A mathematical problem about factorization x^5+x^4+x^3+x^2+x Oh, yes. It's factorization. I forgot to say it. Well, it's better to be more detailed, thank you! Hurry up!

A mathematical problem about factorization x^5+x^4+x^3+x^2+x Oh, yes. It's factorization. I forgot to say it. Well, it's better to be more detailed, thank you! Hurry up!

Can only be decomposed into x (x ^ 4 + x ^ 3 + x ^ 2 + X + 1)
If the decomposition can be continued, then x ^ 4 + x ^ 3 + x ^ 2 + X + 1 = 0 must have a real root
Why is there no real root? Please see the formula for finding the root of quartic equation of one variable
x^5+x^4+x^3+x^2+x
＝x(x^4+x^3+x^2+x+1)
=x(x^5-1)/(x-1)
Ask a mathematical factorization problem
1.a^3-ab^2+4abc-4ac^2
Urgent! Thank you for your help!
A ^ 3-AB ^ 2 + 4abc-4ac ^ 2 = a (a ^ 2-B ^ 2 + 4bc-4c ^ 2) = a [a ^ 2 - (b ^ 2-4bc + 4C ^ 2)] = a [a ^ 2 - (b-2c) ^ 2] = a (a + b-2c) (a-b + 2C) the first step is to extract the public part, the second part is a ^ 2 + 2Ab + B ^ 2 = (a + b) ^ 2, and the last step is a ^ 2-B ^ = (a + b) (a-b)
1+1=2
a((a+2c)(a-2c)+b(4c-b))
Maybe so
a^3-ab^2+4abc-4ac^2
=a(a^2-b^2+4ab-4c^2)
a[a＋2ca-2ac
Why does x2 + 3x-4 = (x-4) (x + 1) not belong to factorization
Because it's wrong
The correct one is (x + 4) (x-1)
Given the equation (M-4) M = x-4 and M is not equal to 1, find the value of 2x ^ 2 - (3x-x ^ 2-2) + 1
If M is not equal to 1, then it is equal to 5. If M is equal to x-4, we can get x = 9. Take it into the formula to get the answer: 220
Positive integer solutions of bivariate linear equation
What is a group of positive integer solutions of the bivariate linear equation x-2y = 3
X = 11, y = 4 is a set~~~~
Given the equation (x-4) M = x-4 and m ≠ 1, find the value of 2x2 - (3x-x2-2) + 1
From (x-4) M = x-4, (x-4) (m-1) = 0, ∵ m ≠ 1, ∵ M-1 ≠ 0, ∵ x-4 = 0, ∵ x = 4, 2x2 - (3x-x2-2) + 1 = 2x2-3x + x2 + 2 + 1 = 3x2-3x + 3 = 3 × 42-3 × 4 + 3 = 48-12 + 3 = 51-12 = 39
The integer solution of the binary equation (x-2013) ^ 2 times (y + 2013) + 13 = 0 is
(x-2013)²(y+2013)=-13=1²×(-13)
So x-2013 = ± 1
y+2013=-13
therefore
x=2012,y=-2026
x=2014,y=-2026
Given the equation (x-4) M = x-4 and m ≠ 1, find the value of 2x2 - (3x-x2-2) + 1
From (x-4) M = x-4, (x-4) (m-1) = 0, ∵ m ≠ 1, ∵ M-1 ≠ 0, ∵ x-4 = 0, ∵ x = 4, 2x2 - (3x-x2-2) + 1 = 2x2-3x + x2 + 2 + 1 = 3x2-3x + 3 = 3 × 42-3 × 4 + 3 = 48-12 + 3 = 51-12 = 39
The quadratic equation 4x-7y = 1 has several pairs of integer solutions
4x-7y=1
x=(7y+1)/4=（4y+3y+1)/4=y+(3y+1)/4
3Y + 1 is a multiple of 4, y = 1,5,9,13,17
x=2,9,16,23,30……
So there are innumerable pairs of integers like this
PS: is the theme right?
Given the quadratic equation 4x-7y = 1, how many pairs of integers satisfy the equation? What is the law of all integer pairs? What is the minimum value of X and y?
The deformation of the equation is x = 7 / 4 * y + 1 / 4
Let y = 4K + m, where k is an integer (can be negative), M = 0,1,2,3
The equation is reduced to
X = 7K + 7/4 * M +1/4
Where 7K is an integer, in order to make x an integer, it must be expanded by 7 / 4 *
Given the quadratic equation 4x-7y = 1, how many pairs of integers satisfy the equation? What is the law of all integer pairs? What is the minimum value of X and y?
The deformation of the equation is x = 7 / 4 * y + 1 / 4
Let y = 4K + m, where k is an integer (can be negative), M = 0,1,2,3
The equation is reduced to
X = 7K + 7/4 * M +1/4
Where 7K is an integer, in order to make x an integer, 7 / 4 * m + 1 / 4 must be an integer
M = 0,2,3 can not satisfy 7 / 4 * m + 1 / 4 as an integer
Only m = 1
So y = 4K + 1
X=7K+2
K is an integer.
This is the law that integer pairs satisfy. You can see that there are infinite pairs.
|X|+|Y| = |7K+2| + |4K+1|
Obviously, when k = 0, take the minimum value of 3, then x = 2, y = 1, obviously (2,1) is a set of solutions of 4x-7y = 1
Let (XO, yo) also be a set of solutions of 4x-7y = 1, then
4xo-7yo=1
4*2-7*1=1
Subtraction of two formulas
4(xo-2)-7(yo-1)=0
(xo-2)/(yo-1)=7/4
Let xo-2 = 7K
yo-1=4k
xo=7k+2
yo=4k+1
K can be any integer, so there are countless (XO, yo)
When | x | + | y | = | 7K + 2 | + | 4K + 1 |, k = 0
So | x | + | y | = | 7K + 2 | + | 4K + 1 | = 3 |
Countless!
Given the equation (x-4) M = x-4 and m ≠ 1, try to find the value of 2x & sup2; - (3x-x & sup2; - 2) + 1
（x-4）m=x-4
If M ≠ 1, then x-4 = 0
X=4
So the original formula = 3x & sup2; - 3x + 3
=48-12+3
=39