There are two situations in this method.

① factorization of x2+(p+q) x+pq formula.

The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of 1: x2+(p+q) x+pq = (x+p) (x+q).

Example 1: x2-2x-8

=(x-4)(x+2)

② Factorization of KX2+MX+N formula.

If k=ab, n=cd and ad+bc=m, then KX2+MX+N = (AX+C) (BX+D).

Example 2: Decomposition 7x2- 19x-6

The diagram is as follows: a= 1 b=7 c=2 d=-3.

Because -3× 7 =-2 1, 1× 2 = 2, and -2 1+2=- 19,

So the original formula = (7x+2) (x-3).

Formula of cross multiplication: divide by quadratic term, divide by constant term, cross multiply and sum to get linear term.

Example 3: 6x2+7x+2

The quadratic term (6X2) of the term 1 is split into 2x3.

The third constant term (2) is divided into: 1×2.

2(X) 3(X)

1 2

Diagonal multiplication: the second linear term of 1×3+2×2 (7X)

Multiply vertically and add horizontally.

Correspondingly, there is also the binary multiplication, which can also be learned.

Removing and adding methods

This method refers to disassembling one term of a polynomial or filling two (or more) terms that are opposite to each other, so that the original formula is suitable for decomposition by improving the common factor method, using the formula method or grouping decomposition method. It should be noted that the deformation must be carried out under the principle of equality with the original polynomial.

For example: bc(b+c)+ca(c-a)-ab(a+b)

=bc(c-a+a+b)+ca(c-a)-ab(a+b)

= BC(c-a)+BC(a+b)+ca(c-a)-ab(a+b)

= BC(c-a)+ca(c-a)+BC(a+b)-ab(a+b)

=(bc+ca)(c-a)+(bc-ab)(a+b)

=c(c-a)(b+a)+b(a+b)(c-a)

=(c+b)(c-a)(a+b)。

Method of completing a square

For some polynomials that cannot be formulated, they can be fitted in a completely flat way, and then factorized by the square difference formula. This method is called matching method. It belongs to the special case of the method of splitting items and supplementing items. It should also be noted that the deformation must be carried out under the principle of equality with the original polynomial.

For example: x2+3x-40

=x2+3x+2.25-42.25

=(x+ 1.5)2-(6.5)2

=(x+8)(x-5)。

Root test method

For the polynomial f(x), if f(a)=0, then f(x) must contain the factor x-a. 。

For example, if f(x)=x2+5x+6 and f(-2)=0, it can be determined that x+2 is a factor of x2+5x+6. (Actually, it is x2+5x+6 = (x+2) (x+3). )

Note: 1. For polynomials whose coefficients are all integers, if x = q/p(p (when p and q are coprime integers), the polynomial value is zero, then q is the divisor of constant terms and p is the divisor of the highest degree.

2. For the polynomial f (a) = 0, where b is the highest coefficient and c is a constant term, then a is the divisor of C/B..

Alternative method

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. This method is called substitution method. Note: don't forget to return the RMB after exchange.

For example, if you decompose (x2+x+1) (x2+x+2)-12, you can make y=x2+x, then

The original formula =(y+ 1)(y+2)- 12.

= y2+3y+2- 12 = y2+3y- 10

=(y+5)(y-2)

=(x2+x+5)(x2+x-2)

=(x2+x+5)(x+2)(x- 1)。

Synthetic division

Let the polynomial f(x)=0 and find its roots as x 1, x2, x3, ..., xn, then the polynomial can be decomposed into f (x) = a (x-x1) (x-x2) (x-x3) ... (x-xn).

For example, when decomposing 2x4+7x3-2x2- 13x+6, let 2x4 +7x3-2x2- 13x+6=0.

By comprehensive division, the roots of the equation are 0.5, -3, -2, 1.

So 2x4+7x3-2x2-13x+6 = (2x-1) (x+3) (x+2) (x-1).

Let y=f(x) be the image of function y=f(x), and find the intersection points of function image and x axis X 1, X2, X3...Xn, then the polynomial can be factorized into f (x) = f (x) = a (x-x1) (x

Compared with ⑼ method, it can avoid the complexity of solving equations, but it is not accurate enough.

Principal component method

For example, in the decomposition of x3+2x2-5x-6, y=x3+2x2-5x-6 can be made.

Make an image, and the intersection with the X axis is -3,-1, 2.

Then x3+2x2-5x-6 = (x+1) (x+3) (x-2).

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.

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.

For example, in the decomposition of x3+9x2+23x+ 15, let x=2, then

x3+9 x2+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 x3+9x2+23x+ 15 may be equal to (x+ 1)(x+3)(x+5), which is true after verification.

method of undetermined coefficients

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.

For example, when we decompose x4-x3-5x2-6x-4, we can see that this polynomial has no primary factor, so it can only be decomposed into two quadratic factors.

So let x4-x3-5x2-6x-4 = (x2+ax+b) (x2+CX+d).

Correlation formula

= x4+(a+c)x3+(AC+b+d)x2+(ad+BC)x+BD

From this, you can get

a+c=- 1，

ac+b+d=-5，

ad+bc=-6，

bd=-4。

The solutions are a= 1, b= 1, c=-2 and d =-4.

Then x4-x3-5x2-6x-4 = (x2+x+1) (x2-2x-4).

You can also see the picture on the right.