Cross multiplication concept
Cross multiplication can decompose some quadratic trinomials. The key of this method is to decompose the coefficient a of the quadratic term into the product of two factors, a 1 and A2, A 1? A2, decompose the constant term c into two factors, the product of c 1 and C2? C2, and make a 1c2+a2c 1 just be the first-order coefficient b, then the result can be written directly: when decomposing factors in this way, we should pay attention to observation, try and realize that it is actually the inverse process of binomial multiplication. When the first coefficient is not 1, it often needs to be tested many times, so be sure to pay attention to the sign of each coefficient.
example
Example 1 Factorization 2x2-7x+3.
Analysis: first write the quadratic coefficient decomposition in the upper left corner and the lower left corner of the crosshair, and then decompose and divide the constant term.
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= 1×3==(-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 2x2-7x+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 1a2+a2c 1
Through diagonal cross multiplication and addition, we get a 1a2+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)。
A method like this helps us to decompose the quadratic trinomial by drawing cross lines. Usually,
This is called cross multiplication.
Example 2 Factorizing 6x2-7x-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 6x2-7x-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 term 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, decompose x2+2x- 15, and cross multiply it.
1 -3
1 5
1×5+ 1×(-3)=2
So x2+2x- 15 = (x-3) (x+5).
Example 3 Factorizing 5x2+6xy-8y2.
Analysis: This polynomial can be regarded as a quadratic trinomial about x, and-8y2 is regarded as a constant term. When decomposing the coefficients of quadratic terms and constant terms, only 5 and -8 need to be decomposed. After the crosshair decomposition, an appropriate group is selected through observation, namely
1 2
5 -4
1×(-4)+5×2=6
Solution 5x2+6xy-8y2 = (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 3:x2+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)
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 of 1: 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 ac=k bd=n
c / - \d ad+bc=m