example
Example 1 Factorization 2x 2-7x+3.
Analysis: first decompose the quadratic coefficient and write it in the upper left corner and the lower left corner of the crosshair, then decompose the constant term and divide it into two parts.
Don't write it in the upper right corner and lower right corner of the crosshair, and then cross multiply to find the algebraic sum to make it equal to the coefficient of the first term.
Quadratic coefficient decomposition (positive factor only):
2= 1×2=2× 1;
Decomposition of constant term:
3= 1×3=3× 1=(-3)×(- 1)=(- 1)×(-3).
Draw a cross line to represent the following four situations:
1 1
╳
2 3
1×3+2× 1
=5
1 3
╳
2 1
1× 1+2×3
=7
1 - 1
╳
2 -3
1×(-3)+2×(- 1)
=-5
1 -3
╳
2 - 1
1×(- 1)+2×(-3)
=-7
After observation, the fourth case is correct, because after cross multiplication, the algebraic sum of the two terms is exactly equal to the coefficient of the first term -7.
Solution 2x 2-7x+3 = (x-3) (2x- 1).
Generally speaking, for the quadratic trinomial ax2+bx+c(a≠0), if the quadratic term coefficient A can be decomposed into the product of two factors, that is, a=a 1a2, the constant term C can be decomposed into the product of two factors, that is, c=c 1c2, and A/KLOC-.
a 1 c 1
? ╳
a2 c2
a 1c2+a2c 1
Cross-multiply diagonally, and then add to get a 1c2+a2c 1. If it is exactly equal to the first term coefficient b of the quadratic trinomial ax2+bx+c, that is, a 1c2+a2c 1=b, the quadratic trinomial can be decomposed into two factors a65438+.
ax2+bx+c =(a 1x+c 1)(a2x+C2)。
The way to help us decompose the quadratic trinomial by drawing cross lines like this is usually called cross multiplication.
Example 2 Factorizing 6x 2-7x-5.
Analysis: According to the method of example 1, the quadratic term coefficient 6 and the constant term -5 are decomposed and arranged respectively, and there are eight different arrangement methods, one of which is
2 1
╳
3 -5
2×(-5)+3× 1=-7
Is correct, so the original polynomial can be factorized by cross multiplication.
Solution 6x 2-7x-5 = (2x+ 1) (3x-5)
It is pointed out that through the examples of 1 and 2, it can be seen that when a quadratic trinomial factor whose quadratic coefficient is not 1 is solved by cross integration, it often needs many observations to determine whether the factor can be solved by cross integration.
For the quadratic trinomial with quadratic coefficient of 1, cross multiplication can also be used to decompose the factors. At this time, you only need to consider how to decompose the constant term. For example, if x 2+2x-15 is decomposed, the cross multiplication is
1 -3
╳
1 5
1×5+ 1×(-3)=2
So x 2+2x- 15 = (x-3) (x+5).
Example 3 Factorization 5x 2+6xy-8y 2.
Analysis: This polynomial can be regarded as a quadratic trinomial about x, and-8y 2 is regarded as a constant term. When we decompose the coefficients of quadratic terms and constant terms, we only need to decompose 5 and -8, and then use the crosshairs to decompose them. After observation, we chose a suitable group, namely
1 2
? ╳
5 -4
1×(-4)+5×2=6
Solution 5x 2+6xy-8y 2 = (x+2y) (5x-4y).
It is pointed out that the original formula is decomposed into two linear formulas about x and y.
Example 4 Factorization (x-y)(2x-2y-3)-2.
Analysis: This polynomial is the product of two factors and the difference of another factor. Only by multiplying the polynomials first can the deformed polynomials be factorized.
Q: What are the characteristics of the factorial of the product of two products? What is the simplest method of polynomial multiplication?
A: If the common factor 2 is proposed for the first two items in the second factor, it will become 2(x-y), which is twice that of the first factor. Then multiply (x-y) as a whole, the original polynomial can be transformed into a quadratic trinomial about (x-y), and the factor can be decomposed by cross multiplication.
Solution (x-y)(2x-2y-3)-2
=(x-y)[2(x-y)-3]-2
=2(x-y) ^2-3(x-y)-2
=[(x-y)-2][2(x-y)+ 1]
=(x-y-2)(2x-2y+ 1)。
1 -2
╳
2 1
1× 1+2×(-2)=-3
It is pointed out that decomposing (x-y) into a whole is another application of holistic thinking method in mathematics.
Example 5 x 2+2x- 15
Analysis: the constant term (-15) < 0 can be decomposed into the product of two numbers with different signs, and can be decomposed into (-1)( 15), or (1)(- 15) or (3).
(-5) or (-3)(5), in which only the sum of -3 and 5 in (-3)(5) is 2.
=(x-3)(x+5)
Summary: ① factorization of x2+(p+q) x+pq formula.
The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).
② Factorization of KX2+MX+N formula
If it can be decomposed into k = AC, n = BD and AD+BC = M, then
kx^2+mx+n=(ax+b)(cx+d)
A b
╳
c d
Popular method
Firstly, the quadratic term is decomposed into (1 X quadratic term coefficient), and the constant term is decomposed into (1 X constant term) and then written in the following format.
1 1
X
Constant term of quadratic coefficient
If the value after cross multiplication is equal to the coefficient of the first term, it holds. If they are not equal, they will be tested as follows. (The general question is very simple, and the correct answer can be worked out at most 3 times. )
The format that needs many experiments is: (Note: abcd at this time does not refer to the coefficient in (AX 2+BX+C), and abcd is preferably an integer).
A b
╳
c d
The first time a= 1 b= 1 c= quadratic coefficient ÷a d= constant term ÷ b.
Seconds a= 1 b=2 c= quadratic term coefficient ÷a d= constant term ÷ b.
The third time a=2 b= 1 c= quadratic term coefficient ÷a d= constant term ÷ B.
The fourth a=2 b=2 c= quadratic coefficient ÷a d= constant term ÷ B.
The fifth a=2 b=3 c= quadratic coefficient ÷a d= constant term ÷ B.
The sixth a=3 b=2 c= quadratic coefficient ÷a d= constant term ÷ B.
Cubic a=3 b=3 c= quadratic term coefficient ÷a d= constant term ÷ B.
......
The rest can be inferred.
Until (ad+cb= coefficient of the first term). The format of the final result is (ax+b)(cx+d).
Sample solution:
2x^2+7x+6
The first time:
1 1
╳
2 6
1x 6+2x 1 = 8 8 & gt； 7 does not hold. Keep trying.
second time
1 2
╳
2 3
1X3+2X2=7, so it is decomposed into: (x+2)(2x+3).
[Edit this paragraph] 1. Cross multiplication (solving the problem of the ratio of the two)
principle
Individuals in a set have only two different values, some of which are A and the others are B. The average value is C. Find the ratio of individuals with value A to individuals with value B.. Suppose a has x and b has (1-X).
AX+B( 1-X)=C
X=(C-B)/(A-B)
1-X=(A-C)/(A-B)
Therefore: x: (1-x) = (c-b): (a-c)
The above calculation process can be abstracted as follows:
a……C-B
……C
b……A-C
This is called cross multiplication.
Pay attention when using cross multiplication.
The first point: it is used to solve the problem of the ratio of the two.
The second point: the obtained proportional relationship is the proportional relationship of cardinal number.
The third point: put the total average in the middle, on the diagonal, reduce the large number and put the result on the diagonal.