1, factorization and algebraic expression multiplication are reciprocal operations.
2. When proposing a common factor, if all the coefficients are integers, the proposed common factor is the product of the greatest common divisor of all the coefficients and the lowest power of the letters contained in all the coefficients.
3. If the first coefficient of a polynomial is negative, it is generally necessary to put forward a "-"sign, so that the first coefficient in brackets is positive. When the "-"sign is put forward, all terms of the polynomial will be changed.
4. Sometimes factors can become common factors after sign transformation or letter rearrangement, for example,-a-b+c =-(a+b-c);
Another example: when n is a natural number, (a-b) 2n = (b-a) 2n; (a-b) 2n-1=-(b-a) 2n-1are commonly used factorizations.
5. Polynomials that can be decomposed by square difference formula a2-b2=(a+b)(a-b) must be binomials or polynomials regarded as binomials, and the signs of these two terms are opposite.
A and B can represent numbers, letters or algebraic expressions, and each term can be written as the complete square of a number (or expression).
5. The polynomial that can be decomposed by the complete square formula A2AB+B2 = (AB) 2 must be a trinomial or a polynomial that is regarded as a trinomial, and the two symbols are the same, which can be written in the complete square form of a number (or formula), and the remaining term is twice the product of these two numbers (or formulas). If there are two complete square terms with negative signs in the three terms, first put forward the negative sign, and then use the complete square formula to decompose the factor. Example 1, factorization -a2-b2+2ab+4.
Answer: -a2-b2+2ab+4
=-(a2-2ab+b2-4)
=-[(a2-2ab+b2)-4]
=-[(a-b)2-4]
=-(a-b+2)(a-b-2)
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 to avoid mistakes. Example 2, factorization factor (a+b) n+2-2 (a+b) n+1+(a+b) n.
Solution: (a+b) n+2-2 (a+b) n+1+(a+b) n.
=(a+b)n[(a+b)2-2(a+b)+ 1]
=(a+b)n(a+b- 1)2
In this problem, the formula of extracting common factor is used first, and then the complete square formula is used.
Example 3. Decomposition factor: x4-8x2+ 16.
Solution: x4-8x2+ 16
=(x2-4)2
=[(x+2)(x-2)]2
=(x+2)2(x-2)2
This problem must be thoroughly decomposed until every polynomial factor can no longer be decomposed. Second, the application of factorization:
Convert the formula into the product of several factors, which can often expand complex operations and convert them into simple addition and subtraction operations in one factor, thus greatly simplifying the operation process. This is the mathematical thinking method of equivalent transformation. Example 1. Calculation:
( 1) ; (2);
(3)2022-542+256×352; (4)62 12-769×373- 1482.
Analysis: This topic includes 18 12-6 12, 3192-2092; 17.52-9.52, 13 1.52-3.52; 2022-542; 62 12- 1482. Let's consider the multiplication formula of polynomials:
(a+b)(a-b)=a2-b2。
Its inverse deformation is a2-b2=(a+b)(a-b).
By applying the above variants, we can transform complex square operations and price reduction into simple addition, subtraction, multiplication and division operations. Solution: (1) = =
(2) = = =.
(3) 2022-542+256×352
=(202+54)×(202-54)+256×352
=256× 148+256×352
=256×( 148+352)
=256×500= 128000.(4)62 12-769×373- 1482.
=(62 1+ 148)×(62 1- 148)-769×373
=769×473-769×373
=769×(473-373)
=769× 100=76900.
Through an example of 1, we can easily get the solution to this kind of problem: (1) find the inverse of the square difference formula and turn the square operation into multiplication operation; (2) Reduced-order differential or extraction factor combined with operation evaluation. At the same time, the example of 1 also shows the importance of factorization in simplifying operations. Example 2. Verification: (1) 710-79-78 = 78× 41; (2) 109+ 108+ 107=5× 106×222; (3) 257-5 12 is divisible by 120; (4)8 17-279-9 13 is divisible by 45.
Analysis: According to the distribution law of multiplication, the polynomial operation is m(a+b+c)=ma+mb+mc.
Conversely, we can get ma+mb+mc=m(a+b+c).
The application of the above conclusion can just achieve the purpose of reducing the number of times and solving the problem in this example. Solution: ∫ (1) 710-79-78 = 78× (72-7-1)
=78×(49-8)=78×4 1,
∴7 10-79-78=78×4 1.(2)∵ 109+ 108+ 107= 107×( 102+ 10+ 1)
= 107×( 100+ 1 1)= 106× 10× 1 1 1
=5× 106×222
∴ 109+ 108+ 107=5× 106×222.(3)∵257-5 12=(52)7-5 12
=5 14-5 12=5 1 1×(53-5)
=5 1 1×( 125-5)=5 1 1× 120,
∴257-5 12 is divisible by 120; (4)∵8 17-279-9 13=(34)7-(33)9-(32) 13
=328-327-326=324×(34-33-32)
=324×(8 1-27-9)=324×45,
∴8 17-279-9 13 is divisible by 45. It can be seen from Example 2 that the main ideas to solve this kind of divisibility problem are: (1) extracting appropriate factors; (2) Simplify the algebraic sum of other numbers after extracting the factor, and get a conclusion that can explain the problem, thus solving the problem. Example 3. Given a = and b =, find the value of (a+b)2-(a-b)2. Solution: (a+b)2-(a-b)2
=[(a+b)+(a-b)][(a+b)-(a-b)]
=2a 2b=4ab，
∴(a+b)2-(a-b)2=4×× =。 Example 4. Solve the equation:
( 1)(65x+63)2-(65x-63)2 = 260； (2)(78x+77)(77x-78)=(78x+77)(77x+78)。
Solution: (1) Find the inverse of the square difference formula and turn the original equation into an equivalent form.
[(65x+63)-(65x-63)][(65x+63)+(65x-63)]= 260，
That is,126x130x = 260, ∴ x=.
(2) The original equation can be changed to (78x+77) (77x-78)-(78x+77) (77x+78) = 0,
That is -78×2×(78x+77)=0,
78x+77=0，∴ x=-。
Through Example 4, it can be seen that applying the idea of equivalent transformation to factorization can often skillfully turn high-order equations into the simplest equations, thus finding the roots of the equations. Example 5. (248- 1) is divisible by two numbers between 60 and 70, which are ().
A, 6 1, 63 B, 6 1, 65 C, 63, 65 D, 63, 67 solution: 248-1= (224+1) (224-1)
=(224+ 1)(2 12+ 1)(2 12- 1)
=(224+ 1)(2 12+ 1)(26+ 1)(26- 1),
∵ 26+ 1=65, 26- 1=63.
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