example
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Example 1 Factorization 2x 2-7x+3.
Analysis: first decompose the quadratic coefficient and write it in the upper left corner and 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= 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 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 1a2+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)。
A method like this helps us to decompose the quadratic trinomial by drawing cross lines. Usually,
This is 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)