Middle School Mathematics Factorization of polynomials

1。Extract the common multiple of 2 to obtain 2 (a -1 4b) = 2 (a + 1 2b) (a-1 2b).

2.The formula for the square difference of the swap position yields (3(2x-y)+2(x+2y))(3(2x-y)-2(x+2y))=(8x+y)(4x-7y).

3.The formula for square difference can be obtained as (x+2y)(x-2y)=-64The same as the third question, the equation (4m+n)(-2m+3n)=-900 can be obtained

1.Square the constant, [(2 under the root) a+b (2 under the root)]*2 under the root) a-b (2 under the root)].

2.[3(2x-y)+2(x+2y)]*3(2x-y)-2(x+2y)]=(8x+y)(4x-7y)

4.（m+2n）²-3m-n）²=[(m+2n)+(3m-n)]*m+2n)-(3m-n)]=(4m+n)(3n-2m)=90*(-10)=-900

3 and 4 can also be done by solving equations directly without factoring.

In the case of unary elements, the polynomial ax 2+bx+c factoring can be factored with two primary factors as (a1x+b1) (a2x+b2).

The coefficient decomposition containing a term is the quadratic term, and the containing b term is the factorization of the constant term, that is, if the sum of the products of a1b2 and a2b1 is equal to b, then this formula can be applied.

This is the cross multiplication, e.g. factoring 2x 2-3x+1

Use the following method of slow airing:

Then 2*(-1)+1*(-1) is exactly equal to the coefficient of the primary term 3, so the original formula is decomposed into (2x-1)(x-1).

2x -1 = 2x-1)

x -1 = x-1)

As for how to see it, it mainly depends on the number sense, but usually the problem will not let the coefficient be greater than 15, when the absolute value of the primary term coefficient is a prime number, the general quadratic term coefficient or the scrambling blind constant term coefficient will be divided into a 1 or -1 out.

The basic way to factor a polynomial is to split the polynomial into two parentheses, each with a factor inside it, and each factor must have a root.

Common technique: Extract the common factor method.

Formula method. Cross multiplication.

Pending coefficient method.

Root-seeking method. Group decomposition method.