Lecture 1: Factorization (1) 1
Lecture 2: Factorization (2) 4
Lecture 3 Some Properties and Applications of Real Numbers 7
Lecture 4 Simplification and Calculation of Fractions 10
Lecture 5 Identification 13
Lecture 6 Evaluation of Algebraic Expressions 16
Lecture 7 radical formula and its operation 19
Lecture 8 Non-negative Numbers 23
Lecture 9 One-variable Quadratic Equation 27
Lecture 10 congruence of triangle and its application 3 1
Lecture 11 Pythagorean Theorem and Its Application 35
Lecture 12 parallelogram 38
Lecture 13 Trapezoid 4 1
Lecture 14 Neutral Line and Its Application 45
Lecture 15 similar triangles (1) 47
Lecture 16 similar triangles (2) 50
Lecture 17 * Set and Simple Logic 54
Lecture 18 induction and discovery 59
Lecture 19 specialization and generalization 63
Lecture 20 Analogy and Association 67
Lecture 2 1 Classification and discussion 70
Lecture 22 Area Problems and Area Methods 74
Lecture 23 Geometric Inequalities 77
Lecture 24: divisibility of * integer 8 1
Lecture 25 * Congruence Formula 84
Lecture 26 integer root problem of quadratic equation with parameters 87
Lecture 27 The Application of 9 1 in Solving the Equation of Sequence of Numbers
Lecture 28 How to Turn Practical Problems into Mathematical Problems 95
Lesson 29 Mathematics in Life (3)- The World in the Mirror 98
Lesson 30 Mathematics in Life (4) ── Knowledge of Buying Fish 99
Lecture 1: Factorization (1)
Factorization of polynomials is one of the basic forms of algebraic identity deformation. 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 the need to master the factorization content, but also the need to cultivate students' problem-solving skills and develop their thinking ability. Both have very unique functions. The methods of extracting common factors, using formulas, grouping decomposition and cross multiplication are mainly introduced in junior high school mathematics textbooks. This lecture and the next lecture will further introduce the methods, skills and applications of factorization on the basis of middle school mathematics textbooks.
1. Use the formula method
In the algebraic multiplication and division method, we have learned several multiplication formulas, and now we use them in reverse, which is the formula commonly used in factorization, such as:
( 1)a2-B2 =(a+b)(a-b)；
(2)a2 2ab+B2 =(a b)2；
(3)a3+B3 =(a+b)(a2-a b+B2)；
(4)a3-b3=(a-b)(a2+ab+b2)。
Here are some commonly used formulas:
(5)a2+B2+C2+2ab+2bc+2ca =(a+b+c)2；
(6)a3+B3+C3-3 ABC =(a+b+c)(a2+B2+C2-a B- BC-ca)；
(7) an-bn = (a-b) (an-1+an-2b+an-3b2+…+ABN-2+bn-1) where n is a positive integer;
(8) an-bn = (a+b) (an-1-an-2b+an-3b2-…+ABN-2-bn-1), where n is an even number;
(9) an+bn = (a+b) (an-1-an-2b+an-3b2-…-ABN-2+bn-1), where n is an odd number.
When using formula to decompose factors, we should choose formulas correctly and appropriately according to the characteristics of polynomials, letters, coefficients, indexes and symbols.
Example 1 decomposition factor:
( 1)-2x5n- 1yn+4x3n- 1yn+2-2xn- 1yn+4；
(2)x3-8y 3-z3-6x yz；
(3)a2+B2+C2-2bc+2ca-2ab；
(4)a7-a5b2+a2b5-b7。
Solution (1) Original formula =-2xn- 1yn(x4n-2x2ny2+y4)
=-2xn- 1yn[(x2n)2-2x2ny 2+(y2)2]
=-2xn- 1yn(x2n-y2)2
=-2xn- 1yn(xn-y)2(xn+y)2。
(2) The original formula =x3+(-2y)3+(-z)3-3x(-2y)(-Z)
=(x-2y-z)(x2+4y2+z2+2xy+xz-2yz)。
(3) The original formula =(a2-2ab+b2)+(-2bc+2ca)+c2.
=(a-b)2+2c(a-b)+c2
=(a-b+c)2。
This small problem can be slightly deformed, and can be solved directly by Formula (5) as follows:
Original formula =a2+(-b)2+c2+2(-b)c+2ca+2a(-b)
=(a-b+c)2
(4) Original formula =(a7-a5b2)+(a2b5-b7)
=a5(a2-b2)+b5(a2-b2)
=(a2-b2)(a5+b5)
=(a+b)(a-b)(a+b)(a4-a3 b+a2 B2-ab3+B4)
=(a+b)2(a-b)(a4-a3 b+a2 B2-ab3+B4)
Example 2 Factorization: A3+B3+C3-3abc.
This problem is actually to prove the formula (6) given above by factorization.
Analyze the formula we already know.
(a+b)3=a3+3a2b+3ab2+b3
The correctness of, now this formula deformation is
a3+b3=(a+b)3-3ab(a+b)。
This formula is also a commonly used formula, and this problem is derived with it.
Solution formula =(a+b)3-3ab(a+b)+c3-3abc.
=[(a+b)3+c3]-3ab(a+b+c)
=(a+b+c)[(a+b)2-c(a+b)+C2]-3ab(a+b+c)
=(a+b+c)(a2+b2+c2-ab-bc-ca)。
It shows that formula (6) is a widely used formula, and many useful conclusions can be drawn by using it. For example, we convert Equation (6) into
a3+b3+c3-3abc
Obviously, when a+b+c=0, then A3+B3+C3 = 3abc; When a+b+c > 0, then a3+b3+c3-3abc≥0, that is, a3+b3+c3≥3abc, and the equal sign holds if and only if a = b = C. 。
If x=a3≥0, y=b3≥0 and z=c3≥0, then there is
The necessary and sufficient condition of equal sign is x = y = z, which is also a common conclusion.
Example 3 factorization: x15+x14+x13+…+x2+x+1.
The characteristic of this polynomial is that it has the term 16. Starting from the highest term of x 15, the degree of x decreases to zero in turn, so it can be decomposed by the formula an-bn.
Solve the cause
x 16- 1 =(x- 1)(x 15+x 14+x 13+…x2+x+ 1)，
therefore
It shows that the decomposition of this problem uses the technique of multiplying by (x- 1) and then dividing by (x- 1), which is very common in equation deformation.
2. Method of splitting and adding projects
Factorization is the inverse operation of polynomial multiplication. In polynomial multiplication, sorting and simplification often combine several similar items into one item, or two similar items with opposite signs cancel each other out to zero. In the factorization of some polynomials, it is necessary to restore those terms that have been merged or cancelled each other, that is, to split one term in a polynomial into two or more terms, or to add two terms, which only meet the opposite requirements. The former is called a split item and the latter is called an add item.
Example 4 factorization: x3-9x+8.
There are many solutions to this problem. Here, we only introduce several solutions using the method of splitting and adding items, and pay attention to the purpose and skills of splitting and adding items.
The solution 1 decomposes the constant term 8 into-1+9.
Original formula =x3-9x- 1+9.
=(x3- 1)-9x+9