Factorization method for solving cubic equation of one variable How can x ^ 3 + x ^ 2-2 = 0 be decomposed into (x-1) (x ^ 2 + 2x + 2) = 0?

Factorization method for solving cubic equation of one variable How can x ^ 3 + x ^ 2-2 = 0 be decomposed into (x-1) (x ^ 2 + 2x + 2) = 0?

X^3+X^2-2
=x^3-1+x^2-1
=(x-1)(x^2+x+1)+(x-1)(x+1)
=(x-1)(x^2+2x+2)
Sometimes it can be done by observation
How can cubic equation of one variable be factorized quickly and effectively?
There is a formula for finding the root of cubic equation of one variable - Cardin's formula, but the homework in books or examinations can usually be decomposed into at least one first-order factor. If there is a rational factor, then the root is the factor of constant term, and 1 or - 1 are the most common two factors. Try to substitute this factor into the equation in advance. If it is 0, then it means that this factor is the root of the equation, In this way, we can get a factor first. Then we can get another quadratic factor by long division
Using factor theorem
How to factorize cubic equation of one variable quickly
Try the root, 1, - 1 and so on into, decompose into a factor, and then divide
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Vedadine's understanding of quartic equation
1. The four roots of the equation x ^ 4-4x ^ 3-24x ^ 2 + 56x + 52 = 0 form an arithmetic sequence to find the solution set of the equation
2. The four roots of the equation x ^ 4-4x ^ 3-34x ^ 2 + ax + B = 0 form an arithmetic sequence, and find the solution set of the equation a, B
Let these four roots be a-3b, A-B, a + B, a + 3B (let b > 0)
Then according to Weida's theorem: the sum of four roots = 4A = 4, we get a = 1
The product of four roots (1-3b) (1-B) (1 + b) (1 + 3b) = 52
That is, (1-9b ^ 2) (1-B ^ 2) = 52
The solution is B ^ 2 = 3, that is, B = √ 3
So the solution set {1-3 √ 3,1 - √ 3,1 + √ 3,1 + 3 √ 3}
The method of the second question is similar
Weida theorem equation
How to use X1 + X2, X1 * X2 to establish an equation in Weida theorem?
X^2-(x1+x2)X+x1*x2
x1+x2=a,x1x2=b
The equation is x ^ 2-ax / 2 + B = 0
A problem about Veda's theorem
Xiao Ming and Xiao Hong do homework together. When solving a quadratic equation of one variable, due to carelessness, Xiao Ming wrongly writes the constant term in the process of simplification, so the two roots of the equation are 8 and 2. Xiao Hong wrongly writes the coefficient of the first term in the process of simplification, so the two roots of the equation are - 9 and - 1. Can you find the original equation?
The coefficient of a term: - (8 + 2) = - 10
Constant term: = (- 9) * (- 1) = 9
Equation: x ^ 2-10x + 9 = 0
Q: why is the coefficient of the first term negative?
Weida's theorem: X1 + x2 = - B / A, where a = 1
So B = - (x1 + x2)
Two unsolvable equations (on Veda's theorem)
1. Given p + q = 198, find the integer root of the equation x2 + PX + q = 0
If we set the roots x1, X2,
So, according to Veda's theorem,
Here we are
X1X2-(X1+X2)=198
X1X2-X1-X2+1=199
(X1-1)(X2-1)=199
X1 and X2 are positive integers,
So how to find the solution?
X1=?X2=?
And, (x1 + 1) (x2 + 1) = 12
(x1－1)(x2－1)=3
What's the solution,
2. It is known that the image of quadratic function y = - x2 + PX + Q intersects with X axis at (α, 0) and (β, 0), and α > 1 > β
[1] The first question goes on to your: (x1-1) (x2-1) = 199,
X1 = [199 / (x2-1)] + 1, x1, X2 are integer solutions, 199 are prime numbers, then x2-1 = 1, 199, X2 = 2200;
When x2 = 2, X1 = 200; when x2 = 200, X1 = 2;
The integer solution is 2200
[2] In fact, α β is the root of y = 0, the opening is downward, 1 is between two, f (1) > 0
-1+p+q>0
p+q>1
1、(X1-1)(X2-1)=199
(x1-1) and (x2-1), one equals 1 and the other equals 199
So X1 and X2, one equals 2 and the other equals 200
2. Because α > 1 > β, α - 1 > 0, β - 1
On the proof of Vader's theorem
When we prove Vader's theorem:
f(X)=An(X-X1)(X-X2)...(X-Xn)
Why is it equal to
An[X^n - (X1+X2+..+Xn)X^(n-1) + (X1X2+X1X3+...+Xn-1Xn)X^(n-2) +
...+ (-1)^(n)X1X2..Xn]
(x-x1)(x-x2)…… How does (x-xn) open
The root formula of quadratic equation of one variable is: x = (- B ± √ B ^ 2-4ac) / 2a, then X1 = (- B + √ B ^ 2-4ac) / 2a, X2 = (- B - √ B ^ 2-4ac) / 2a, X1 + x2 = (- B + √ B ^ 2-4ac / 2a) + (- B - √ B ^ 2-4ac / 2a) X1 + x2 = - B / a X1 * x2 = (- B + √ B ^ 2-4ac / 2a) * (- B - √ B ^ 2-4ac / 2a) X1 * x2 = C /
This problem proves Veda's theorem
It is known that X1 and X2 are two heels of the quadratic equation AX squared + BX + C = 0 (a is not equal to 0, B squared - 4ac is greater than or equal to 0)
From the root formula X1 = [- B - √ (B & # 178; - 4ac)] / 2ax2 = [- B + √ (B & # 178; - 4ac)] / 2a, so X1 + x2 = [- B - √ (B & # 178; - 4ac) - B + √ (B & # 178; - 4ac)] / 2A = - 2b / 2A = - B / ax1x2, so x1x2 = [(- b) &# 178; - (B & # 178; - 4ac)] / (2a * 2A) = 4ac / 4A & # 178; = C / A
Use the root formula to express two, then add and multiply
How can a quadratic equation of one variable have two roots in Weida's theorem?
Weida's theorem (known in current textbooks as the relationship between the root and coefficient of a quadratic equation with one variable): the sum of two elements of a quadratic equation with one variable is equal to the opposite number of the quotient obtained by dividing its coefficient of the first term by its coefficient of the second term; the product of two elements is equal to the quotient obtained by dividing its constant term by its coefficient of the second term
ax^2+bx+c=0 ,a≠0
There are: X1 + x2 = - B / A; x1 × x2 = C / A
When Δ = B ^ 2-4ac ≥ 0, there is a real root,
If there is no conditional restriction, there must be roots. Only the range of numbers is extended to the plural
If Δ = 0.. then there is only one root=