Factorization of group decomposition method 1.(a+2)(a+3)+(a+1)(a+4)+(a+1)(a+3)+(a+2)(a+4) 2.36x3-12x2-x + 1 / 3

Factorization of group decomposition method 1.(a+2)(a+3)+(a+1)(a+4)+(a+1)(a+3)+(a+2)(a+4) 2.36x3-12x2-x + 1 / 3

1.(a+2)(a+3)+(a+1)(a+4)+(a+1)(a+3)+(a+2)(a+4)=(a+1)[(a+3)+(a+4)]+(a+2)[(a+3)+(a+4)] =(2a+7)(2a+3)2.36x³-12x²-x+1/3=36x²(x-1/3)-(x-1/3) =(x-1/3)(6x+1)(6x-1)
1. The original formula is 4A ^ 2 + 20A + 21
2 3
2 7
Cross method
Original formula = 2A + 3) (2a + 7)
2. Original formula = 36x ^ 2 (x-1 / 3) - (x-1 / 3)
Original formula = (36x ^ 2-1) (x-1 / 3) = [(6x) ^ 2-1] (x-1 / 3) = (6x-1) (6x + 1) (x-1 / 3)
=[(a+2)(a+3)+(a+1)(a+3)]+[(a+1)(a+4)+(a+2)(a+4)]
=(2a+3)(a+3)+(2a+3)(a+4)
=(2a+3)(2a+7)
=4a^2+20a+21
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Second question, "what does" - x + one third "mean?
It is known that f (x) = x & # 178; - 3x, X belongs to [1,4], the maximum and minimum of which are
f(x)=x²-3x+9/4-9/4
]=(x-3/2)²-9/4
Axis of symmetry x = 3 / 2
therefore
X = 3 / 2, the minimum is - 9 / 4
X = 4, max
It is known that f (x) = x & # 178; - 3x, X belongs to [1,4], the maximum is 4, and the minimum is 9 / 4 of negative
Using the general form of quadratic equation AX & sup2; + BX + C = 0 (a ≠ 0, B C is a real number) and the collocation method, the root formula △ = B & sup2; - 4ac is proved
Using the general form of quadratic equation AX & sup2; + BX + C = 0 (a ≠ 0, B C is a real number)
Proving the root formula △ = B & sup2; - 4ac by collocation method
ax²+bx=-c
x²+bx/a=-c/a
x²+bx/a+b²/4a²=-c/a+b²/4a²
(x+b/2a)²=(b²-4ac)/4a²=[±√(b²-4ac)/2a]²
x+b/2a=±√(b²-4ac)/2a
x=-b/2a±√(b²-4ac)/2a
x=[-b±√(b²-4ac)]/2a
Factorization by grouping
2ax-10ay+5by-bx
3ax+4by+4ay+3bx
M2 + 5n-mn-5m
20(x+y)+x+y
X square - y square + ax + ay
A square - 2Ab + b square - C square
X cube + x square y-xy2-y cube
4A square-b square + 6a-3b
20 (x + y) + X + y = 20 (x + y) + (x + y) = 21 (x + y) x square - y square + ax + ay = (x + y) (X-Y) + a (x + y) = (x + y) (X-Y + a) a square - 2Ab + b square - C square = (a-b) &# 178; - C & # 178; = (a-b + C) (a-b-c) x cube + x square y-xy2-y cube = x & # 178; (x + y) - Y & # 178; (x + y) = (x + y) (X & # 178; -
2ax-10ay+5by-bx
=2a(x-5y)-b(x-5y)
=(2a-b)(x-5y)
3ax+4by+4ay+3bx
=3x(a+b)+4y(a+b)
=(3x+4y)(a+b)
M2 + 5n-mn-5m
=m(m-n)-5(m-n)
=(m-5)(m-n)
20(x+y)+x+y
=(x+y)(20+1)
Given the square of a = 3x - 2x + 1, B = - 2x + X - 2, compare the size of a and B
a-b=(3x²-2x+1)-(-2x+x²-2)
=3x²-2x+1+2x-x²+2
=2x²+3≥3＞0
So a > b
a=3x^2-2x+1
b=-2x+x^2-2
a-b=2x^2+3
Because 2x ^ 2 is always greater than or equal to 0, so A-B > = 3
So a > 3
a＞b
The relation between the root and coefficient of quadratic equation of one variable!
For one dollar quadratic
(1) Two reciprocal conditions are as follows:
(2) If two numbers are opposite to each other:
(3) The conditions of two different signs are as follows:
(4) The conditions of two identical numbers are as follows:
(5) If both are positive numbers:
(6) The condition of two negative numbers is:
Ax ^ 2 + BX + C = 0 (1) two reciprocal conditions are: X1 * x2 = 1, C / a = 1, a = C (2) two reciprocal conditions are: X1 + x2 = 0, - B / a = 0, B = 0 (3) two different sign conditions are: X1 * X20 (5) two are positive conditions are: X1 * x2 > 0, X1 + x2 > 0C / a > 0, - B / a > 0 (6) two negative conditions are: X1 * x2 > 0
Suppose the equation is x ^ 2 + BX + C = 0
1 c=1
2 b=0
3 c＜0
4 c＞0
5 -b＞0 c＞0
6 -b＜0 c＜0
Factorization by grouping method
1. The square of 4x - the square of 4xy-a + the square of Y
Square of x-2y-4y + X
The square of 5x - 10xy + the square of 5Y - 2x + 2Y
The square of 4A + the square of 9b - the square of C - 12ab + 2c-1
4x^2-4xy-a^2+y^2=(2x-y)^2-a^2=(2x-y-a)(2x-y+a)x^2-2y-4y^2+x=(x+1/2)^2-(2y+1)^2=(x-2y-1/2)(x+2y+3/2)5x^2-10xy+5y^2-2x+2y=5(x-y)^2-2(x-y)=(x-y)(5x-5y-2)4a^2+9b^2-c^2-12ab+2c-1=(2a-3b)^2-(c-1)^2=(2a-3b-c...
Let's know the square of a = 3x - 2x + 1, and the square of B = 2x + X - 2
A-B
=3x²-2x+1-2x-x²+2
=2x²-4x+3
=2x²-4x+2+1
=2(x-1)²+1≥1>0
So a > b
Let m = √ X
Then M > = 0
y=m2-m=(m-1/2)2-1/4
So m = 1 / 2, minimum = - 1 / 4
Then a is y ≥ - 1 / 4
If a and B = B, then a is a subset of B
That is, x > A contains ≥ - 1 / 4
So a
How to find the root of quadratic equation with one variable by Weida theorem?
Help~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
Why say X1 + x2 = - B / A, and x1 × x2 = C / a?
How did he get his roots?
Thank you!
If you are a junior three, you should know how to find the root formula
So X1 = - B + {root B squared - 4ac] / 2A
X2 = - B - {root square b-4ac} / 2A
So X1 + x2 = - 2b / 2A = - B / A
X1X2=A/C
My problem is the factorization method
How to use grouping decomposition method to factorize?
Please give an example!
According to the basic requirements, there are two kinds: 1. There are common factors after grouping. For example, x ^ 3-3x + 3y-x ^ 2Y = (x ^ 3-x ^ 2Y) - 3 (X-Y) = x ^ 2 (X-Y) - 3 (X-Y) = (X-Y) (x ^ 2-3) 2. Although there is no common factor after grouping, it can be solved by multiplication formula. X ^ 2-y ^ 2 + 2y-1 = x ^ 2 - (y ^ 2-2y +)