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)? Solution: 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. Decomposition factor A+4A B+4B (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, factorization 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

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 7x2-19x-6 analysis: 1 -3 7 2

2-2 1=- 19

Solution: 7x2-19x-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 x2 +3x-40 solves x2+3x-40 = x+3x+()-()-40 = (x+)-() = (x+) (x+) = (x+) (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)