How to factorize a cubic equation of one variable?

How to factorize a cubic equation of one variable?

For Ax ^ 3 + BX ^ 2 + CX + D = 0
(1) Let y = X-B / (3a), take into the above formula, and get
The form of Y ^ 3 + py + q = 0
For (M + n) ^ 3 = m ^ 3 + n ^ 3 + 3MN (M + n), we get
（M+N）^3-3MN(M+N)-（M^3+N^3）=0
Compared with y ^ 3 + py + q = 0, there are
M+N=y;-3MN=p;-M^3-N^3=q
Take n = - P / (3m) into - m ^ 3-N ^ to solve m ^ 3 〉 to solve m --- > to solve N --- > to solve y --- > to solve X
The concrete solution of factorization of cubic equation of one variable
How to factorize? What are the solutions, such as X3 + 3x2-24x + 28 = 0
And when is this knowledge content? Which textbook? Which semester?
Is this knowledge point from high school or junior high school? Which semester? I'll go to the bottom
The root formula of cubic equation of one variable can not be made by ordinary deductive thinking. The matching method similar to the root formula of quadratic equation of one variable can only formalize the standard cubic equation of one variable of type ax ^ 3 + BX ^ 2 + CX + D + 0 into a special type of x ^ 3 + PX + q = 0
Only a few can be factorized
I'm a freshman in high school
Algebra, like this one
You can try it instead of plus or minus 1, 2, 4, 7, 28
Calculus, what do you want to see
You can divide this equation by x-a. if you want to divide it, a can be found
More attention should be paid to polynomial division
It should be possible to decompose factors, based on experience
The factorization of quadratic equation of one variable uses the cross factorization method.
The factorization of cubic equation of one variable can be done by double cross factorization.
I won't tell you if you copy the one in front of me, but there are still mistakes,
P and P ^ 3 are both wrong?
If we derive the function of the equation, we can do the pulling
In high school, you can only calculate by algebraic value..
Let's say simple numbers like 1, 2,.. As you said, this problem can replace 1..
Then we propose X-1 and decompose the rest..
Wait until the University, learned advanced mathematics, you can use other methods to solve, but also a little difficult..
High school stage, can only be observed equation, through factorization to complete.
For example, X3 + 3x2-24x + 28 = 0 is changed into (x + 7) (X-2) (X-2) = 0. If this problem is changed: x ^ 3 + 4 * x ^ 2-17 * x + 28 = 0, (x + 7) * (x ^ 2-3 * x + 4) = 0, its solution is as follows:
[ -7]
[ 3/2+1/2*i*7^(1/2)]
[ 3/2-1/2*i*7^(1/2)]
1. Unfold
High school stage, can only be observed equation, through factorization to complete.
For example, X3 + 3x2-24x + 28 = 0 is changed into (x + 7) (X-2) (X-2) = 0. If this problem is changed: x ^ 3 + 4 * x ^ 2-17 * x + 28 = 0, (x + 7) * (x ^ 2-3 * x + 4) = 0, its solution is as follows:
[ -7]
[ 3/2+1/2*i*7^(1/2)]
[ 3/2-1/2*i*7^(1/2)]
A real root, a pair of imaginary roots. Where I = (- 1) ^ 0.5
Point C is the golden section point (AC > BC) of line ab. try to explain that the golden ratio is about 0.618
According to the golden ratio, x = BC / AC = AC / ab
Let AC = 1 have BC = X
There is x / 1 = 1 / (1 + x)
There is x ^ 2 + X-1 = 0
The solution is x = (√ 5-1) / 2
Approximately equal to 0.618
Pro, talk to me!
It is known that point C is the golden section point of line AB, AB / AC is about 0.618a, and the approximate value of CB / AC is obtained
0.618a
It is known that C is the golden section of line AB, and AC > BC, CB is equal to 2 cm, then AB is equal to 2 cm
1:0.618
First of all, you need to know that the golden ratio is 1:0.618
CB = 2 cm, and AC > BC
Then AC: BC = 1:0.618
Then AC = 3.236 cm
The golden ratio is 0.618. According to the title, AC = 0.618 * AB, BC = 0.382 * AB, so AB = 2 / 0.382 = 5.236cm.
A point C on the line AB divides the line AB into AC and BC, and then the line AB is called the golden section point by point C. The ratio of AC and ab is（
A point C on the line AB divides the line AB into AC and BC, and then the line AB is called golden section point by point C. The ratio of AC and ab is (approximately equal to) called golden ratio.
A point C on AB divides the line segment AB into AC and BC, and AC ^ 2 = AB * BC, then the line segment AB is called the golden section point of point C, and the ratio of AC and ab is 0.618（
The root of half minus one is about 0.618
AC＞BC
How to deduce Weida theorem
Ax & # 178; + BX + C = 0, (a ≠ 0), that is, two of a (X & # 178; + BX / A + C / a) = 0 -------- 1) are x1, then the original equation is equivalent to the equation: a (x-x1) (x-x2) = 0, that is, a [x & # 178; - (x1 + x2) x + x1x2] = 0 -------- 2) contrast 1)
Finding the general solution of cubic equation and Weida's theorem
The general solution of cubic equation
The content of Vieta's theorem
In the univariate quadratic equation AX ^ 2 + BX + C = 0 (a ≠ 0 and △ = B ^ 2-4ac ≥ 0)
Let two roots be X1 and x2
Then X1 + x2 = - B / A
X1*X2=c/a
Generalized Weida theorem
Generally, for an equation of degree n with one variable ∑ AIX ^ i = 0
Its roots are recorded as x1, X2 Xn
We have
∑Xi=(-1)^1*A(n-1)/A(n)
∑XiXj=(-1)^2*A(n-2)/A(n)
...
∏Xi=(-1)^n*A(0)/A(n)
Where ∑ is the sum and Π is the product
If the quadratic equation of one variable
The root in the complex set is, then
Weida, a French mathematician, first discovered this relationship between the roots and coefficients of algebraic equations. Therefore, people call this relationship Weida's theorem. History is interesting. Weida obtained this theorem in the 16th century. The proof of this theorem depends on the basic theorem of algebra. However, the basic theorem of algebra was first made by Gauss in 1799
From the basic theorem of algebra, we can deduce: any equation of degree n with one variable
Therefore, the left end of the equation can be decomposed into a product of factors in the complex range
Where is the root of the equation. By comparing the coefficients at both ends, we obtain the Veda theorem
Weida's theorem is widely used in equation theory
How to build equation with Veda theorem?
For example, X1 + x2 = - 4 / 17, X1 * x2 = 5
You'd better tell me how to deduce it
What I mean is to know the result of Vader's formula and use it to build a quadratic equation of one variable
This theorem actually says
In a quadratic equation of one variable
For example, ax + BX + C = 0 (this side must be 0)
A, B and C are three constants respectively
And x1x2 is the two solutions of this equation. Of course, X1 = x2
How can I reason and forget "it's too long."
In a word, when we find the solution of the equation by solving the three numbers ABC
It's very convenient to do this. You just need to remember the formula
Establish the equation
If we knew x1, x2
Suppose X1 = 1, X2 = 2‘
Then X1 + x2 = 1 + 2 = - B / a = 3
X1X2＝1*2＝C／A＝2
If you know one of the constants, you can know what the three numbers are
X1+X2＝－B／A
X1X2＝C／A
Do you understand?
x1+x2=-b/a
x1*x2=c/a
X1 and X2 are two solutions respectively!
aX2+bX+c=0（a≠0）
Weida theorem: X1 + x2 = - B / A
X1×X2=c/a
The derivation is based on the formula: X1 + x2 = - B + √ (b2-4ac) / 2A + [- B - √ (b2-4ac)] = - B / A
The other is the same
Because x1 × + x2 = - 4 / 17 = - B / A
x1×x2=5 =c/a
And then it can be solved
Veda's theorem for equation of degree n with one variable
To find the formula of the Veda theorem for the equation of degree n with one variable, it is better to be more detailed,
For example, a cubic equation is x ^ 3 + PX ^ 2 + QX + r = 0
a. B and C are its three roots, then the equation can also be written as:
（x-a）（x -b）（x-c）=0
Expand to get x ^ 3 - (a + B + C) x ^ 2 + (AB + BC + Ca) x - ABC = 0
According to the corresponding equality of the coefficients, the following results are obtained:
Thus & nbsp; - (a + B + C) = P
（ab+bc+ca）=Q
 -abc=R
In addition, to a higher level:
For details, please refer to relevant information