How to factorize a cubic equation of one variable? Let me give you a cubic equation of one variable. How to factorize it? It's better to be easy to read and understand, rather than paste it casually.

How to factorize a cubic equation of one variable? Let me give you a cubic equation of one variable. How to factorize it? It's better to be easy to read and understand, rather than paste it casually.

The formula for finding roots of cubic equation with one variable can not be made by ordinary deductive thinking. The matching method similar to the formula for finding roots of quadratic equation with one variable can only formalize the standard cubic equation with one variable of type ax ^ 3 + BX ^ 2 + CX + D + 0 into a special type of x ^ 3 + PX + q = 0
The solution of the formula for solving the cubic equation of one variable can only be obtained by inductive thinking, that is, the form of the formula for finding the root of cubic equation of one variable can be concluded according to the form of the formula for finding the root of linear equation of one variable, quadratic equation of one variable and special higher order equation. The form of the formula for finding the root of cubic equation of one variable, such as x ^ 3 + PX + q = 0, should be x = a ^ (1 / 3) + B ^ (1 / 3), It is the sum of two open cubes. The form of the formula for finding the root of cubic equation with one variable is summed up. The next step is to find out the content in the open cube, that is, to use P and Q to express a and B
(1) Let x = a ^ (1 / 3) + B ^ (1 / 3) be cubic at the same time
（2）x^3=(A+B)+3(AB)^(1/3)(A^(1/3)+B^(1/3))
(3) Because x = a ^ (1 / 3) + B ^ (1 / 3), so (2) can be changed into
X ^ 3 = (a + b) + 3 (AB) ^ (1 / 3) X
(4) Compared with the univariate cubic equation and the special type of x ^ 3 + PX + q = 0, we can see that x ^ 3-3 (AB) ^ (1 / 3) x - (a + b) = 0
(5) - 3 (AB) ^ (1 / 3) = P, - (a + b) = Q
（6）A+B＝－q,AB＝-（p/3）^3
(7) In this way, the root formula of cubic equation with one variable is transformed into the root formula of quadratic equation with one variable, because a and B can be regarded as the two roots of quadratic equation with one variable, and (6) is the Veda theorem about the two roots of quadratic equation with one variable in the form of ay ^ 2 + by + C = 0
（8）y1＋y2＝－（b/a）,y1*y2=c/a
(9) Comparing (6) and (8), we can make a = Y1, B = Y2, q = B / A, - (P / 3) ^ 3 = C / A
(10) Because the root formula of quadratic equation of type ay ^ 2 + by + C = 0 is
y1＝－（b＋（b^2－4ac）^(1/2)）/(2a)
y2＝－（b－（b^2－4ac）^(1/2)）/(2a)
Can be transformed into
(11)y1＝－(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2＝－(b/2a)+((b/2a)^2-(c/a))^(1/2)
Substituting a = Y1, B = Y2, q = B / A, (P / 3) ^ 3 = C / A in (9) into (11), we can get
(12)A＝－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)
B＝－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)
(13) Substituting a and B into x = a ^ (1 / 3) + B ^ (1 / 3), we get
(14)x=（－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)+（－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)
Equation (14) is only a real root solution of one variable cubic equation. According to Weida's theorem, one variable cubic equation should have three roots, but according to Weida's theorem, as long as one of the roots is solved, the other two roots can be easily solved
How to factorize cubic equation of one variable
Equation 1: - x ^ 3 + 7x ^ 2-16x + 12 = 0
How to become
Equation 2: - (x-3) (X-2) ^ 2 = 0
Please help!
-x^3+7x^2-16x+12=0
-x³+3x²+(4x²-16x+12)=0
-x²(x-3)+4(x-1)(x-3)=0
(x-3)[-x²+4(x-1)]=0
(x-3)(-x²+4x-4)=0
-(x-3)(x²-4x+4)=0
-(x-3)(x-2)²=0
x³-7x²+16x-12 ＝x³-3x²－4x²+16x-12
=(x³-3x²)－(4x²-16x+12)
=x²(x-3)-(2x-2)(2x-6)
... unfold
x³-7x²+16x-12 ＝x³-3x²－4x²+16x-12
=(x³-3x²)－(4x²-16x+12)
=x²(x-3)-(2x-2)(2x-6)
=x²(x-3)-4(x-1)(x-3)
=(x-3)(x²-4x+4)
=(x-3) (X-2) &
Several factoring problems, please
Known: A ^ 2 + B ^ 2-A + 6B + 91 / 4 = 0, find the value of 4A + 3B
(a-b)^2-8(a^2-b^2)+16(a-b)^2
32a(x^2+x)^2-2a
5.64^2+2×5.64×4.36+4.36^2
5.9^2-2×5.9×0.9+0.9^2
3xy(a+b)^2-9x^2(a+b)
a(x-2y+3z)-b(2y-x-3z)
x(x-y)(a-b)-y(y-x)(b-a)
2x_ 2y+3xy(x-y)
I can't even understand the questions. Are you inputting the questions or am I a college student in vain
Question 8 (a-b) (X-Y) ^ 2
Question 2: (a-b) ^ 2-8 (a ^ 2-B ^ 2) + 16 (a-b) ^ 2 input error? Just enter 17 (a-b) ^ 2. Why (a-b) ^ 2 -. + 16 (a-b) ^ 2
I'm sorry,
WangHan 1224 is doing very well. Just write some personal opinions on two small topics.
1. 91 / 4 means 9.25. Strictly speaking, this kind of writing is wrong. It can be used in a period of time in primary school. So a = 1 / 2. B = - 3, so the answer is - 7
3.32a(x^2+x)^2-2a
=2a[16x^2(x+1)^2-1]
= 2a[4x^2+4x-1][4x^2+4x+1]
=2A [4x ^ 2 +... Expanded
WangHan 1224 is doing very well. Just write some personal opinions on two small topics.
1. 91 / 4 means 9.25. Strictly speaking, this kind of writing is wrong. It can be used in a period of time in primary school. So a = 1 / 2. B = - 3, so the answer is - 7
3.32a(x^2+x)^2-2a
=2a[16x^2(x+1)^2-1]
= 2a[4x^2+4x-1][4x^2+4x+1]
=2A [4x ^ 2 + 4x-1] (2x + 1) ^ 2] fold up
Your first question is 9.25? If a = 0.5 and B = - 3, the final result is - 7
If it's not, it's related to the plural, which seems difficult to solve. The rest of them are very good.
Proof: there is a solution to the equation (a-b) x square + (B-C) x + C-A = 0 of X, which is a mathematical equation of degree 2 with 1 elementary and 2 variables
Proof: for the solution of the equation (a-b) x square + (B-C) x + C-A = 0 of X, one solution is 1
Mathematics of 1-variable quadratic equation in elementary two
a≠b
be careful
Mathematics problems in grade two of junior high school
Because when x = 1, the left side of the equation is A-B + B-C + C-A = 0 = the right side
So x = 1 is the solution of the original equation
When x = 1, left = A-B + B-C + C-A = 0
Right = 0
Left = right
X = 1 is the solution of the original equation
Using Weida theorem
aX^2+bX+c=0
The sum of the two is - B / A
The product of two is C / A
But your question
Nonsense, of course x = 1 holds
I think there is something wrong with this question
(X & # 178; - 1) &# 178; - 6 (X & # 178; - 1) + 9 is not bad in math. But factorization is not very good. How can I learn factorization well?
﹙x²－1﹚²－6﹙x²－1﹚＋9
=[（x^2-1）-3]^2
=[(x-1)(x+1)-3]^2
Be proficient in complete square formula, cross formula and square difference formula
If a = 4x ^ 2-6x + 3, B = 5x ^ 2-3x + 4, then 9x ^ 2-9x + 7 is equal to
9x²-9x+7
=(4x²-6x+3)+(5x²-3x+4)
=A+B
Answer: a + B
A+B
Mathematical master to solve several one variable quadratic equations
1、 Formula solution
1,x^2+2x-2=0
2,2x ^ 2 + 1 = 2 and root 3
2、 Using factorization to solve the problem
1,（x+2)(x-4)=0
2,4x(2x+1)=3(2x+1)
3,(4x-1)(5x+7)=0
4,3x(x-1)=2-2x
5,(2x+3)^2=4(2x+3)
6,2(x-3)^2=x^2-9
Solve the following equation
1,5(x^2-x)=3(x^2+x)
2,(x-2)^2=(2x+3)^2
3,2y^+4y=y+2
4,2x+6=(x+3)^2
5,(x-2)(x-3)=12
It will be in + minutes
1.x^2+2x-2=0
△=b^2-4ac=4+8=12
B ^ 2 addition and subtraction
-------=- 2-radical 3 or - 2 + radical 3
2A
2.2x ^ 2 + 1 = 2 and root 3
2X ^ 2 + (1-2 and root 3) = 0
A = B ^ 2-4ac = 16 and radical 3-8
B ^ 2 addition and subtraction
-------Approximately equal to - 1.1
2A
2、 Using factorization to solve the problem
1,（x+2)(x-4)=0
x=-2 x=4
2,4x(2x+1)=3(2x+1)
4x(2x+1)-3(2x+1)=0
(2x-1)(4x-3)=0
X = half x = three quarters
3,(4x-1)(5x+7)=0
X = quarter x = minus seven fifths
4,3x(x-1)=2-2x
3x(x-1)=-2（x-1）
3x(x-1)+2（x-1）=0
(x-1)(3x+2)=0
X = 1 x = minus two thirds
5,(2x+3)^2=4(2x+3)
4x^2+12x+9=8x+12
4x^2+4x-3=0
X = half x = minus three-thirds
6,2(x-3)^2=x^2-9
(x-3)(x-9)=0
x=3 x=9
Solve the following equation
1,5(x^2-x)=3(x^2+x)
Using the common factor to solve the problem
2,(x-2)^2=(2x+3)^2
3,2y^2+4y=y+2
4,2x+6=(x+3)^2
5,(x-2)(x-3)=12
x=2 x=3
4xy+1-4x²-y²
8x³+4x²-2x-1
4xy+1-4x²-y²=1 - (4x²+y²-4xy)= 1 ² - (2x-y)²=(1+2x-y)(1-2x+y)8x³+4x²-2x-1=(8x³+4x²)- (2x+1)=4x²(2x+1) - (2x+1)= (4x²-1)(2x+1)=(2x+1)² (...
In the first question, we combine 4xy with the last two terms and get the following result: 1 - (4x2-4xy + Y2) = 1 - (2x-y) 2 = (1 + 2x-y) (1-2x + y)
In the second question, the first two items and the last two items are combined to get = (8x3 + 4x2) - (2x + 1) = 4x2 (2x + 1) - (2x + 1) = (4x2-1) (2x + 1) = (2x-1) (2x + 1) (2x + 1) = (2x-1) (2x + 1) 2
　　4xy+1-4x²-y²
=1 - (4x²+y²-4xy)
= 1 ² - (2x-y)²
=(1+2x-y)(1-2x+y)
8x³+4x²-2x-1
=(8x³+4x²)- 2x-1
=4x²(2x+1) - (2x+1)
= (4x²-1)(2x+1)
=(2x+1)(2x+1)(2x-1)
=(2x+1)²(2x-1)
Remember to adopt!
15x = 23x-24 12 + 5x = 3x-8 3 (x-3) = 9 - (x + 6) solution 4 + x = 4x-17 solution
solve equations
Hurry
.
15x=23x-24
8x=24
X=3
12+5x=3x-8
2x=-20
x=-10
3(x-3)=9-(x+6)
4x=9+9-6=12
X=3
4+x=4x-17
3x=21
X=7
How do you do in the exam!
15X-23X=-24 X=3 2X=-20 X=-10 3x-9=9-x-6 4x=12 x=3 21=3x x=7
A problem of quadratic equation with one variable in elementary 2 mathematics
One side of the wall and the other three sides of the wall are used to form a rectangular flower bed with an area of 130m ^ 2. The length of the wall is 15cm. What is the length and width of the flower bed?
Let length x, width (33-x) / 2
The area is x * (33-x) / 2 = 130
(33X-X^2)/2=130
-X^2+33X-260=0
X^2-33X+260=0
（X-13)(X-20)=0
So x = 13 or x = 20
Because of X