That is, the square difference of two numbers is equal to the product of the sum of these two numbers and the difference of this number.
For example: 4x2-9 = (2x) 2-32 = (2x+3) (2x-3).
(2) Complete square formula: A2AB+B2 = (AB) 2.
Among them, A2 A2 AB+B2 is called the completely flattened mode.
That is to say, the sum of squares of two numbers plus (or minus) twice the product of these two numbers is equal to the square of the sum (or difference) of these two numbers.
For example: 4x2-12xy+9y2 = (2x) 2-2 2x3y+(3y) 2 = (2x-3y) 2.
Inquiry communication
Are the following deformations correct? Why?
( 1)x2-3 y2 =(x+3y)(x-3y)；
(2)4x 2-6xy+9 y2 =(2x-3y)2；
(3)x2-2x- 1=(x- 1)2。
The instruction (1) is incorrect, so it cannot be decomposed in rational numbers at present.
(2) Incorrect, 4x2-6xy+9y2 is not completely flat and cannot be decomposed.
(3) Incorrect. X2-2x- 1 is not completely flat, so it cannot be decomposed by the complete square formula, nor can it be decomposed within a rational number.
Knowledge point 4 grouping decomposition method
(1) shape: am+an+bm+bn=(am+an)+(bm+bn)
=a(m+n)+b(m+n)
=(m+n)(a+b)
(2) Form: x2-y2+2x+1= (x2+2x+1)-y2.
=(x+ 1)2-y2
=(x+y+ 1)(x-y+ 1)。
Polynomials are grouped properly, and there can be common factors or formulas after grouping. This factorization method is called grouping decomposition.
The general grouping method of knowledge rule summary (1) grouping decomposition is not unique.
For example, there are two ways to decompose am+an+bm+bn:
Method1:am+an+BM+bn = (am+an)+(BM+bn) = a (m+n)+b (m+n) = (m+n) (a+b).
Method 2: am+an+BM+bn = (am+BM)+(an+bn) = m (a+b)+n (a+b) = (m+n) (a+b).
(2) Grouping is not only tentative, but also purposeful, or common factors can appear after grouping, or formulas can be used after grouping.
For example: am+an+bm+bn has a common factor after grouping; X2-y2+2x+ 1 The grouped formula can be used.
Group decomposition is the basic method of factorization, which embodies the idea of turning the whole into parts and taking the whole as the whole. How to group reasonably is the key to solve the problem. Common grouping methods are:
(1) grouped by letter;
(2) grouping by frequency;
(3) Grouping by coefficient.
For example, decompose the following factors.
( 1)am+BM+an+bn；
(2)x2-y2+x+y；
(3)2ax-5by+2ay-5bx。
Factorization of knowledge point 5 x2+(p+q)x+pq quadratic trinomial.
x2+(p+q)x+pq=(x+p)(x+q)。
In fact, it is: x2+(p+q)x+pq.
=x2+px+qx+pq
=(x2+px)+(qx+pq)
=x(x+p)+q(x+p)
=(x+p)(x+q)。
∴x2+(p+q)x+pq=(x+p)(x+q).
Using this formula, the quadratic trinomial factor can be decomposed. When p=q, this formula is transformed into x2+2px+p2 or x2+2qx+q2, which is completely flat and can be used to decompose factors.
Such as factorization x2+3x+2.
(Analytic) Because the quadratic term coefficient of quadratic trinomial x2+3x+2 is 1, the constant term 2 =1x2, and the linear term coefficient 3= 1+2, this is a formula of x2+(p+q)x+pq.
Solution: x2+3x+2=(x+ 1)(x+2)
Analysis of typical examples of teacher-student interaction
Application of basic knowledge
The application of the basic knowledge in this section mainly includes: (1) mastering the common factor method, formula method and group decomposition method of factorization; (2) The quadratic trinomial of x2+(p+q)x+pq type can be decomposed.
Example 1 decompose the following factors by extracting common factors.
( 1)ax-ay； (2)6 XYZ-3xz 2； (3)-x3z+x4y；
(4)36 aby- 12 abx+6ab； (5)3x(a-b)+2y(b-a)；
(6)x(m-x)(m-y)-m(x-m)(y-m)。
(Analytic) Problems (1) ~ (4) can be directly decomposed by extracting common factors, and problems (5) and (6) should be properly deformed first, in which (5) the problem transforms b-a into-(a-b) and (6) the problem transforms (X-M) into (Y-M).
Solution: (1)ax-ay=a(x-y)
(2)6xyz-3xz2=3xz(2y-z)。
(3)-x3z+x4y=x3(-z+xy)。
(4)36 aby- 12 abx+6ab = 6ab(6y-2x+ 1)。
(5)3x(a-b)+2y(b-a)= 3x(a-b)-2y(a-b)=(a-b)(3x-2y)。
(6)x(m-x)(m-y)-m(x-m)(y-m)
=x(m-x)(m-y)-m(m-x)(m-y)
=(m-x)(m-y)(x-m)
=-(m-x)2(m-y)。
Summarize the following problems that should be paid attention to when using factorization:
(1) factorization results, if there are similar items in each bracket and each bracket cannot be decomposed.
Such as: (7m-8n)(x+y)-(3m-2n)(x+y)
=(x+y)[(7m-8n)-(3m-2n)]
=(x+y)(4m-6n)。
=2(x+y)(2m-3n)。
(2) If small items like (5) and (6) need to be unified, they should be unified first, and the number of times of unification should be as few as possible, so as to reduce the probability of mistakes in unified calculation. It should be noted that (a-b)n=(b-a)n(n is an even number).
For example, the factorization factor A (x-y) 2+B (y-x) 3+C (y-x) 2.
This topic can unify (x-y) into (y-x) and (y-x) into (x-y), but it is simpler to unify (x-y) into (y-x) because (x-y) 2 = (y-x) 2.
a(x-y)2+b(y-x)3+c(y-x)2
=a(y-x)2+b(y-x)3+c(y-x)2
=(y-x)2[a+b(y-x)+c]
=(y-x)2(a+ times -bx+c).
(3) Factorization should be written as a product if it finally has the same base power.
For example: (7a-8b)(a-2b)+(a-8b)(a-2b)
=(a-2b)[(7a-8b)+(a-8b)]
=(a-2b)(8a- 16b)
=8(a-2b)(a-2b)
=8(a-2b)2。
Students do the following factorization.
( 1)am+an； (2)(xy+ay-by)；
(3)(2a+b)(2a-3b)+(2a+5b)(2a+b)； (4)3x(a-b)-2y(b-a)；
(5)4p( 1-q)3+2(q- 1)2； (6)ab2(x-y)m+a2b(x-y)m+ 1。
Teacher's comments (1) Original formula =a(m+n) (2) Original formula = y (x+a-b);
(3) The original formula = 2 (2a+b) 2; (4) The original formula = (a-b) (3x+2y);
(5) The original formula = (1-q) 2 (4p-4pq+2); (6) The original formula = ab (x-y) m (b+ax-ay).
Example 2 breaks down the following categories.
( 1)m2+2m+ 1； (2)9 x2- 12x+4；
(3) 1- 10x+25 x2； (4)(m+n)2-6(m+n)+9。
(Analysis) The purpose of this question is to investigate factorization with complete square formula.
Solution: (1) m2+2m+1= (m+1) 2.
(2)9x2- 12x+4=(3x-2)2。
(3) 1- 10x+25 x2 =( 1-5x)2。
(4)(m+n)2-6(m+n)+9=(m+n-3)2。
Students do the following factorization.
( 1)(x2+4)2-2(x2+4)+ 1； (2)(x+y)2-4(x+y- 1)。
Teacher's comment (1) Original formula = (x2+3) 2; (2) The original formula = (x+y-2) 2.
Example 3 breaks down the following categories.
( 1)x2+7x+ 10； (2)x2-2x-8；
(3)y2-7y+ 10； (4)x2+7x- 18。
(Analytic) The quadratic term coefficient of quadratic trinomial x2+7x+ 10 is 1, the constant term is 10=2×5, and the linear term coefficient is 7=2+5, so it is a formula of x2+(p+q)x+pq type, which can be used as a formula.
Solution: (1) x2+7x+10 = (x+2) (x+5).
(2)x2-2x-8=(x-4)(x+2)。
(3)y2-7y+ 10=(y-2)(y-5)。
(4)x2+7x- 18=(x+9)(x-2)。
X2+(p+q)x+pq type quadratic trinomial factorization summary, ① PQ > 0, then the signs of P and Q are the same, if P+Q > 0, then P > 0, Q > 0; If q+p < 0, then p < 0, q+p < 0; ② If PQ < 0, P and Q are different symbols; If P+Q > 0, the absolute value is positive; If P+Q < 0, the absolute value is negative.
Students do the following factorization.
( 1)m2-7m+ 12； (2)x2 y2-3xy- 10；
(3)(m-n)2-(m-n)- 12； (4)x2-xy-2y2。
Teacher's comments (1) Original formula = (m-3) (m-4); (2) The original formula = (xy-5) (xy+2);
(3) The original formula = (m-n-4) (m-n+3); (4) The original formula = (x-2y) (x+y).
Comprehensive application problem
The comprehensive application of knowledge in this section mainly includes: (1) group decomposition and factorization; (2) Comprehensive application of the equation; (3) Comprehensive application of geometry knowledge; (4) Comprehensive application of several factorization methods.
Example 4 Factorization.
( 1)x3-2 x2+x； (2)(a+b)2-4a 2； (3)x4-8 1x2y 2；
(4)x2(x-y)+y2(y-x)； (5)(a+b+c)2-(a-b-c)2。
(Analysis) The purpose of this question is to investigate the comprehensive application of common factor method and formula method to decompose factors.
Solution: (1) x3-2x2+x = x (x2-2x+1) = x (x-1) 2.
(2)(a+b)2-4a 2 =(a+b+2a)(a+b-2a)=(3a+b)(b-a)。
(3)x4-8 1x2y 2 = x2(x2-8 1 y2)= x2(x+9y)(x-9y)。
④x2(x-y)+y2(y-x)= x2(x-y)-y2(x-y)
=(x-y)(x2-y2)=(x-y)(x+y)(x-y)
=(x+y)(x-y)2。
(5)( a+b+c)2-(a-b-c)2
=[(a+b+c)(a-b-c)][(a+b+c)-(a-b-c)]
=2a (2b+2c)
=4a(b+c)。
Example 5 The following factors are decomposed by grouping decomposition method.
( 1)a2-B2+a-b； (2)a2+B2-2ab- 1；
(3)(ax+by)2+(ay-bx)2； (4)a2-2ab+b2-c2-2c- 1。
(Analytic) The grouping decomposition method is generally the factorization of four or more polynomials. Grouping has two purposes: first, common factors can appear after grouping; second, formulas can be applied after grouping, in which (1) questions have common factors after grouping; (3) questions need to be grouped after removing brackets; (2) and (4) questions can be used after grouping.
Solution: (1) A2-B2+A-B = (A2-B2)+(A-B)
=(a+b)(a-b)+(a-b)=(a-b)(a+b+ 1)。
(2)a2+B2-2ab- 1 =(a2-2ab+B2)- 1
=(a-b)2- 1 =(a-b+ 1)(a-b- 1)。
(3)(ax+by)2+(ay-bx)2
= a2 x2+2 abxy+B2 y2+a2 y2-2 abxy+B2 x2
=a2x2+b2y2+a2y2+b2x2
=(a2x2+a2y2)+(b2y2+b2x2)
=a2(x2+y2)+b2(x2+y2)
=(a2+b2)(x2+y2)。
(4)a2-2ab+b2-c2-2c- 1
=(a2-2ab+b2)-(c2+2c+ 1)
=(a-b)2-(c+ 1)2
=[(a-b)+(c+ 1)][(a-b)-(c+ 1)]
=(a-b+c+ 1)(a-b-c- 1)。
When solving the factorization problem, first consider whether there is a common factor, and if there is, first mention the common factor; If there is no common factor or the common factor is extracted, the following situations are usually considered:
(1) If there are four or more items, consider using the group decomposition method;
(2) If it is quadratic trinomial or completely flat, we can consider using x2+(p+q)x+pq formula or complete square formula to decompose the factor;
(3) If it is two terms, consider whether the factor can be decomposed by the square difference formula.
Finally, until each factor can no longer be decomposed.
Example 6 Solving Equation
(Analytic) This problem is a system of binary quadratic equations. As far as the current level of knowledge is concerned, it is difficult to solve it by substitution elimination or addition and subtraction elimination. But we found that the equation x2-4y2=5 can be factorized into (x+2y)(x-2y)=5, and then x-2y= 1 is substituted into the equation (.
Solution: (x+2y)(x-2y)=5 from ①, ③.
Substituting ② into ③ gives x+2y=5, ④.
∴ The original equation is converted into
②+④ 2x=6，∴ x = 3。
②-④ 4y=4，∴ y = 1。
∴ The solution of the original equation is
Students do and solve equations.
The teacher commented.
Example 7 If A, B and C are three sides of a triangle and satisfy the relationship a2+b2+c-ab-ac-bc=0, try to judge the shape of this triangle.
Solution: ∫a2+B2+C2-a b-AC-BC = 0,
∴2a2+2b2+2c2-2ab-2ac-2bc=0.
That is, (a2-2ab+B2)+(B2-2bc+C2)+(C2-2ac+A2) = 0,
(a-b)2+(b-c)2+(c-a)2=0。
According to the nonnegativity of the square,
∴a=b=c.
This triangle is an equilateral triangle.
Example 8 Calculate the following problems by factorization.
( 1)234×265-234×65; (2)992+ 198+ 1.
(Analytic) Common factor method and formula method are mainly used to decompose factors for calculation.
Solution: (1) 234× 265-234× 65 = 234× (265-65)
=234×200=46800.
(2)992+ 198+ 1=992+2×99× 1+ 1
=(99+ 1)2= 1002
= 10000.
Students do and calculate the following questions by factorization.
( 1)7.6× 199.9+4.3× 199.9- 1.9× 199.9;
(2)20022-4006×2002+20032;
(3)5652× 1 1-4352× 1 1;
(4)(5)2-(2)2.
Teacher's comment (1) Original formula =1999; (2) The original formula =1;
(3) The original formula =143000 o; (4) The original formula = 28.
Example 9 If 9x2+kxy+36y2 is completely flat, then k =.
(Analytical) The completely flat form is as follows: a2 2ab+b2, that is, the sum (or difference) of the sum of the squares of two numbers and twice the product of these two numbers.
∫9 x2+kxy+36 y2 =(3x)2+kxy+(6y)2，
∴ kxy=2 3x 6y=36xy。
∴k= 36。
If x2+(k+3)x+9 is completely flat, then k =.
The teacher evaluates k=3 or k =-9.
Exploration and innovation issues
Example 10 calculation.
(Analysis) The purpose of this question is to examine the flexible application of factorization, namely = A-B (A+B ≠ 0).
Solution: Original formula =+…
+
=( 1-2)+(3-4)+(5-6)+…+(2003-2004)
=(- 1)×(2004÷2)
=- 1002.
Example 1 1 If x2+kx+20 can be factorized within an integer range, then the integer value that k can take is ().
A.2 B.3 C.4 D.6
(Analysis) If x2+kx+20 is factorized within the integer range, consider the factorization factor of 20 with formula x2+(p+q)x+qq, and 20 can be decomposed into: 20× 1, (-20)×(- 1),1. (-20)+(- 1), 10+2, (-10)+(-2), 5+4, (-5)+(-4), so k has six possible values, so the correct answer is D.
Example 12 factorization factor (x4+x2-4) (x4+x2+3)+ 10.
(Analysis) Take x4+x2 as a whole and replace it with a new letter, thus simplifying the structure of the formula.
Solution: Let x4+x2=m, then the original formula can be simplified as follows
(m-4)(m+3)+ 10
=m2-m- 12+ 10
=m2-m-2
=(m-2)(m+ 1)
=(x4+x2-2)(x4+x2+ 1)
=(x2+2)(x2- 1)(x4+x2+ 1)
=(x2+2)(x+ 1)(x- 1)(x4+x2+ 1)。
Students do a test: the product of four consecutive natural numbers plus 1 must be a complete square number.
The teacher commented that four consecutive natural numbers are n, n+ 1, n+2, n+3, then
n(n+ 1)(n+2)(n+3)+ 1
=(n2+3n)(n2+3n+2)+ 1
=(n2+3n)2+2(n2+3n)+ 1
=(n2+3n+ 1)2
∴ n (n+1) (n+2) (n+3)+1must be a complete square number.
Example 13 if x2+7xy+my2-5x+43y-24 can be decomposed into the product of two linear factors, x and y, try to find the value of m 。
(Analysis) Make x2+7xy+My2-5x+43y-24 = (x+ay+b) (x+cy+d) by the method of undetermined coefficient, and then compare the coefficients to get m. 。
Solution: Let x2+7xy+My2-5x+43y-24 = (x+ay+b) (x+cy+d) = x2+(a+c) xy+(b+d) x+(ad+BC) y+BD.
Compare the coefficients of polynomials
B=-8, d=3, or b=3, D =-8 can be obtained from ③ and ⑤.
(1) When b=-8, d=3, a=9 and c=-2, ⑥.
(2) When b=3 and d=-8, a=-2 and c = 9. ⑦.
∴m=- 18.
Students know that the polynomial 2x3-x2+m has a factor (2x+ 1) and find the value of m. 。
Teacher's evaluation can be set as 2x3-x2+m=(2x+ 1)(x2+ax+b), then 2x3-x2+m=2x3+(2a+ 1)x2+
(a+2b)x+b。
The coefficient of that comparison polynomial can be obtained.
Looking forward to the senior high school entrance examination, click the senior high school entrance examination.
Summary and prospect of the proposition of senior high school entrance examination
The content of this chapter mostly appears in the form of fill-in-the-blank questions and multiple-choice questions in the senior high school entrance examination, and is rarely put forward directly by factorization, but its combination with equations, binary linear equations, quadratic functions and fractional operations is common and should be paid full attention to in learning.
Prediction of test questions in senior high school entrance examination
Example 1 (1) (Fuzhou, 2004) Decomposition factor: A2-25 =；;
(2) (Changsha, 2004) Decomposition factor: xy2-x2y =；;
(3) (2004 Guiyang) Decomposition factor: x2-1=;
(4) (Nanjing in 2004) Decomposition factor: 3 x2-3 =；;
(5) (2004 Hubei) Decomposition factor: x2+2xy+y2-4 =;
(6) (2004 Shaanxi) Decomposition factor: x3y 2-4x =；;
(7) (Guangzhou, 2004) Decomposition factor: 2 x2-2 =；;
(8) (2004 Guilin) Decomposition factor: A3+2 a2+A =；;
(9) (2004 Qinghai) Decomposition factor: x3y-4xy+4y =;
(10) (Harbin, 2004) Decomposition coefficient: A2-2ab+B2-C2 =.
(Analytic) (1) can be directly decomposed by the square difference formula. (2) It can be directly decomposed by extracting common factors. (4)3 x2-3 = 3(x2- 1)= 3(x+ 1)(x- 1)。
= (x+y) 2-4 = (x+y+2) (x+y-2)。 (6) First, the common factor is extracted, and then the factor is decomposed by formula method. x3y 2-4x = x(x2 y2-4)= x
x(xy+2)(xy-2)。
Answer: (1) (a+5) (a-5) (2) xy (y-x) (3) (x+1) (x-1) (4) 3 (x+1.
(6)x(xy+2)(xy-2)(7)2(x+ 1)(x- 1)(8)a(a+ 1)2(9)y(x-2)2( 10)(a-b+c)(a-b-c)
Example 2 (Anhui, 2004) Among the following polynomials, () can be used to decompose factors.
A.x2-y B.x2+2y C.x2+y2 D.x2-xy+y2
Answer: b
Example 3 (Qinghai, 2004) factorized polynomial a2-ab+ac-bc, with different grouping methods.
(Analysis) One is: A2-AB+AC-BC = (A2-AB)+(AC-BC);
The other is: a2-ab-ac-bc=(a2+ac)-(ab+bc),
There are two grouping methods.
Example 4 (Hubei, 2004) x2-y2-x-y factorization results are as follows.
Answer: (x+y)(x-y- 1)
Example 5 (Hohhot, 2004) factorized the following formula: x-x2-y+y2 =.
Answer: (x-y)( 1-x-y)
Example 6 (Linfen, 2004) Solving Equation
(Analytic) Convert a binary quadratic equation into a binary linear equation by factorization.
Solution: (x-2y)(x+y)=0 from ①, ③.
Substituting ② into ③ gives x-2y=0, ④.
The original equation is reduced to
②-④ 3y=2，∴ y =。
Substitute y= into ④ to get x =.
∴ The solution of the original equation is
Example 7 (Gansu, 2004) In order to factorize x2-7x+b within an integer range, the possible values of b are. (Write any one).
(Analytic) This is an open question with different answers, based on the formula X2+(P+Q) X+PQ.
Answer: -8
Example 8 (Ningxia, 2004) The result of factorization of polynomial 1-x2+2xy-y2 is ().
A.( 1-x-y)( 1+x-y)b .( 1+x-y)( 1-x+y)
C.( 1-x-y)( 1-x+y)d .( 1+x-y)( 1+x+y)
(Analytic) Solving problems by group decomposition.
1-x2+2xy-y2 = 1-(x2-2xy+y2)
= 1-(x-y)2 =( 1+x-y)( 1-x+y)。
Therefore, the correct answer is item B.
Course Summary This section summarizes
1. This part mainly studies: decomposing factors by extracting common factors; Decomposition of factors by formula method; Group decomposition method decomposes factors; Factorization of quadratic trinomial x2+(p+q)x+pq.
2. Factorization can be used to solve calculation problems.
Exercise selection and solution of textbook exercises
Textbook page 200 ~ 20 1
Exercise 15.5
1.( 1) Original formula = 5a2 (3a+2); (2) Original formula = 3bc (4a-c);
(3) The original formula = 2 (p+q) (3p-2q); (4) The original formula = (a-3) (m-2).
2.( 1) Original formula = (1+6b) (1-6b); (2) The original formula = 3 (2x+y) (2x-y);
(3) The original formula = (0.7p+12) (0.7p-12); (4) The original formula = 3 (x+y) (x-y).
3.( 1) Original formula = (1+5t) 2; (2) The original formula = (m-7) 2;
(3) The original formula = (y+) 2; (4) Original formula = (3m+n) 2;
(5) Original formula = (5a-8) 2; (6) The original formula = (a+b+c) 2.
4.( 1) Original formula = 314; (2) The original formula = 508000.
5.( 1) Original formula = (a+b) 2; (2) The original formula = (p+2) (p-2);
(3) The original formula =-y (2x-y) 2; (4) The original formula = 3a (x+y) (x-y).
6. Solution: When v = ir 1+ir2+ir3, r 1 = 19.7, R2=32.4, R3=35.9 and I = 2.5.
v = 19.7×2.5+32.4×2.5+35.9×2.5
=2.5( 19.7+32.4+35.9)
=2.5×88
=220.
7. Solution: When r = 7.8 cm, r = 1. 1 cm, π = 3. 14,
πR2-4πr2
=3. 14×7.82-4×3. 14× 1. 12
=3. 14×(7.82-4× 1. 12)
=3. 14×(7.8+2× 1. 1)(7.8-2× 1. 1)
=3. 14× 10×5.6
= 175.84(cm2)。
∴ Shadow area is 175.84 square kilometers.
8. Tip: There are two methods. One is to subtract the area of the intersecting square from the area of two roads; The other is to use the area of one road plus the area divided into two sections.
As shown in figure 15- 19, if the left side of the transverse tunnel is m meters long and the right side is (x-m-2) meters long, the tunnel area is
2x+m 2+2(x-m-2)
=2x+2m+2x-4-2m
=(4x-4) (m2)。
9. Tip: ∫4 y2+my+9 is completely flat.
Mine is 2× 2y× 3 = 12y.
∴m= 12。
10. solution: the conclusion is n (n+2)+ 1 = (n+ 1) 2.
The proof process is as follows:
∫n(n+2)+ 1 = N2+2n+ 1 =(n+ 1)2。
∴n(n+2)+ 1=(n+ 1)2