Use parentheses when grouping: there is a "+"sign in front of parentheses, and all items in parentheses remain unchanged; Parentheses are preceded by a "-",and everything in parentheses changes sign.
When there are many terms in polynomials, polynomials can be grouped reasonably to achieve the purpose of smooth decomposition. Of course, other sub-methods can also be integrated, and the grouping method is not necessarily unique.
Example 1 decomposition factor: x15+m12+M9+M6+m3+1.
The formula = (x15+m12)+(M9+M6)+(m3+1).
= m 12(m3+ 1)+M6(m3+ 1)+(m3+ 1)
=(m3+ 1)(m 12+M6 ++ 1)
=(m3+ 1)[(M6+ 1)2-M6]
=(m+ 1)(m2-m+ 1)(M6+ 1+m3)(M6+ 1-m3)
Example 2 Factorization: x4+5x3+ 15x-9
Analysis can be grouped according to coefficient characteristics.
The solution formula =(x4-9)+5x3+ 15x.
=(x2+3)(x2-3)+5x(x2+3)
=(x2+3)(x2+5x-3)
Attachment: For reference only.
Lesson 4 Factorization
[knowledge point]
The definition of factorization, the general steps of extracting common factors, the application of formula method, grouping decomposition method, factorization of quadratic trinomial (cross multiplication, finding roots) and factorization.
[Outline requirements]
Understand the concept of factorization, master the factorization methods such as extracting common factors, formulas and grouping decomposition, and master the method of decomposing quadratic binomial by using the root formula of quadratic equation, so that simple polynomials can be factorized.
[Examination Highlights and Frequently Asked Questions]
Examining the factorization ability, factorization often appears in the senior high school entrance examination questions. This paper focuses on the fractional extraction of common factors, the application formula method, the grouping decomposition method and their comprehensive applications. The types of exercises are mostly fill-in-the-blank questions, as well as multiple-choice questions and solutions.
Decompose knowledge points
Polynomial factorization is to turn a polynomial into the product of several algebraic expressions. Factorization should be carried out until each factor can no longer be decomposed. The common methods of factorization are:
(1) common factor method
Such as polynomials
Where m is called the common factor of this polynomial, and m can be a monomial or a polynomial.
(2) using the formula method, that is, using
Write the results.
(3) Cross multiplication
For a quadratic trinomial with a quadratic term coefficient of L, find A and B that satisfy ab=q and a+b=p, and if so, find the general quadratic trinomial.
A 1a2=a, c 1c2 = C, A1= A1,a2, C 1, C2 of B, if any, then
(4) Grouping decomposition method: properly group the projects, so that they can be decomposed by grouping first and then by grouping.
Use parentheses when grouping: there is a "+"sign in front of parentheses, and all items in parentheses remain unchanged; Parentheses are preceded by a "-",and everything in parentheses changes sign.
(5) Root formula method: If there are two roots X 1, X2, then
Examination questions:
1. Which of the following factorizations is correct ()?
(A) 1- 14 x2 = 14(x+2)(x-2)(B)4x–2 x2–2 =-2(x- 1)2
(C)(x-y)3-(y-x)=(x-y)(x-y+ 1)(x-y- 1)
(D)x2–y2–x+y =(x+y)(x–y– 1)
2. The following equation (1) a2-b2 = (a+b) (a–b), (2) x2–3x+2 = x (x–3)+2.
(3) 1 x2–y2- 1(x+y)(x–y)，(4 )x2 + 1 x2 -2-( x - 1x )2
The number of factorizations from left to right is ()
1 (B) 2 (C) 3 (D) 4。
3. If x2+MX+25 is a completely flat mode, the value of m is ().
(A) 20 (B) 10 (C)
4. If x2+MX+N can be decomposed into (x+2) (x–5), then m=, N =；;
5. If the quadratic trinomial 2x2+x+5m can be factorized in the real number range, then m =；;
6. If a factor of x2+kx-6 is (x-2), then the value of k is;
7. Break down the following factors:
( 1)a3-a2-2a(2)4 m2-9 N2-4m+ 1
(3)3a2+bc-3ac-ab (4)9-x2+2xy-y2
8. Factorization in the range of real numbers:
( 1)2 x2-3x- 1(2)-2 x2+5xy+2 y2
Examination center training:
1. Break down the following factors:
( 1). 10a(x-y)2-5b(y-x)(2)。 An+1-4an+4an- 1
(3).x3(2x-y)-2x+y (4)。 x(6x- 1)- 1
(5)2ax- 10ay+5by+6x(6)
*(7).a4+4 (8)。 (x2+x)(x2+x-3)+2
(9).x5y-9xy5 ( 10)。 -4x2+3xy+2y2
( 1 1). 4a-a5( 12). 2x 2-4x+ 1
( 13). 4 y2+4y-5( 14)3 x2-7X+2
Problem solving guide:
1. The following operation: (1) (a-3) 2 = A2-6A+9 (2) X-4 = (X+2) (X-2)
(3) ax2+a2xy+a = a (x2+ax) (4)116x2-14x+14 = x2-4x+4 = (x-2) 2.
1 (B)2 (C)3 (D)4
2. No matter what the value of a is, the algebraic expression -A2+4A-5 is ().
(a) greater than or equal to 0 (B)0 (C) greater than 0 (D) less than 0.
3. If x2+2 (m-3) x+ 16 is a completely flat mode, then the value of m is ().
(a)-5 (b) 7 (c)- 1 (d) 7 or-1
4. (x2+y2) (x2-1+y2)-12 = 0, then the value of x2+y2 is;
5. Break down the following factors:
( 1). 8xy(x-y)-2(y-x)3 *(2). X6-y6
(3).x3+2xy-x-xy2 *(4)。 (x+y)(x+y- 1)- 12
(5). 4ab-( 1-a2)( 1-B2)(6)。 -3 square meters -2 meters +4
*4。 Given A+B = 1, find the value of A3+3ab+B3.
5.a, B and C are ⊿ABC trilateral, and the symbols of B2-A2+2ac-C2 are explained by factorization.
6.0 < A ≤ 5, where a is an integer. If 2x2+3x+A can be factorized by cross multiplication, qualified A can be found.
Independent training:
1. The common factor of polynomials x2-y2, x2-2xy+y2, x3-y3 is.
2. Fill in the appropriate numbers or formulas, so that the left can be decomposed into the right results:
( 1)9 x2-()2 =(3x+)(- 15y)，(2).5x2+6xy-8y2=(x )( -4y)。
3. The area of rectangle is 6x2+13x+5 (x >); 0), where one side is 2x+ 1 and the other side is.
4. Factorizing A2-A-6, the correct one is ()
(A)A(A- 1)-6(B)(A-2)(A+3)(C)(A+2)(A-3)(D)(A- 1)(A+6)
5. Among the polynomials A2+4ab+2b2, A2-4ab+ 16b2, A2+A+ 14, 9A2- 12ab+4b2, () can be decomposed by the complete square formula.
1 (B) 2 (C) 3 (D) 4。
6. Let (x+y) (x+2+y)- 15 = 0, then the value of x+y is ().
(A)-5 or 3 (B) -3 or 5 (C)3 (D)5
7. Regarding the quadratic trinomial x2-4x+c can be decomposed into the product of two integral coefficients, then c can take () of the following four values.
(A) -8 (B) -7 (C) -6 (D) -5
8. If x2-mx+n = (x-4) (x+3), then the values of m and n are ().
(A) m=- 1，n=- 12 (B)m=- 1，n= 12 (C) m= 1，n=- 12 (D) m= 1，n= 12。
9. The algebraic expression Y2+My+254 is completely flat, so the value of m is.
10. Given 2x2-3xy+y2 = 0 (neither x nor y is zero), the value of xy+yx is.
1 1. Decomposition factor:
( 1). x2(y-z)+8 1(z-y)(2). 9m 2-6m+2n-N2
*(3).ab(C2+D2)+CD(a2+B2)(4). a4-3 a2-4
*(5). x4+4y 4 *(6). a2+2ab+B2-2a-2 b+ 1
12. Factorization in the range of real numbers
( 1)x2-2x-4(2)4x 2+8x- 1(3)2x 2+4xy+y2
Math factorization test questions in the second day of junior high school
Liu jinzhen
First, multiple-choice questions:
The common factor of 1. polynomial15x3y2m2-35x4y2m2+20x3ym is ().
A 5x3y B 5x3ym C 5x3m D5x3m2y
2. The following deformation from left to right is decomposed into ()
a(a+B)2 = a2+2ab+B2 B x2-4x+5 =(x-2x)2+ 1
c x2-5x-6 =(x+6)(x- 1)D x2- 10x+25 =(x-5)2
3. If the polynomial x2+kxy+9y2 is a completely flat path, then the value of k is ().
A6b3c-6 d-6 or 6
4. Polynomial a2+a-b2-b is decomposed by group decomposition method. Different grouping methods are ().
A 1 species B 2 species C 3 species D 4 species
5. Among the polynomials a2+b2, x2-y2, -x2-y2, -a2+b2, there is () which can decompose the factors.
A 4 B 3 C 2 D 1
6. If the polynomial x2-MX- 15 can decompose the factor, the value of m is ().
A 2 or -2b 14 or-14c2 or-14d 2 or 14.
7. The following polynomial without the factor (x- 1) is ().
a x3-x2-x+ 1 B x2+y-xy-x C x2-2x-y2+ 1D(x2+3x)2-(2x+2)2
8. If X is ()
A 1 B - 1 C D -2
9. If the factor of the polynomial 4ab-4a2-b2-m is (1-2a+b), the value of m is ().
A 0 B 1 C - 1 D 4
10. If (A2+B2-3)-10 = 0, then the value of a2+b2 is ().
A-2b5c2d-2 or 5
Second, the decomposition is as follows:
1 、- m2–N2+2mn+ 1 2 、( a+b)3d–4(a+b)2cd+4(a+b)c2d
3.(x+a)2 –( x–a)2 4。
5.–x5y–xy+2x3y 6。 X6–x4–x2+ 1
7.(x+3)(x+2)+x2–9 8。 (x–y)3+9(x–y)–6(x–y)2
9.(a2+B2– 1)2–4a2b 2 10。 (ax+by)2+(bx–ay)2
Third, a simple calculation method:
1.2.
Fourth, simplify the evaluation:
1.2 ax2–8axy+8ay 2–2 a2。 Known: A2–B2–5 = 0c2–D2–2 = 0.
Where x–2y =1a = 3, find the value of: (AC+BD)2-(AD+BC)2.
5. Observe the following factorization process: The factorization method is called collocation method.
X2+2ax–3aa2 Please use the matching method to decompose the factors:
= x2+2ax+a2–a2–3a2 (add a2 first, then subtract a2) m2–4mn+3N2.
= (x+a) 2–4a2 (using the complete square formula)
= (x+a+2a) (x+a–2a) (using the square difference formula)
=(x+3a)(x–a)
The quadratic trinomial is completely flat by adding and subtracting the above items.
/Chu/upload files _ 8875/200607/20060727002 . doc
http://www . Kao Shi . ws/html/2005/03 1 1/ 1320 10 . html
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(1)(2m+n)(2m-n)=4m2-n2 This operation belongs to.
(2)x2-2x+ 1=(x- 1)2 This operation belongs to.
(3) Use the complete flat die 49x2+y2+ =( -y)2.
Autonomous learning:
Can 1。 Can 993-99 be divisible by 100? what do you think? Communicate with peers.
Is this how hours are calculated?
993-99
=99×992-99× 1
=99(992- 1)
=99×9800
=98×99× 100
So 993-99 can be divisible by 100.
(1) How does Xiao Ming judge whether 993-99 can be divisible by 100?
(2) What other positive integers can be divided by 993-99?
Answer: (1) Xiao Ming divides 993-99 by factorization, and it is concluded that 993-99 is a multiple of 100, so 993-99 can be divisible by 100.
(2) It can also be divisible by positive integers such as 98, 99, 49, 1 1.
2. Calculate the following categories:
( 1)(m+4)(m-4)=；
(2)(y-3)2 =；
(3)3x(x- 1)=；
(4)m(a+b+c)=。
Fill in the blanks according to the above formula:
( 1)3x2-3x=()()
(2)m2- 16=()()
(3)ma+mb+mc=()()
(4)y2-6y+9=()()
Excuse me, what do you think is the relationship between the above two groups of exercises?
Answer: the first group:
( 1)m2- 16； (2)y2-6y+9； (3)3 x2-3x； (4)ma+m b+ MC；
The second group:
( 1)3x(x- 1)； (2)(m+4)(m-4)； (3)m(a+b+c)； (4)(y-3)2 .
The first group is the result of multiplying a polynomial by a polynomial, and the second group is to write a polynomial as the product of several three-dimensional forms, which happens to be a reciprocal relationship.
3. The deformation of equal sign from left to right in the following categories is decomposed into ()
A.(x+3)(x-3)= x2-9b . x2+x-5 =(x-2)(x+3)+ 1
C.a2b+ab2=ab(a+b) D。
Answer: c
4. Prove that if a hundred digits of a three-digit number are exchanged with a single digit number, the difference between the new number and the original number can be divisible by 99.
It is proved that if the original hundredth digit is X, the decimal digit is Y, and the unit digit is Z, then the original digit can be expressed as 100x+ 10y+z, and the digits after swapping positions are100z+10y+X. ..
Then: (100z+10y+x)-(100x+10y+z)
= 100 z- 100x+x-z
= 100(z-x)-(z-x)
=99(z-x)
Then the original conclusion is valid.
5. as shown in fig. 3- 1①, dig a small square with a side length of B (A > B) from the square with a side length of a, and cut and splice the rest into a rectangle (as shown in fig. ②). The education department uses the area of two figures (shaded parts) to verify an equation, which is ().
A.(a+2b)(a-b)= a2+a b-2 B2 b .(a+b)2 = a2+2ab+B2
C.(a-b)2 = a2-2ab+B2 d . a2-B2 =(a+b)(a-b)
Answer: D.
2.2 common factor method
Teaching purposes and requirements: go through the process of exploring the common factors of polynomials and determine the common factors of polynomials in specific problems; Will use the common factor method to decompose polynomials (the letter index in polynomials is limited to positive integers); Further understand the significance of factorization, strengthen students' intuitive thinking and infiltrate the thinking method of regression.
Teaching emphases and difficulties:
Key points: Let students understand the significance and principles of putting forward common factors.
Difficulty: The common factor of polynomial term can be determined.
The key is to make students understand the significance and principle of putting forward common factors.
Quick response:
Common factor of 1 2m2x+4mx2 _ _ _ _ _ _ _ _。
2. The common factor of A2B+AB2+A3B3 _ _ _ _ _ _ _ _.
Common factor of 3.5m (a-b)+10n (b-a) _ _ _ _ _.
4.-5xy- 15 XYZ-20x2y =-5xy(_ _ _ _ _ _ _ _ _ _ _ _ _ _ _)。
Autonomous learning:
1. Teacher Zhang will present prizes to the students who won the space modeling competition. He came to the stationery store and decided to buy 10 pens, 10 notebooks in 5 yuan and 10 bottles of ink in 4 yuan at the unit price of10 in 6 yuan. Because there were many things to buy, the salesman decided to give a 10% discount and asked about the price.
On this issue, two students gave their own methods.
Method 1:16×10× 90%+5×10× 90%+4×10× 90% =144+45+36 = 225 (.
Method 2:16×10× 90%+5×10× 90%+4×10× 90% =10× 90% (1.
Excuse me: which is the better calculation method of the two students? Why?
Answer: The second student (the second method) is better, because the second method puts the factor 10×90% in brackets and only calculates it once, which obviously reduces the amount of calculation.
2. Do all terms of (1) polynomial ab+bc contain the same factor? What about the polynomial 3x2+x? How about polynomial mb2+nb?
(2) Write the polynomial above as the product of several factors, explain your reasons and communicate with your peers.
Answer: (1) All terms of polynomial ab+bc contain the same factor B, all terms of polynomial 3x2+x contain the same common factor X, and all terms of polynomial mb2+nb contain the same common factor B.
3. Break down the following categories:
3x+6； 7x 2-2 1x； 8a3b 2- 12ab 3c+ABC； a(x-3)+2b(x-3)； 5(x-y)3+ 10(y-x)2 .
Answer: (1) 3x+6 = 3x+3x2 = 3 (x+2) (2) 7x2-21x = 7x? x-7x？ 3=7x(x-3)
(3)8a3b2- 12ab3c+abc=ab？ 8a2b-ab？ 12b2c+ab？ c=ab(8a2b- 12b2c+c)
(4)a(x-3)+2b(x-3)=(x-3)(a+2b)
(5)5(x-y)3+ 10(y-x)2 = 5(x-y)3+ 10[-(x-y)]2 = 5(x-y)3+ 10(x-y)2 = 5(x-y)2(x-y+2)
4. Break down the following factors:
( 1)3 x2-6xy+x(2)-4 m3+ 16 m2-26m
Answer: (1) 3x2-6xy+x = x (3x-6y+1) (2)-4m3+16m2-25m =-2m (2m2-8m+13).
5. Factorize the factors.
Answer: =
6. Break down the following categories:
( 1)4q( 1-p)3+2(p- 1)2
(2) 3m(x-y)-n(y-x)
(3)m(5ax+ay- 1)-m(3ax-ay- 1)
Answer: (1) 4q (1-p) 3+2 (p-1) 2 = 2 (1-p) 2 (2q-2pq+1).
(2)3m(x-y)-n(y-x)=(x-y)(3m+n)
(3)m(5ax+ay- 1)-m(3ax-ay- 1)= 2am(x+y)
calculate
(1) Given A+B = 13 and AB = 40, find the value of a2b+ab2;
(2) 1998+ 19982- 19992
Answer: (1)a2b+ab2=ab(a+b). When a+b= 13, the original formula =40× 13=520.
(2) 1998+ 19982- 19992=- 1999
8. Compare the dimensions of 2002×20032003 and 2003×20022002.
Answer: let 2002 = X.
∫2002×20032003-2003×20022002 = x？ 1000 1(x+ 1)-(x+ 1)？ 1000 1 x=0
∴2002×20032003=2003×20022002
2.3 using the formula method
Teaching purposes and requirements: Through the process of square difference formula and complete square formula of algebraic expression multiplication, the method of factorization of formula method is obtained, so as to develop students' reverse thinking and reasoning ability; Decomposition factor (index is a positive integer) by formula method (directly using formula no more than twice)
Teaching emphases and difficulties:
Emphasis: to cultivate students' reverse thinking and reasoning ability.
Difficulties: I can understand and summarize the characteristics of factorization deformation, and at the same time I can fully feel the process of reciprocal deformation and the integrity of mathematical knowledge.
Quick response:
1. Decomposition factor:1x2-y2 =; x2-4 =； ②a2 B2-2ab+ 1 =； = ;
2. In the following polynomials, the factor that can be decomposed by the square difference formula is ().
a . 16 a2-25 B3 b .- 16 a2-25 B2 c . 16 a2+25 B2 d .( 16 a2-25 B2)
3. The following can't be completely square formula decomposition is ()
a . x2+y2+2xy b .-x2+y2+2xy c .-x2-y2-2xy d .-x2-y2+2xy
4. Break down the following factors:
( 1)9a2m 2- 16 B2 N2； (2) ; (3)9(a+b)2- 12(a+b)+4 (4)
Autonomous learning:
1.( 1) Observe the polynomial x2-25.9x-y2. What are their characteristics?
(2) Write them as the product of two factors, explain your reasons and communicate with your peers.
Answer: (1) Every term of a polynomial can be written as a square. For example, in x2-25, x2 itself is square, and 25=52 is also square; The same is true for 9x-y2.
(2) The inverse multiplication formula (a+b)(a-b)=a2-b2 shows that X2-25 = X2-52 = (X+5), 9X2-Y2 = (3x) 2-Y2 = (3x+y) (3x-y).
2. Put the multiplication method
(a+b) 2 = A2+2ab+B2, (a-b) 2 = A2-2ab+B2, conversely, a2+2ab+b2=(a+b)2, a2-2ab+b2=(a-b)2.
Is the above change process a factorization factor? State your reasons.
Answer: A2AB+B2 = (AB) 2 is the factorization factor. Because (a+b)2 is the product of factors, and (a-b)2 is also the product of factors.
3. Break down the following factors:
( 1)25- 16 x2； (2)(3)9(m+n)2-(m-n)2； (4)2x 3-8x；
(5)x2+ 14x+49； (6)(m+m)2-6(m+n)+9(7)3 ax2+6 axy+3 ay2； (8)-x2-4y2+4xy
Answer:
( 1)25- 16 x2 =(5+4x)(5-4x)(2)= 1
(3)9(m+n)2-(m-n)2=4(2m+n)(m+2n)
(4)2x 3-8x = 2x(x2-4)= 2x(x2-2x)= 2x(x+2)(x-2)
(5)x2+ 14x+49 = x2+2×7x+72 =(x+7)2
(6)(m+m)2-6(m+n)+9 =[(m+n)-3]2 =(m+n-3)2
(7)3 ax2+6 axy+3 ay2 = 3a(x2+2xy+y2)= 3a(x+y)2
(8)-x2-4y2+4xy=-(x-2y)2
4. Break down the following factors:
( 1) ; (2)(a+b)2- 1； (3)-(x+2)2+ 16(x- 1)2；
(4)
Answer: (1); (2)(a+b)2- 1 =(a+b+ 1)(a+b- 1)
(3)-(x+2)2+ 16(x- 1)2 = 3(x-2)(5x-2)；
(4)
5. Break down the following factors:
( 1)m2- 12m+36； (2)8a-4a 2-4；
(3) ; (4) 。
Answer: (1) m2-12m+36 = (m-6) 2; (2)8a-4a 2-4 =-4(a- 1)2；
(3) ;
(4)
6. Prove that (x+1) (x+2) (x+3) (x+4)+1is a completely flat road.
Prove 1: The original formula =(x2+5x+4)(x2+5x+6)+ 1.
=(x2+5x)2+ 10(x2+5x)+25
=(x2+5x+5)2 ∴ The original proposition holds.
Proof 2: The original formula = [(x+1) (x+4)] [(x+2) (x+3)]+1.
=(x2+5x+4)(x2+5x+6)+ 1
Let a=x2+5x+4, then x2+5x+6=a+2.
The original formula =a(a+2)+ 1=(a+ 1)2.
That is (x+1) (x+2) (x+3) (x+4)+1= (x2+5x+5) 2.
Proof 3: The original formula =(x2+5x+4)(x2+5x+6)+ 1.
manufacture
The original formula = (x2+5x+5-1) (x2+5x+5+1)+1.
=(m- 1)(m+ 1)+ 1 = m2 =(x2+5x+5)2
7. Knowing that A, B and C are three sides of △ABC, judge whether the shape of △ABC is a2+b2+c2-ab-bc-ca=0.
Answer: ∫a2+B2+C2-a b-BC-ca = 0.
∴2a2+2b2+2c2-2ab-2bc-2ac=0
That is, A2-2ab+B2+B2-2bc+C2+A2-2ac+C2 = 0.
∴(a-b) 2+(b-c) 2+(a-c) 2=0
∫(a-b)2≥0，(b-c) 2≥0，(a-c) 2≥0
∴a-b=0,b-c=0,a-c=0
∴a=b,b=c,a=c
This triangle is an equilateral triangle.
8. Let x+2z=3y, and try to judge whether the value of x2-9y2+4z2+4xz is a fixed value.
A: When x+2z=3y, the value of x2-9y2+4z2+4xz is a fixed value of 0.
6. Prove that (x+1) (x+2) (x+3) (x+4)+1is a completely flat road.
Prove 1: The original formula =(x2+5x+4)(x2+5x+6)+ 1.
=(x2+5x)2+ 10(x2+5x)+25
=(x2+5x+5)2 ∴ The original proposition holds.
Proof 2: The original formula = [(x+1) (x+4)] [(x+2) (x+3)]+1.
=(x2+5x+4)(x2+5x+6)+ 1
Let a=x2+5x+4, then x2+5x+6=a+2.
The original formula =a(a+2)+ 1=(a+ 1)2.
That is (x+1) (x+2) (x+3) (x+4)+1= (x2+5x+5) 2.
Proof 3: The original formula =(x2+5x+4)(x2+5x+6)+ 1.
manufacture
The original formula = (x2+5x+5-1) (x2+5x+5+1)+1.
=(m- 1)(m+ 1)+ 1 = m2 =(x2+5x+5)2
1. According to the concept of factorization, judge which of the following equations is factorization, which is not, and why.
( 1)6abxy=2ab？ 3xy
(2)
(3)(2x- 1)？ 2=4x-2
(4)4 x2-4x+ 1 = 4x(x- 1)+ 1。
fill (up) a vacancy
(1)(2m+n)(2m-n)=4m2-n2 This operation belongs to.
(2)x2-2x+ 1=(x- 1)2 This operation belongs to.
(3) Use the complete flat die 49x2+y2+ =( -y)2.