What is the cross thought multiplication of factorization?

In some quadratic trinomials, the coefficients of the first term and the third term can be decomposed into the product of two numbers respectively, and then the quadratic trinomial formula can be factored with the help of the method of drawing cross lines, which is called cross multiplication

1 1 = 1 (quadratic coefficient).

ab=ab (constant term).

1 a+1 b = a+b (primary term coefficient).

It is necessary to put a quadratic trinomial formula with a coefficient of quadratic term that is not 1.

When factoring the equation: if the constant term q is positive, then decompose it into two homogeneous factors whose sign is the same as the sign of the primary term coefficient p

If the constant term q is negative, then decompose it into two heterogeneous factors, where the factor with the greater absolute value has the same sign as the coefficient p of the primary term

For the two factors that are decomposed, it is also necessary to see whether their sum is equal to the coefficient p of the primary term: cross multiplication.

1)x2-6x-7

2)x2+6x-7

3)x2-8x+7

4)x2+8x+7

5)x2-5x+6

6)x2-5x-6

7)x2+5x-6

8)x2+5x+6

Solution: (1)x2-6x-7=(x-7)(x+1)2)x2+6x-7=(x+7)(x-1).

3)x2-8x+7=(x-7)(x-1)

4)x2+8x+7=(x+7)(x+1)

5)x2-5x+6=(x-2)(x-3)

6)x2-5x-6=(x-6)(x+1)

7)x2+5x-6=(x+6)(x-1)

8)x2+5x+6=(x+2)(x+3)

The square of x + ax + ab can be broken down into the form of (x + a) (x + b), which is the cross multiplication.

Cross multiplication.

It is a special method among several methods of factorization, under certain conditions, the speed of solving the problem by cross multiplication is relatively fast, saving time and avoiding a large number of operations, and it is not easy to make mistakes.

1. The concept of cross multiplication.

The method of cross decomposition is simply as follows: the multiplication of the left side of the cross equals the quadratic coefficient.

The right multiplication equals the constant term.

Cross multiplication and then addition equals the primary term coefficient. In fact, it is the use of multiplication formulas.

x+a)(x+b)=x2+(a+b)x+ab.

2. The general steps of cross-multiplication factorization.

1) Decompose the coefficient of the quadratic term and the constant term into factors respectively;

2) Try the cross diagram to make a cross over the cross line.

The sum of the numbers obtained after multiplication is the coefficient of the primary term;

3) Determine a suitable cross diagram and write the results of factorization;

4) Inspection. Factorization of a polynomial cross multiplication with a quadratic term coefficient of 1.

Factorization of a polynomial cross multiplication with a quadratic term coefficient of 1.

Factorization of polynomial cross-multiplication with a coefficient of less than 1.

Factorization of polynomial cross-multiplication with a coefficient of less than 1.

3. The mantra of cross multiplication.

Decompose from beginning to end, cross multiply, sum and make up, and write in parallel. The vertical division constant is cross-checked, and the horizontal factor cannot be messed up.

1) Cross-examination of vertical division constants:

Vertically divide quadratic terms and constant terms, that is, write the coefficients of quadratic terms and constant terms vertically;

Cross multiplication, and addition, that is, oblique multiplication and then addition, to obtain a term coefficient;

The test is confirmed, and the coefficient of the term is checked to be correct.

2) The horizontal factor should not be messed with.

That is, write the factor horizontally, not crosswise, and you can't mess with it here.

Cross multiplication is one of the fourteen methods in factorization, which is mainly used for factoring polynomials, and the basic formula: x + (p + q) x + pq = (x + p) (x + q).

The method of cross decomposition method is simply as follows: the left side of the cross is multiplied equals the quadratic term coefficient, the right side is multiplied equals the constant term, and the cross multiplication and then added is equal to the primary term coefficient, in fact, it is the inverse operation of the multiplication formula (x+a)(x+b)=x +(a+b)x+ab to carry out factorization.

Decompose with the coefficient characteristics of the factor.

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That's a question. I don't know how many ...

For example, (x+2)(2x-1)=0 2x +3x-2=0x: multiply the coefficient of x by 1 2=2

Coefficient of x: 2 2+1 (-1)=3

The last constant: 2 (-1) = -2

Just do a few more.

Cross multiplication: The left side of the cross is multiplied equals the quadratic coefficient, the right side is multiplied equals the constant term, and the cross multiplication and then added is equal to the primary term coefficient.

For example: 6x

11x-10=0 handfuls.

6 is divided into 2 and.

3. Divide -10 into 5 and.

3-2 crosses multiplied: 3

4=11 satisfies the equation. i.e.: 6

x²+11x-10=

2x+5(3x-

So, there is: 2x

x1=-5/

23x-2=0x2

The above is the detailed process.

That's a question. I don't know how many ...

For example, (x+2)(2x-1)=0 2x +3x-2=0x: multiply the coefficient of x by 1 2=2

Coefficient of x: 2 2+1 (-1)=3

The last constant: 2 (-1) = -2

Just do a few more.

The method of cross multiplication is simply as follows: the left side of the cross is multiplied equals the quadratic term, the right side multiplied is equal to the constant term, and the cross multiplication and addition are equal to the primary term. In fact, it is to use the inverse operation of the multiplication formula (x+a)(x+b)=x + (a+b)x+ab to factorize.

Cross multiplication can factor certain quadratic trinomials. For an integer of the form ax +bx+c=(a1x+c1)(a2x+c2), the key to the method is to decompose the quadratic term coefficient a into the product a1·a2 of the two factors a1 and a2, and decompose the constant term c into the product c1·c2 of the two factors c1 and c2, and make a1c2+a2c1 exactly the coefficient b of the primary term, then the result can be written directly.

ax²+bx+c=(a1x+c1）(a2x+c2）。When using this method to decompose a factor, it is important to observe, try, and realize that it is essentially the inverse of binomial multiplication.

When the first coefficient is not 1, it often takes several tests, and it is important to pay attention to the symbols of each coefficient. The basic formula: x +(p+q) +pq=( +p)( q).

Cross multiplication.

x^2＋（p

Factorization of the formula of type q) x pq.

The characteristics of this type of quadratic trinomial are that the coefficient of the quadratic term is 1; The constant term is the product of two numbers; The coefficient of the primary term is the sum of the two factors of the constant term. Therefore, it is possible to factor a quadratic trinomial with a coefficient of 1 for some quadratic terms:

x^2＋（p

q）x＋pq＝（x＋p）（x＋q）

Such as x-square-14+48x-8

x-6 so = (x-8) (x-6) the last two numbers multiplied = 48 added = -14

Take a look.

For example, x +7x+12

Observe that the coefficients in front of x are 17 .12

Split 1 into 1 1

12 split into 3 4

The cross-multiplication and addition should be equal to the coefficient before the first term, here it is 7 just 3+4=7, and the one written on the left is expressed in x, and the positive number on the right is added, and the negative number is subtracted, that is, (x+3) (x+4).

It's done.

The solution to that problem is much clearer, and it is difficult to say without examples.

It's a matter of rules. If you can divide 12 into other numbers and multiply them and follow the rules, then that's fine.

m²+4m-12=(m-2)(m+6)

The multiplication of the two 1s in the first column of this array is equal to the coefficient of the quadratic term of the equation, and the multiplication of -2 and 6 in the second column is equal to the constant term, and the cross-multiplication and addition, i.e., 1 6+1 (-2), if equal to the coefficient of the primary term, then such a distribution is correct, and the original formula can be factored to obtain (m-2) (m+6).

If so distributed.

In this way, both the quadratic coefficient and the constant term can be satisfied, but 1 4+1 (-3)=1 cannot meet the coefficient of the primary term, so the original formula cannot be decomposed into (m-3) (m+4).

This kind of problem requires you to practice for a long time, which can save time to factorize.

= cannot be matched into 3 and 4 The equation becomes (m+3)(m+4)=m +7m+12

Cross multiplication is actually to take out the constant terms and match.

And make the sum of the numbers equal to the coefficient of the primary term. For example, 12 is split into -2 and 6 because -2+6=4

To express it alphabetically, m squared + n + x squared, and m and x are split into their respective factors m1, m2, x1, x2 is equal to.

m1 times m2) + (m1 times x2 + m2 times x1) + (x1 times x2).