must contain factorial (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: 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. Prove that for any number x, y, the value of the following formula will not be 33.

x^5+3x^4y-5x^3y^2+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.

Twelve factorization methods

Transforming a polynomial into the product of several algebraic expressions is called factorization of this polynomial. There are many methods of factorization, which are summarized as follows:

1, Public Welfare Law

If every term of a polynomial contains 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 3-2x2-x (the senior high school entrance examination in Huai 'an in 2003)

x^3 -2x^2 -x=x(x^2 -2x- 1)

2. Application of formula method

Because factorization and algebraic expression multiplication have reciprocal relationship, if the multiplication formula is reversed, it can be used to decompose some polynomials.

Example 2, factorization factor A 2+4A B+4B 2 (Nantong City, 2003)

Solution: A 2+4A B+4B 2 = (A+2B)

3. Grouping decomposition method

Factorizing the polynomial am+an+bm+bn, we can first divide the first two terms into a group and propose the common factor A, then divide the last two terms into a group and propose the common factor B, so as to get a(m+n)+b(m+n), and we can also propose the common factor M+N, so as to get (a+b) (m+).

Example 3: Decomposition Factor M 2+5N-Mn-5M

Solution: m 2+5 n-Mn-5 m = m 2-5 m-Mn+5 n.

= (m^2 -5m )+(-mn+5n)

=m(m-5)-n(m-5)

=(m-5)(m-n)

4. Cross multiplication

For a polynomial in the form of MX 2+PX+Q, if a×b=m, c×d=q and ac+bd=p, the polynomial can be factorized into (ax+d)(bx+c).

Example 4, factorization factor 7x 2- 19x-6.

Analysis:

1 -3

7 2

2-2 1=- 19

Solution: 7x 2- 19x-6 = (7x+2) (x-3)

5. Matching method

For those polynomials that cannot be formulated, some can use it to make a completely flat way, and then factorize it with the square difference formula.

Example 5, factorization factor x 2+3x-40

Solution x 2+3x-40

=x^2+3x+2.25-42.25

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

=(x+8)(x-5)

6. Removal and addition methods

Polynomials can be divided into several parts and then factorized.

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)-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)

7. Alternative methods

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.

Example 7, factorization factor 2x^4 -x^3 -6x^2 -x+2 2-x+2-x+22-x+2.

8. Root method

Let the polynomial f(x)=0 and find its roots as x 1, x2, x3, …xn, ... xn, then the polynomial can be factorized into f (x) = (x-x1) (x-x2) (x-x3) ...

Example 8, factorization factor 2x 4+7x 3-2x 2- 13x+6.

Solution: Let f (x) = 2x4+7x3-2x2-13x+6 = 0.

According to the comprehensive division, the roots of f(x)=0 are 1/2, -3, -2, 1.

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

9. Mirror 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.

Example 9, Factorization X 3+2x 2-5x-6

Solution: Let y y y= x^3 +2x^2 -5x-6 5x-6-65x-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).

10, 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.

Example 10, factorization factor a (b-c)+b (c-a)+c (a-b)

Analysis: this question can choose a as the main element, arranged from high to low.

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

=(b-c) [a -a(b+c)+bc]

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

1 1, using the 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.

Example 1 1, factorization factor X 3+9X 2+23x+ 15.

Solution: let x=2, then x3+9x2+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 =(x+ 1)(x+3)(x+5), which is true after verification.

12, 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.

Example 12, factorization factor x 4-x 4-x 3-5x 2-6x-4-4.

Analysis: It is easy to know that this polynomial has no first factor, so it can only be decomposed into two quadratic factors.

Solution: 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

So the solution is

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

"Four Notices" of Factorization for Beginners

Factorization first appeared in the second volume of Algebra, a three-year junior high school textbook for nine-year compulsory education, and was taught in the second semester of junior high school, but its content penetrated into the whole middle school mathematics textbook. Learning it can not only review the four operations of algebraic expressions in senior one, but also lay a good foundation for the next chapter of this book. Learning it well can not only cultivate students' observation ability, attention and calculation ability, but also improve students' comprehensive analysis and problem-solving ability. Four of them must be highly valued by teachers and students.

Four points in factorization are scattered on page 5 and 15 of the textbook, which can be summarized as follows in four sentences: the first term is often negative, the first term is "public", the first term is "public" and the last term is 1. Here are a few 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)?

For example, the three sides A, B and C of △ ABC in Example 2 have the following relationship: -C2+A2+2ab-2bc = 0, which proves 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.

Example 3 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. Prevent students from appearing such as 6p (x-1) 3-8p2 (x-1) 2+2p (1-x) 2 = 2p (x-1) 2 [3 (x-1)

Example 4 Factorizing X4-5x2-6 in the real number range.

Solution: x4-5x2-6 = (x2+1) (x2-6) = (x2+1) (x+6) (x-6).

The "bottom" here refers to factor decomposition, which 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).