Methods and techniques for factoring, what are methods and techniques for factoring?

xy+y-9x-9

y(x＋1)－9(x＋1)

y－9)(x＋1)

The above is a grouped factoring method.

Group Decomposition Method Group decomposition is a complex method of factorization, let's learn about it.

There are four or six terms or greater than four terms of equations that can be grouped and decomposed, and there are two forms of general grouping decomposition: binary and trianth.

For example: dichotomy:

ax+ay+bx+by

ax+ay)+(bx+by)

a(x+y)+b(x+y)

a+b)(x+y)

We put ax and ay into a group, bx and by into a group, and used the multiplicative distributive property to match the two pairs, which immediately solved the difficulty.

Again, this question can do the same. Exchange the other two identical ones for :

ax+ay+bx+by

ax+bx)+(ay+by)

x(a+b)+y(a+b)

a+b)(x+y)

Trinity division: 2xy-x 2+1-y 2

x^2+2xy-y^2+1

x^2-2xy+y^2)+1

1-(x-y)^2

1+x-y)(1-x+y)

Let's do a few practice questions in this paragraph:

Solution: =5x(a+b)+3y(a+b).

5x+3y)(a+b)

Explanation: The coefficients can be grouped and decomposed as the same, as above, 5ax and 5bx are regarded as a whole, and 3ay and 3by are regarded as a whole, which can be easily solved by using the multiplicative distributive property.

2. x^3-x^2+x-1

Solution: =(x 3-x 2)+(x-1).

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

x-1)(x^2+1)

Using the dichotomy method, the common factor method proposes x 2, and then conjuncts to solve it easily.

3. x^2-x-y^2-y

Solution: =(x 2-y 2)-(x+y).

x+y)(x-y)-(x+y)

x+y)[(x-y)-1]

x+y)(x-y-1)

The binary dichotomy is used, and then the formula method A 2-b 2=(a+b)(a-b), and then the combination is solved.

Exercises: 1) 18a 2-32b 2-18a+24b

2） x^2-25+y^2-2xy

3） y^4-4y^3+4y^2-1

4） 4a^2-b^2-4c^2+4bc

2） (x-y+5)(x-y-5)

3） (y^2-2y-1)(y-1)^2

4）(2a+b-2c)(2a-b+2c)

To decompose a factor, it is to find a common factor in the variable, so that it can come up with a common factor, like this one, it is not difficult to see that the common factor is (y-1).

What are the ways to defactor factors?

1. Mention the common factor method.

The common factors that are contained in the terms of several polynomials are called the common factors of the terms of the polynomial. If the items of a polynomial have a common factor, you can propose this common factor to reduce the polynomial into the form of the product of two factors, and this method of factoring is called the common factor method.

2. Formula method.

If you reverse the multiplication formula, you can factor some polynomials, which is called the formula method.

Precautions. 1. The left side of the equation must be a polynomial;

2. The result of factoring must be expressed in the form of a product;

3. Each factor must be an integer, and the number of each factor must be lower than the number of the original polynomial;

4. Decompose the factors until each polynomial factor can no longer be decomposed.

Cross multiplication, undetermined coefficient method, double cross multiplication, symmetric polynomial, rotational symmetric polynomial method, coincidence theorem method, and there is no universally applicable method for finding the common factor decomposition. In the competition, there are also splitting and adding and subtracting terms, changing the element method, long division, short division, division, etc.

Note three principles:

1. The decomposition should be thorough (whether there is a common factor and whether a formula can be used).

2. The final result is only parentheses.

3. In the final result, the coefficient of the first term of the polynomial is positive (for example: -3x2+x=x(-3x+1)), but the first term is not necessarily positive, such as -2x-3xy-4xz=-x(2+3y+4z).

1）（a+b）³＝a³＋3a²b＋3ab²＋b³

2）a³+b³=a³+a²b-a²b+b³=a²（a+b）-b（a²-b²）=a²（a+b）-b（a+b）（a-b）

a+b）[a²-b（a-b）]=a+b）（a²-ab+b²）

3）a³-b³=a³-a²b+a²b-b³=a²（a-b）+b（a²-b²）=a²（a-b）+b（a+b）（a-b）

a-b）[a²+b（a+b）]=a-b）（a²+ab+b²）

1. Mention the common factor method

If all the terms of a polynomial contain a common factor, then the common factor can be proposed and the polynomial can be transformed into the product of two factors.

Example: Decompose the factor x2 -2x -x, x -2x -x=x (x -2x-1).

Second, the application of the formula method

Since factorization has an inverse relationship with integer multiplication, if the multiplication formula is reversed, then it can be used to factor certain polynomials. For example, the square of the sum and the square of the difference.

Example: Decompose the factor a +4ab+4b, a +4ab+4b = a+2b).

3. Group decomposition method

To decompose the factored pose of the polynomial am+an+bm+bn, you can first divide its first two terms into a group and propose the common factor a, divide its last two into a group, and propose the common factor b, so as to obtain a(m+n)+b(m+n), and you can propose the common factor m+n, so as to obtain (a+b)(m+n).

Example: Decompose the factor m2+5n-mn-5m, m2+5n-mn-5m= m2-5m-mn+5n, (m-5m)+mn+5n), m(m-5)-n(m-5), (m-5)(m-n).

The methods and techniques of factorization are as follows:

Factorization is not difficult, the decomposition method should be remembered, if there is a common factor for each item, the first extraction is not slow, if there is no common factor for each item, apply the formula to test.

If it's a binomial, the squared difference formula is ahead, and if it's a trinomial, it's perfectly squared.

Whole, none of the above methods will work, use the group to take a look, in the face of quadratic trinomials, cross multiplication to find convenience, can be decomposed and then divided.

Solution, which cannot be decomposed is the answer.

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

This sub-deformation of the formula is called factorization of this polynomial, also known as factoring this polynomial.

Break down the general steps.

1. If the first term of the polynomial is negative, the negative sign should be extracted first;

The "negative" here refers to the "negative sign". If the first term of the polynomial is negative, a negative sign is generally proposed so that the coefficient of the first term in parentheses is positive.

2. If each item of the polynomial contains a common factor, then the common factor is extracted first, and then the factor is further decomposed;

Note: When a whole term of a polynomial is a common factor, after proposing the common factor first, do not omit 1 in parentheses; Mention the common factor should be cleaned at once, and the polynomial in each parenthesis can no longer be decomposed.

3. If there is no common factor for each item, then you can try to use formulas and cross multiplication to decompose them;

4. If the above methods cannot be decomposed, try to decompose by grouping, splitting items, and supplementing items.

Formula: first mention the first negative sign, then see if there is a common factor, and then see if you can set the formula, cross multiplication and try it, and the group decomposition should be appropriate.

1.Extract the common factor.

This is the most basic. It's just that if there is a common factor, it will be brought up, and everyone will know this, so I won't say much.

2.Perfectly squared.

a^2+2ab+b^2=(a+b)^2

a^2-2ab+b^2=(a-b)^2

If you see that there are two numbers squared in the formula, you should pay attention to it, find out if there is twice the product of the two numbers, and if so, follow the above formula.

3.Square Difference Formula.

a^2-b^2=(a+b)(a-b)

This should be memorized, because it is possible to add terms when matching perfect squares, and if the front is perfectly squared, and then subtract a number, you can use the square difference formula to break it down.

4.Cross multiplication.

x^2+(a+b)x+ab=(x+a)(x+b)

This one is very practical, but it is not easy to use.

When the above method cannot be used to decompose, the lower cross multiplication method can be used.

Example: x 2 + 5 x + 6

First of all, it is observed that there are quadratic terms, primary terms, and constant terms, which can be multiplied by crosses.

The coefficient of the primary term is 1So it can be written as 1*1

The constant term is 6It can be written as 1*6, 2*3, -1*-6, -2*-3 (decimals are not recommended).

Then arrange it like this.

The positions of the following columns can be reversed, as long as the product of these two numbers is a constant term).

Then multiply diagonally, 1*2=2, 1*3=3Add the product again. 2+3=5, which is the same as the coefficient of the primary term (it may not be equal, so you should try another time at this time), so it can be written as (x+2) (x+3) (at this time, it will be done horizontally).

I'll write a few more formulas, and the landlord will figure it out for himself.

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

2x^2+5x-12=(2x-3)(x+4)

In fact, the most important thing is to use it yourself, the above methods can actually be used together, and practice is always better than teaching others.

By the way. If the b 2-4ac of an equation is less than 0, the formula cannot be decomposed in any way (in the range of real numbers, b is the coefficient of the first term, a is the coefficient of the quadratic term, and c is the constant term).

These methods are generally applicable when the highest order is secondary!

What are the ways to defactor factors?