Who s going to break down the factoring problem?

It is to take the sum or difference of several algebraic formulas, to and fro, to and fro, and to make a product.

Here are a few simple examples.

If it is broken down, it is a*(b+1).

2.Using the formula a2-b2=(a+b)*(a-b), note that it must be broken down to the point that it can no longer be decomposed.

I'll add you, this is the safest way!

Formula method (sum of squares formula, square difference formula), cross multiplication (the most important), matching method (troublesome, beginners can give it a try).

Use the perfect square formula for the difference of two numbers.

A 2-2AB+B 2=(A-B) 2

e x + e (a x) a 2

e (x 2)] 2 + [e (a x 2)] 2 a 2e (x 2) xe (a 2).

e (x 2) e (one x 2)] 2.

Here's how, please refer to:

This is the same as the general division, the dividend and the divisor are written according to the power of the unknown number, and then the highest order of the dividend is divided by the highest order of the divisor as the quotient, and so on. As shown below:

<> by assembling the terms with the highest power, subtracting them, and calculating them one level at a time, the factorization is completed.

A polynomial is decomposed in a range (e.g., in a range of real numbers, i.e., all terms are real) into the form of the product of several integers.

You can keep getting rid of it. This is followed by +x+1

Even if you can't divide it, x-x-2 can be factored (x+1) (x-2), and the later ones can be calculated.

As a very important identity deformation problem in middle school mathematics, factorization is used to solve a complex problem by factoring the :i l:x,-l- polynomials.

The problem of factorization, I think that in general, factorization may be some junior high school problems, and this must be in the formula. It's very simple to put the common factor first, and then set the formula later.

The process of solving the problem is as follows:

Original = x 2+2x + 1-3-3x

x+1)^2-3-3x

x+1)^2-3(1+x)

x+1)(x+1)-3(x+1)

x+1)(x-2)

For the sake of visual clarity, let an=x,a(n+1)=y, then the original equation becomes x -(2y-1)x-2y=0

x²-2xy+x-2y=0

x²+x-(2xy+2y)=0

x(x+1)-2y(x+1)=0

x-2y)(x+1)=0

So (an-2a(n+1))(an +1)=0

Solution: (m+2n) -3m-n) =(m+2n+3m-n)(m+2n-3m+n)=(4m+n)(3n-2n)(this is the result of the squared difference formula) and it is known that 4m+n=90,2m-3n=10, so 3n-2n=-10, so (m+2n)-3m-n) =(4m+n)(3n-2n)=90 (-10)=-900

Multiply the cross to get (a 2-9)(a 2+3)=(a-3)((a+3)(a 2+3) If you still don't understand, you can consult me online.

Cross multiplication is a very useful mathematical tool that must be learned.