1, the method of cross multiplication: the left multiplication of the cross is equal to the quadratic term coefficient, the right multiplication is equal to the constant term, and the cross multiplication is equal to the linear term coefficient.
2. Use of cross multiplication: (1) Cross multiplication is used to decompose factors. (2) Solving the quadratic equation with one variable by cross multiplication.
Advantages of cross multiplication: it is faster to solve problems with cross multiplication, which can save time and is not easy to make mistakes.
4. Defects of cross multiplication: 1. Some problems are relatively simple to solve by cross multiplication, but not every problem is simply solved by cross multiplication. 2. Cross multiplication is only applicable to quadratic trinomial type problems. 3. Cross multiplication is more difficult to learn.
5. Examples of cross multiplication problem solving:
1), using cross multiplication to solve some simple and common problems.
Example 1 M? 0? 5+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 2.
Solution: Because 1 -2
1 ╳ 6
So m? 0? 5+4m- 12=(m-2)(m+6)
Example 2 Handle 5x? 0? 5+6x-8 factorization factor
Analysis: In this question, 5 can be divided into 1× 5, and -8 can be divided into-1×8, -2×4, -4×2, -8× 1. When the coefficient of quadratic term is divided into 1×5 and the constant term is divided into -4×2, it is consistent with this question.
Solution: Because 1 2
5 ╳ -4
So 5x? 0? 5+6x-8=(x+2)(5x-4)
Example 3 solving equation x? 0? 5-8x+ 15=0
Analysis: put x? 0? 5-8x+ 15 is regarded as a quadratic trinomial about x, so 15 can be divided into 1× 15 and 3×5.
Solution: Because 1 -3
1 ╳ -5
So the original equation can be transformed into (x-3)(x-5)=0.
So x 1=3 x2=5.
Example 4. Solve equation 6x? 0? 5-5x-25=0
Analysis: put 6x? 0? If 5-5x-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.
Solution: Because 2 -5
3 ╳ 5
So the original equation can be changed to (2x-5)(3x+5)=0.
So x 1=5/2 x2=-5/3.
2) Use cross multiplication to solve some difficult problems.
Example 5 14x? 0? 5-67xy+ 18y？ 0? 5-factor factorization factor
Analysis: put 14x? 0? 5-67xy+ 18y？ 0? 5 is regarded as a quadratic trinomial about X, so 14 can be divided into 1× 14, 2×7, 18y? 0? 5 can be divided into y. 18y, 2y.9y and 3y.6y
Solution: Because 2-9 years old
7 ╳ -2y
So 14x? 0? 5-67xy+ 18y？ 0? 5 =(2 to 9 years) (7 to 2 years)
Example 6 10x? 0? 5-27xy-28y？ 0? 5-x+25y-3 factorization factor
Analysis: This question should organize this polynomial into a quadratic trinomial form.
Solution 1, 10x? 0? 5-27xy-28y？ 0? 5 x+25y-3
= 10x？ 0? 5-(27y+ 1)x -(28y？ 0? 5-25 years old +3 years old) 4 -3 years old
7y ╳ - 1
= 10x？ 0? 5-(27y+ 1)x-(4y-3)(7y- 1)
=[2x-(7y- 1)][5x+(4y-3)]2-(7y– 1)
5 ╳ 4y - 3
=(2x -7y + 1)(5x +4y -3)
Note: this question, first put 28y? 0? 5-25y+3 is decomposed into (4y-3)(7y-1) by cross multiplication, and then decomposed into 10x by cross multiplication. 0? 5-(27y+1) x-(4y-3) (7y-1) is decomposed into [2x -(7y-1)][5x +(4y -3)].
Solution 2, 10x? 0? 5-27xy-28y？ 0? 5 x+25y-3
=(2x-7y)(5x+4y)-(x-25y)-3 ^ 2-7y
=[(2x-7y)+ 1][(5x-4y)-3]5╳4y
=(2x-7y+ 1)(5x-4y-3)2x-7y 1
5 x - 4y ╳ -3
Note: For this question, put 10x first. 0? 5-27xy-28y？ 0? Use cross multiplication to decompose 5 into (2x -7y)(5x +4y), and then use cross multiplication to decompose (2x-7y)-(x-25y)-3 into [(2x -7y)+ 1] [(5x -4y)-3].
Example 7: Solve the equation about x: x? 0? 5- 3ax + 2a？ 0? 5–a B- b？ 0? 5=0
Analysis: 2a? 0? 5–a B- b？ 0? Cross multiplication can be used for factorization.
Solution: x? 0? 5- 3ax + 2a？ 0? 5–a B- b？ 0? 5=0
x？ 0? 5- 3ax +(2a？ 0? 5–a B- b？ 0? 5)=0
x？ 0? 5- 3ax +(2a+b)(a-b)=0 1 -b
2 ╳ +b
[x-(2a+b)][x-(a-b)]= 0 1-(2a+b)
1 ╳ -(a-b)
So x1= 2a+bx2 = a-b.