Transforming a polynomial into the product of several algebraic expressions is called factorization of this polynomial. There are many methods of factorization, which are summarized as follows:
1, Public Welfare Law
If every term of a polynomial contains a common factor, then this common factor can be put forward, so that the polynomial can be transformed into the product of two factors.
Example 1, factorization factor x-2x-x (Huai' an senior high school entrance examination in 2003)
x -2x -x=x(x -2x- 1)
2. Application of formula method
Because factorization and algebraic expression multiplication have reciprocal relationship, if the multiplication formula is reversed, it can be used to decompose some polynomials.
Example 2, factorization factor a +4ab+4b (2003 Nantong senior high school entrance examination in 2003)
Solution: a +4ab+4b =(a+2b)
3. Grouping decomposition method
Factorizing the polynomial am+an+bm+bn, we can first divide the first two terms into a group and propose the common factor A, then divide the last two terms into a group and propose the common factor B, so as to get a(m+n)+b(m+n), and we can also propose the common factor M+N, so as to get (a+b) (m+).
Example 3. Decomposition factor m +5n-mn-5m
Solution: m +5n-mn-5m= m -5m -mn+5n.
= (m -5m )+(-mn+5n)
=m(m-5)-n(m-5)
=(m-5)(m-n)
4. Cross multiplication
For a polynomial in the form of mx +px+q, if a×b=m, c×d=q and ac+bd=p, the polynomial can be factorized into (ax+d)(bx+c).
Example 4, factorization factor 7x-19x-6
Analysis: 1 -3
7 2
2-2 1=- 19
Solution: 7x-19x-6=(7x+2)(x-3)
5. Matching method
For those polynomials that cannot be formulated, some can use it to make a completely flat way, and then factorize it with the square difference formula.
Example 5, Factorization Factor x +3x-40
Solution x +3x-40=x +3x+() -() -40
=(x+ ) -()
=(x++)(x++)
=(x+8)(x-5)
6. Removal and addition methods
Polynomials can be divided into several parts and then factorized.
Example 6: Decomposition factor bc(b+c)+ca(c-a)-ab(a+b)
Solution: BC (B+C)+CA (C-A)-AB (A+B) = BC (C-A+A+B)+CA (C-A)-AB (A+B).
= BC(c-a)+ca(c-a)+BC(a+b)-ab(a+b)
=c(c-a)(b+a)+b(a+b)(c-a)
=(c+b)(c-a)(a+b)
7. Alternative methods
Sometimes in factorization, you can choose the same part of the polynomial, replace it with another unknown, then factorize it and finally convert it back.
Example 7, factorization factor 2x -x -6x -x+2
Solution: 2x-x-6x-x+2 = 2 (x+1)-x (x+1)-6x.
=x [2(x + )-(x+ )-6
Let y=x+, x [2(x+)-(x+ )-6.
= x [2(y -2)-y-6]
= x (2y -y- 10)
=x (y+2)(2y-5)
=x (x+ +2)(2x+ -5)
= (x +2x+ 1) (2x -5x+2)
=(x+ 1) (2x- 1)(x-2)
8. Root method
Let the polynomial f(x)=0 and find its roots as x, x, x, …x, ... x, then the polynomial can be decomposed into f (x) = (x-x) (x-x)...(x-x).
Example 8, factorization factor 2x +7x -2x-13x+6.
Solution: Let f(x)=2x +7x -2x-13x+6=0.
According to the comprehensive division, the roots of f(x)=0 are -3, -2, 1.
Then 2x+7x-2x-13x+6 = (2x-1) (x+3) (x+2) (x-1).
9. Mirror image method
Let y=f(x), make the image of function y=f(x), and find the intersection points x, x, x, …x, ... x between the image of function and the X axis, then the polynomial can be decomposed into f (x) = f (x) = (x-x) (x-x) ... (ten
Example 9, Factorization x +2x -5x-6
Solution: Let y= x +2x -5x-6.
Make it image, as shown on the right, and the intersection with the X axis is -3,-1, 2.
Then x+2x-5x-6 = (x+1) (x+3) (x-2).
10, principal component method
First, choose a letter as the main element, then arrange the items from high to low according to the number of letters, and then factorize them.
Example 10, factorization factor a (b-c)+b (c-a)+c (a-b)
Analysis: this question can choose a as the main element, arranged from high to low.
Solution: a (b-c)+b (c-a)+c (a-b) = a (b-c)-a (b-c)+(b c-c b)
=(b-c) [a -a(b+c)+bc]
=(b-c)(a-b)(a-c)
1 1, using the special value method.
Substitute 2 or 10 into x, find the number p, decompose the number p into prime factors, properly combine the prime factors, write the combined factors as the sum and difference of 2 or 10, and simplify 2 or 10 into x, thus obtaining factorization.
Example 1 1, factorization factor x +9x +23x+ 15.
Solution: let x=2, then x+9x+23x+15 = 8+36+46+15 =105.
105 is decomposed into the product of three prime factors, namely 105=3×5×7.
Note that the coefficient of the highest term in the polynomial is 1, while 3, 5 and 7 are x+ 1, x+3 and x+5, respectively, when x=2.
Then x+9x+23x+15 = (x+1) (x+3) (x+5).
12, undetermined coefficient method
Firstly, the form of factorization factor is judged, then the letter coefficient of the corresponding algebraic expression is set, and the letter coefficient is calculated, thus decomposing polynomial factor.
Example 12, decomposition factor x -x -5x -6x-4.
Analysis: It is easy to know that this polynomial has no first factor, so it can only be decomposed into two quadratic factors.
Solution: let x -x -5x -6x-4=(x +ax+b)(x +cx+d)
= x+(a+c)x+(AC+b+d)x+(ad+BC)x+BD
So the solution is
Then x-x-5x-6x-4 = (x+x+1) (x-2x-4)