F(x) = AP1k1(x) p2k2 (x) … piki (x) *, where α is the coefficient of the highest term of f (x), P 1 (x), P2 (x) … PI (x) is
(*) or the typical decomposition of polynomial f(x). Proof: See Gao Dai P52-53.
In elementary mathematics, the decomposition of polynomials is called factorization, and its general steps are: one mention, two sets, three groups and so on.
The requirement is: until we can't separate.
2. Method introduction
2. 1 common factor method:
If every term of a polynomial has a common factor, we can first consider putting forward the common factor and factorizing it. Note that every term must have a common factor.
Example 15x3+ 10x2+5x
Obviously, each term contains a common factor 5x, so we can consider extracting the common factor 5x, and the remaining x2+2x+ 1 can still be decomposed.
Solution: Original formula =5x(x2+2x+ 1)
=5x(x+ 1)2
2.2 formula method
That is, if a polynomial satisfies the structural characteristics of a special formula, it can be factorized by a set of formulas, so it is required to be familiar with some commonly used formulas. In addition to the basic formulas in textbooks, some basic formulas that often appear in math competitions are summarized as follows:
a2-b2=(a+b)(a-b)
a2 2ab+b2=(a b)2
a3+b3=(a+b)(a2-ab+b2)
a3-b3=(a-b)(a2+ab+b2)
a3 3a2b+3ab2 b2=(a b)3
a2+b2+c2+2ab+2bc+2ac=(a+b+c)2
a 12+a22+…+an2+2a 1 a2+…+2an- 1an =(a 1+a2+…+an)2
a3+B3+C3-3 ABC =(a+b+c)(a2+B2+C2-a B- AC-BC)
An+bn = (a+b) (an-1-an-2b+…+bn-1) (n is an odd number)
Explain the factorial theorem, that is, for unary polynomial f(x), if f(b)=0, it must contain linear factorial X-B. It can be judged that when n is even, when a = b and a =-b, there is an-bn=0, so an-bn must contain a+b and a-b factors.
Example 2 factorization: ① 64x6-y12②1+x+x2+…+x15.
Formulas can be used to analyze various minor problems.
Solution ① 64x6-y12 = (8x3-y6) (8x3+y6)
=(2x-y2)(4x 2+2x 2+y4)(2x+y2)(4x 2-2x 2+y4)
② 1+x+x2+…+x 15=
=( 1+x)( 1+x2)( 1+x4)( 1+x8)
When we pay attention to polynomial decomposition, we should first construct the formula and then decompose it.
2.3 Grouping decomposition method
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)
2.4 cross multiplication
Cross multiplication can be considered for quadratic trinomial with the structural characteristics of ax2+bx+c,
That is, x2+(b+c)x+bc=(x+b)(x+c) When the coefficient of the x2 term is not 1, cross multiplication can also be performed.
Example 3 decomposition factor: ①x2-x-6②6x2-x- 12.
Solution ① 1x2
1x-3
Original formula =(x+2)(x-3)
②2x-3
3x4
Original formula =(2x-3)(3x+4)
Note: "ax4+bx2+c" type can also consider this method.
2.5 Pairwise cross multiplication
Cross multiplication is a basic method commonly used in quadratic trinomial decomposition. For more complex polynomials, especially some quadratic sextuples, such as 4x2-4xy-3y2-4x+ 10y-3, cross multiplication factorization can also be used. The specific steps are as follows:
(1) The quadratic trinomial composed of the first three times is decomposed by cross multiplication, and a cross multiplication graph is obtained.
(2) Divide the constant term into two factors and fill them in the right side of the second cross, so that the sum of the products of the intersection of the two factors in the second cross is equal to the primary term containing Y in the original formula, and the sum of the products of the intersection of the two factors at the left side of the first cross must be equal to the primary term containing X in the original formula.
Example 5 Factorization
①4x 2-4xy-3 y2-4x+ 10y-3②x2-3xy- 10y 2+x+9y-2
③a b+ B2+a-b-2④6x 2-7xy-3 y2-xz+7yz-2z 2
Solution ① Original formula =(2x-3y+ 1)(2x+y-3)
2x-3y 1
2xy-3
② Original formula =(x-5y+2)(x+2y- 1)
x-5y2
x2y- 1
③ Original formula =(b+ 1)(a+b-2)
0ab 1
ab-2
④ Original formula =(2x-3y+z)(3x+y-2z)
2x-3yz
3x-y-2z
Note: Type ③ can be supplemented by oa2, and double cross multiplication can be used. Of course, this question can also be grouped.
Such as (ab+a)+(B2-b-2) = a (b+1)+(b+1) (b-2) = (b+1) (a+b-2).
(4) The three letters of the formula satisfy the quadratic six-term formula, and -2z2 can be regarded as a constant decomposition:
2.6 Demolition method and addition method
For some polynomials, if factorization cannot be performed directly, one of them can be decomposed into the difference or sum of two terms. Then the factors are decomposed by grouping method and formula method, among which the method of splitting and adding items is not unique, and there are many different ways to solve them. Be sure to analyze the problem in detail and choose a simple decomposition method.
Example 6 Decomposition factor: x3+3x2-4
Analysis method 1: -4 can be decomposed into-1, and -3 is (x3- 1)+(3x2-3).
Method 2: Add x4 and subtract x4, that is, (x4+3x2-4)+(x3-x4).
Method 3: Add 4x and subtract 4x, that is, (x3+3x2-4x)+(4x-4).
Method 4: Divide 3x2 into 4x2-x2, namely (x3-x2)+(4x2-4).
Method 5: Break x3 into 4x2-3x3 (4x3-4)-(3x3-3x2) and so on.
Solution (Option 4) Original formula =x3-x2+4x2-4
= x2(x- 1)+4(x- 1)(x+ 1)
=(x- 1)(x2+4x+4)
=(x- 1)(x+2)2
2.7 alternative methods
The substitution method introduces a new letter variable, replaces the letter variable in the original formula, and simplifies the formula. Use this
This method can simplify the factorization of some special polynomials.
Example 7 factorization:
(x+ 1)(x+2)(x+3)(x+4)- 120
It will be very boring to analyze this, but we notice that
(x+ 1)(x+4)=x2+5x+4
(x+2)(x+3)=x2+5x+6
Therefore, the problem can be decomposed by substitution method.
Solution formula =(x2+5x+4)(x2+5x+6)- 120.
Let y=x2+5x+5, then the original formula = (y-1) (y+1)-120.
=y2- 12 1
=(y+ 1 1)(y- 1 1)
=(x2+5x+ 16)(x2+5x-6)
=(x+6)(x- 1)(x2+5x+ 16)
Note: You can also use x2+5x+4=y or x2+5x+6=y or x2+5x=y here. Please compare carefully which method is simpler?
2.8 undetermined coefficient method
The undetermined coefficient method is an important method to solve the deformation of algebraic constants. If the letter frame after algebraic deformation can be determined, but the letter coefficient is too high to be determined, we can first express the letter coefficient with unknowns, and then list n equations (groups) with special coefficients according to the properties of polynomial constants, and solve this equation (group) to get the undetermined coefficients. The undetermined coefficient method is widely used, and only some applications of its factorization are studied here.
Example 7 Factorization: 2a2+3ab-9b2+ 14a+3b+20.
The analysis belongs to the quadratic six-term formula, and double cross multiplication can also be considered. Here we use the undetermined coefficient method.
Firstly, decompose 2a2+3ab+9b2=(2a-3b)(a+3b).
The solution can be set to the original formula =(2a-3b+m)(a+3b+n).
= 2 a2+3 ab-9 B2+(m+2n)a+(3m-3n)b+ Mn…………
Compare the coefficients of two polynomials (original formula and * formula)
m+2n= 14( 1)m=4
3m-3n=-3(2)= >
mn=20(3)n=5
∴ Original formula =(2x-3b+4)(a+3b+5)
Note that for formula (*), because the equations in which a and b take arbitrary values are valid, it is also possible to find m and n by the special value method.
Let a= 1, b=0, m+2n= 14m=4.
= & gt
Let a=0, b= 1, m=n=- 1n=5.
2.9 factorial theorem, comprehensive division factorization
For integral coefficient unary polynomial f (x) = anxn+an-1xn-1+…+a1x+A0.
According to the factorial theorem, it can be judged whether there is a first-order factorial (x-) (where P and Q are coprime), where P is the divisor of the first coefficient an and Q is the divisor of the last coefficient a0.
If f()=0, there must be (x-) to divide the polynomial synthetically.
Example 8 Factorization Factor x3-4x2+6x-4
This is an integer coefficient unary polynomial, because the positive divisor of 4 is 1, 2,4.
The possible factors are x1,x 2, x 4,
∫f( 1)≠0，f( 1)≠0
But f(2)=0, so (x-2) is the factor of this polynomial, and then divide by synthesis.
2 1-46-4
2-44
1-220
So the original formula =(x-2)(x2-2x+2).
Of course, this problem can also be decomposed, such as x3-4x2+4x+2x-4.
=x(x-2)2+(x-2)
=(x-2)(x2-2x+2)
There are many ways to decompose factors, and their methods are interrelated. A problem is likely to be solved by multiple methods at the same time. So after knowing these methods, we must pay attention to the flexible use of various methods and firmly grasp them!