Cross factorization can decompose quadratic trinomial (not necessarily in integer range). For the shape of the axe? +bx+c = (a 1 x+c 1) (a2x+C2). The key of the method is to decompose the quadratic coefficient A into the product of two factors A1,a2, and the constant term C into two factors C1,a2. +bx+c = (a1x+c1) (a2x+C2). When using this method to decompose factors, we should pay attention to observation and try to understand that its essence is 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. Basic formula: x? +(p+q)x+pq=(x+p)(x+q).
Example of general operation method: editing
Answer? +a-42
First, let's look at the first number. Is it an a? , representing the multiplication of two a, inferring (a+? )×(a -？ ),
Then let's look at the second item. Formula +a is the result of merging similar terms, so it is inferred as binomial × binomial.
The last term is -42, and -42 is -6×7 or 6×7, which can also be decomposed into -2 1×2 or 2/kloc-0 /× 2.
First of all, 2 1 and 2, whether positive or negative, can't be 1 after any addition or subtraction, but only-19 or 19, so the latter is excluded.
Then, determine whether it is -7×6 or 7×6.
(a+(-7))×(a+6)=a？ x？ -ax-42 (the calculation process is omitted)
The result is inconsistent with the original result, and the original formula +a becomes-a.
Recalculate:
(a+7)×(a+(-6))=a？ +a-42
Correct, so a? +a-42 is decomposed into (a+7)×(a-6), which is a popular factorization method now.
Specific application
The double cross decomposition method is a factorization method. For Ax and other types? +Bxy+Cy？ The polynomial factorization of +Dx+Ey+F usually uses the undetermined coefficient method. The operation process of this method is complicated. For this problem, if the "double cross decomposition method" (principal component method) is adopted, this kind of polynomial is easy to factorize.
For example: 3x? +5xy-2y？ +x+9y-4=(x+2y- 1)(3x-y+4)
Because 3= 1×3, -2=2×(- 1), -4 = (- 1) × 4,
And1× (-1)+3x2 = 5, 2x4+(-1) = 9,1× 4+3x (-1) =/kloc-0.
Tip: Take the missing item as a coefficient of 0, and multiply 0 by any number to get 0.
For example: ab+b? +a-b-2
=0× 1×a？ +ab+b？ +a-b-2
=(0×a+b+ 1)(a+b-2)
=(b+ 1)(a+b-2)
Tip: set x? =y, cx? Split it into mx? Make peace with ny
For example: 2x 4+ 13x 3+20x? + 1 1x+2
=2y？ + 13xy+ 15x？ +5y+ 1 1x+2
=(2y+3x+ 1)(y+5x+2)
=(2x？ +3x+ 1)(x？ +5x+2)
=(x+ 1)(2x+ 1)(x？ +5x+2)
When we decompose the quadratic trinomial, we often use the cross decomposition method. For some binary quadratic sextuples (ax? +bxy+cy？ +dx+ey+f), we can also use the cross decomposition method to decompose the factors.
For example, factorization factor 2x? -7xy-22y？ -5x+35y-3。 Let's arrange the above formula in descending order of X and take Y as a constant, then the above formula can be transformed into
2x？ -(5+7y)x-(22y？ -35y+3)，
It can be regarded as a quadratic trinomial about X.
For the constant term, it is a quadratic trinomial about y, which can also be decomposed into
that is
-22y？ +35y-3 =(2y+3)(- 1 1y- 1)。
Then the quadratic trinomial about x is decomposed by cross decomposition method.
therefore
Original formula = [x+(2y-3)] [2x+(-11y+1)]
=(x+2y-3)(2x- 1 1y+ 1)。
(x+2y)(2x- 1 1y)= 2 x2-7xy-22 y2；
(x-3)(2x+ 1)= 2 x2-5x-3；
(2y-3)(- 1 1y+ 1)=-22y？ +35y-3。
This is the so-called double cross decomposition method, which is also commonly known as the "principal component method"
Analysis of polynomial ax by double cross decomposition method +bxy+cy？ The steps of factorization of +dx+ey+f are:
(1) Decompose ax? +bxy+cy？ , get a cross plot (with two columns);
⑵ The constant term f is decomposed into two factors and filled in the third column. It is required that the sum of the cross products of the second and third columns in the original formula is equal to ey, and the sum of the cross products of the first and third columns in the original formula is equal to DX.
We call the algebraic expression of anx n+a (n-1) x (n-1)+…+a1x+A0 (n is a non-negative integer) as a univariate polynomial about x, and use f(x), g(x), ….
f(x)=x？ -3x+2,g(x)=x^5+x？ +6,…,
When x=a, the value of the polynomial f(x) is represented by f(a), just like the polynomial f(x) above.
f⑴= 12-3× 1+2 = 0；
f(-2)=(-2)？ -3×(-2)+2= 12.
If f(a)=0, a is called the root of the polynomial f(x).
Theorem 1 (factorial theorem) If a is the root of unary polynomial f(x), that is, f(a)=0 holds, then polynomial f(x) has at least one factor x-a. 。
According to factorial theorem, the key to find the first factorial of unary polynomial f(x) is to find the root of polynomial f(x). For any polynomial f(x), there is no general method to find the root. However, when all the coefficients of the polynomial f(x) are integers, that is, integer coefficient polynomials, the following theorem is often used to determine whether it has a rational root.
How to decompose factors
Example 7x+(-8x) =-x
Solution: Original formula =(x+7)(x-8)
Example 2
-2x+(-8x)=- 10x
Solution: Original formula =(x-2)(x-8)
Example 3,
Analysis: Although the quadratic coefficient of this problem is not 1, it can also be factorized by cross decomposition.
because
9y + 10y= 19y
Solution: Original formula =(2y+3)(3y+5)
Example 4, factorization.
Analysis: Because
2 1x + (- 18x)=3x
Solution: Original formula =(2x+3)(7x-9)
Example 5, factorization.
Analysis: This problem can be factorized by taking (x+2) as a whole.
because
-25(x+2)+[-4(x+2)]= -29(x+2)
Solution: Original formula =[2(x+2)-5][5(x+2)-2]
=(2x- 1)(5x+8)
Example 6, factorization.
Analysis: This problem can be decomposed by cross decomposition method as a whole, and then cross multiplication can be applied once.
because
-2+[- 12]=- 14 a+(-2a)=-a 3a+(-4a)=-a
Solution: Original formula =[-2][-12]
=(a+ 1)(a-2)(a+3)(a-4)
Sample analysis and editing
Example 1
Put 2x? -7x+3 factorization factor.
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 (only take positive factor because the result of taking negative factor is the same as that of taking positive factor! ):
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 3
╳
2 1
1× 1+2×3=7 ≠-7
1 1
╳
2 3
1×3+2× 1=5 ≠-7
1 - 1
╳
2 -3
1×(-3)+2×(- 1)=-5 ≠-7
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? -7x+3=(x-3)(2x- 1)
Usually, for quadratic trinomial ax? +bx+c(a≠0), if the quadratic coefficient a can be decomposed into the product of two factors, namely a=a 1a2, then the constant term c can be decomposed into the product of two factors, namely c=c 1c2, and a 1, a2, c/kl.
a 1 c 1
╳
a2 c2
a 1c2 + a2c 1
Cross-multiply and add diagonally to get a 1c2+a2c 1. If it is exactly equal to the quadratic trinomial ax? The linear coefficient b of +bx+c is a 1c2+a2c 1=b, then the quadratic trinomial can be decomposed into the product of two factors a 1x+c 1 and a2x+c2, namely
ax^2+bx+c=(a 1x+c 1)(a2x+c2)
The method of drawing cross lines like this to help us decompose the quadratic trinomial is usually called cross decomposition.
Example 2
Put 5x? +6xy-8y？ Factorization.
Analysis: this polynomial can be regarded as a quadratic trinomial about x, while -8y? As a constant term, when we decompose the coefficients of the quadratic term and the constant term, we only need to decompose 5 and -8, and after the decomposition with the crosshair, we choose a suitable group through observation, namely
1 2
╳
5 -4
1×(-4)+5×2=6
Solution 5x? +6xy-8y？ =(x+2y)(5x-4y)。
It is pointed out that the original formula is decomposed into two linear formulas about x and y.
Example 3
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 factors of the above product? 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, if (x-y) is taken as a whole and multiplied by it, the original polynomial can be transformed into a quadratic trinomial about (x-y), and the factor can be decomposed by cross-decomposition.
Solution (x-y)(2x-2y-3)-2
=(x-y)[2(x-y)-3]-2
=2(x-y)？ -3(x-y)-2
1 -2
╳
2 1
1× 1+2×(-2)=-3
=[(x-y)-2][2(x-y)+ 1]
=(x-y-2)(2x-2y+ 1)。
It is pointed out that if the elements X and Y are replaced by (x+y) and (x+y) is taken as the element, this is the method of substitution.