Ninth grade unary quadratic equation

Factorization 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. 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. Junior high school mathematics textbooks mainly introduce the methods of extracting common factors, using formulas and grouping.

(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 each polynomial factor 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-2x-x (Huai' an senior high school entrance examination in 2003)

x -2x -x=x(x -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 +4ab+4b (2003 Nantong senior high school entrance examination in 2003)

Solution: a +4ab+4b =(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 +5n-mn-5m

Solution: m +5n-mn-5m= m -5m -mn+5n.

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

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

=(m-5)(m-n)

4. Cross multiplication

For a polynomial in the form of mx +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-19x-6

Analysis: 1 -3

7 2

2-2 1=- 19

Solution: 7x-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 +3x-40

Solution x +3x-40=x +3x+() -() -40

=(x+ ) -()

=(x++)(x++)

=(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 -x -6x -x+2

Solution: 2x-x-6x-x+2 = 2 (x+1)-x (x+1)-6x.

=x [2(x + )-(x+ )-6

Let y=x+, x [2(x+)-(x+ )-6.

= x [2(y -2)-y-6]

= x (2y -y- 10)

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

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

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

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

8. Root method

Let the polynomial f(x)=0 and find its roots as x, x, x, …x, ... x, then the polynomial can be decomposed into f (x) = (x-x) (x-x)...(x-x).

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

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

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

Then 2x+7x-2x-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 points x, x, x, …x, ... x between the image of function and the X axis, then the polynomial can be decomposed into f (x) = f (x) = (x-x) (x-x) ... (ten

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

Solution: Let y= x +2x -5x-6.

Make it image, as shown on the right, and the intersection with the X axis is -3,-1, 2.

Then x+2x-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 +9x +23x+ 15.

Solution: let x=2, then x+9x+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+9x+23x+15 = (x+1) (x+3) (x+5).

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, decomposition factor x -x -5x -6x-4.

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

Solution: let x -x -5x -6x-4=(x +ax+b)(x +cx+d)

= x+(a+c)x+(AC+b+d)x+(ad+BC)x+BD

So the solution is

Then x-x-5x-6x-4 = (x+x+1) (x-2x-4)

References:

/question/3623 16 1 1 . html？ ansup 1

Interviewee: Essays-Scholars Level 2 7-22 15:46

Revision reply: Jane, the reply you want to revise is as follows: the integral rule is closed.

factoring

Factorization (factorization)