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How to calculate the cross multiplication in mathematics? Thank you.
1, the method of cross multiplication:

For the unary quadratic equation about x: ax? +bx+c=0。 (a≠0)

The method of cross multiplication is to separate the quadratic coefficient A from the constant term C. 。

That is, the quadratic term (x? ) is divided into m and s,

Decomposition of constant term c into n and t,

So that the left multiplication of the split cross is equal to the quadratic coefficient, that is, ms=a,

The multiplication on the right is equal to the constant term, that is, nt=c,

The cross product plus equals the first-order coefficient, that is, mt+ns = b.

(hence the name crossover method)

Common formats are:

After being taken apart.

Constant term of quadratic coefficient

m n

╳ t

How to divide the coefficient depends on the coefficient of the first term. As long as you can get the coefficient of the first term, there is no formula. )

Like x? +2x+ 1=0,

Quadratic equation is not ax? +bx+c=0?

Corresponding to this equation, a= 1, b=2, c= 1.

The so-called crossing is to dismantle a and C.

Here a= 1 is relatively easy to calculate, so it is naturally split into 1× 1.

C is also divided into 1× 1.

Can be written as:

1 1

1 ╳ 1

From/kloc-0 /×1+/kloc-0 /×1= 2.

That is, the cross multiplication and addition (/kloc-0 /×1+/kloc-0 /×1) is equal to the linear term coefficient of 2,

So the original equation can be transformed into (x+ 1)(x+ 1)=0.

Another example is:

Example 1 M? +4m- 12 factorization factor

Analysis: The constant term-12 in this question can be divided into-1× 12, -2×6, -3×4, -6×2,-12× 1.

When-12 is divided by -2×6, this problem is met.

Solution: Because

1 -2

1 ╳ 6

From 1×6+ 1×(-2)=4.

That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.

So m? +4m- 12=(m-2)(m+6)

Example 2 Handle 5x? +6x-8 factorization factor

Analysis: 5 in this question can be divided into 1×5.

-8 can be divided into-1×8, -2×4, -4×2, -8× 1.

When the quadratic term coefficient 5 is divided by 1×5, and the constant term -8 is divided by -4×2, this problem is met.

Solution: Because

1 2

5 ╳ -4

From 1×(-4)+5×2=6.

That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.

So 5x? +6x-8=(x+2)(5x-4)

Example 3 solving equation x? -8x+ 15=0

Analysis: put x? -8x+ 15 is regarded as a quadratic trinomial about x, then1× 15, 3×5, (-1)×(- 15), (-)

When 15 is divided by (-3)×(-5), this problem is met.

Solution: Because

1 -3

1 ╳ -5

From 1×(-5)+ 1×(-3)=-8.

That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.

So the original equation can be transformed into (x-3)(x-5)=0.

So x 1=3 x2=5.

Example 4. Solve equation 6x? -5x-25=0

Analysis: put 6x? -5x-25 is regarded as a quadratic trinomial about x,

Then 6 can be divided into 1×6, 2×3,

-25 can be divided into-1×25,

-5×5,-25× 1。

When the quadratic coefficient 6 is divided by 2×3 and the constant term -25 is divided by -5×5, this problem is met.

Solution: Because

2 -5

3 ╳ 5

From 2×5+3×(-5)=-5.

That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.

So the original equation can be changed to (2x-5)(3x+5)=0.

So x 1=5/2 x2=-5/3.

To put it bluntly, it is to make up four numbers and meet the following requirements: cross left multiplication equals quadratic term coefficient, right multiplication equals constant term, and cross multiplication plus equals linear term coefficient.

2, the use of cross multiplication:

(1) uses cross multiplication to decompose the factor. (2) Solving the quadratic equation with one variable by cross multiplication.

3. Advantages of cross multiplication:

Using cross multiplication to solve problems is faster, can save time, and the calculation is not large, so it is not easy to make mistakes.

4, cross method using conditions:

The coefficient is not complicated and easy to decompose.

(but not all of them can be used, such as 4x? +2x+ 1, however matched, is not equal to 2.

At this time, the root formula should be used for the corresponding factorization. )