Factorization: (x2-1) 2 + 6 (1-x2) + 9

Factorization: (x2-1) 2 + 6 (1-x2) + 9

（x2-1）2+6（1-x2）+9=（x2-1）2-6（x2-1）+9=（x2-1-3）2=（x-2）2（x+2）2．
X ^ 2-6x-614 = 0, any one of "square root", "factorization", "matching method" or "formula method"
X & # 178; - 6x-614 = 0 was solved by formula method
Formula X & # 178; - 6x = 614
x²-6x+9=614+9
(x-3) & 178; = 623 square root
X-3 = ± √ 623 √ 623 is the simplest quadratic radical
X-3 = √ 623 or x-3 = - √ 623
x1=3+√623 ,x2=3-√623
If the variables X and y satisfy the constraint condition {x > = - 1; Y > = x; 3x + 2Y
Using image method, draw x > = - 1; Y > = x; 3x + 2Y
(X & # 178; - 1) &# 178; - 6 (X & # 178; - 1) + 9 factorization
(x²-1)²-6(x²-1)+9
=(x²-1+3)²= (x²+2)²
The equation 2x ^ 2-3x = 2 is solved by formula method and collocation method respectively
2x^2-3x-2=0
Formula method:
△=9+16=25
X = [3 ± 5] / 4 = 2 or - 1 / 2
Preparation method:
x^2-3/2*x-1=0
(x-3/4)^2=25/16
x-3/4=±5/4
X = 2 or - 1 / 2
If the variables X, y satisfy the constraint conditions - 1 ≤ X: X ≤ y; 3x + 2Y ≤ 5, then the maximum value of Z = 2x + y is
3x+2y
How to decompose ^ 2 + 3x-4?
How do you think it can be decomposed?
It cannot be decomposed in the range of rational numbers
Using formula method to solve quadratic equation of one variable: 2x ^ 2-2 √ 2 + 1 = 0
Is there an X in the equation?
Is this the original problem
If so: (√ 2x - 1) ^ 2 = 0
√2X -1=0
X=√2/2
Let the variables X and y satisfy the constraint condition x ≥ x − y ≥ 02x − y − 2 ≤ 0, then the maximum value of Z = 3x-2y is 0______ .
When the answer is shown in the diagram, the maximum value of the objective function (z = 4.2) is drawn
Finding x ^ 3-3x ^ 2-9x-5 = 0; factorization process
x^3-3x^2-9x-5
=x^3+x^2-4x^2-4x-5x-5
=x^2(x+1)-4x(x+1)-5(x+1)
=(x+1)(x^2-4x-5)
=(x+1)(x+1)(x-5)
=[(x+1)^2](x-5)
=0
So (x + 1) ^ 2 = 0, or X-5 = 0
The solution is: x = - 1, x = 5
x^3-3x^2-9x-5=x^3+1-3(x^2+3x+2)
=(x+1)(x^2+1-x)-3(x+2)(x+1)
=(x+1)(x^2+4x-5)
=(x+1)(x+1)(x-5)
It is not difficult to find that when x = - 1, x ^ 3-3x ^ 2-9x-5 = 0,
So x ^ 3-3x ^ 2-9x-5 = (x + 1) * a
If you divide by one, you get
A=X^2-4*X-5
So x ^ 3-3x ^ 2-9x-5 = (x + 1) (x + 1) (X-5)
x^3-3x^2-9x-5=0
x^3-3x^2-4x-5x-5=0
x(x^2-3x-4)-5(x+1)=0
x(x+1)(x-4)-5(x+1)=0
(x+1)[x(x-4)-5]=0
(x+1)(x^2-4x-5)=0
(x+1)(x+1)(x-5)=0
(x-5)(x+1)^2=0
∵x=5 x=-1
The possible roots of higher order equations are generally the ratio of the divisor of the highest order coefficient to the divisor of the constant term
So x = 5 is obviously a root of the equation
There are many ways. X-5 can be removed with x ^ 3-3x ^ 2-9x-5
In addition, it can be decomposed into X-5
x^3-3x^2-9x-5＝（x^3-5x^2）＋（2x^2－10x）＋（x－5）
＝x^2（x－5）＋2x（x－5）＋（x－5）
... unfold
The possible roots of higher order equations are generally the ratio of the divisor of the highest order coefficient to the divisor of the constant term
So x = 5 is obviously a root of the equation
There are many ways. X-5 can be removed with x ^ 3-3x ^ 2-9x-5
In addition, it can be decomposed into X-5
x^3-3x^2-9x-5＝（x^3-5x^2）＋（2x^2－10x）＋（x－5）
＝x^2（x－5）＋2x（x－5）＋（x－5）
＝。。。。。。。。 Put it away
Which symbol is a division sign?
X^3-3X^2-9X-5=0
One
---X - 1.5X-9X-5=0
Three
1... Unfold
Which symbol is a division sign?
X^3-3X^2-9X-5=0
One
---X - 1.5X-9X-5=0
Three
1 4.5 27
---X - ---X - ---X-5=0
3 3 3
1-4.5-27
------------X - 5=0
Three
-30.7
----------X=5
Three
Three
X=5* ---------
-30.7
One hundred and fifty
X= - ----------
307 put away
x^3-3x^2-9x-5
=x^3+x^2-4x^2-4x-5x-5
=x^2(x+1)-4x(x+1)-5(x+1)
=(x+1)(x^2-4x-5)
=(x+1)(x+1)(x-5)
=[(x+1)^2](x-5)
=0
x=-1 x=5