If the quadratic polynomial x2 + 2x + K can be divided by X-1, then the value of K is?

I think we should first split x2 + 2x + K into X-1 related ones
Become (x-1) 2 + 4x-1 + K
Then it becomes (x-1) 2 + 4 (x-1) + 3 + K
When k = - 3, it can be divisible
It seems so

If X3 + AX2 + BX + 8 has two factors x + 1 and X + 2, then a + B = ()
A. 7B. 8C. 15D. 2l

Let X3 + AX2 + BX + 8 = (x + 1) (x + 2) (x + C) = X3 + (3 + C) x2 + (2 + 3C) x + 2c, ∧ C = 4, so a = 7, B = 14, ∧ a + B = 21, so D

If the value of AX2 + BX + C is equal to 0 when x = m, then AX2 + BX + C contains a factor (x-m). Please use this method to determine a factor of X3 + 4x2 + X-2

Because when x = - 1, x ^ 3 + 4x ^ 2 + X-2 = 0, there is a factor (x + 1) for x ^ 3 + 4x ^ 2 + X-2

Mathematical proof: when n is a positive integer, the value of n ^ 3-N must be a multiple of 6

Mathematical induction
(1) When n = 1, 1 ^ 3-1 = 0 can be divided by 6
When n = 2, 2 ^ 3-2 = 6 can be divided by 6
(2) Suppose that when n = K (k is a positive integer), K ^ 3-K can be divided by 6
Then when n = K + 1, (K + 1) ^ 3 - (K + 1) = (K + 1) [(K + 1) ^ 2-1] = (K + 1) (K + 2) k
K (K + 1) (K + 2) is the product of three consecutive positive integers
There must be an even multiple of 3 in three consecutive positive integers
So there are 2 and 3 factors in K (K + 1) (K + 2) that can be determined by 6 integers
Synthesis (1) (2) shows that for any positive integer n ^ 3-N must be a multiple of 6

Proof: when n is a positive integer, the value of n ^ 3-N must be a multiple of 6
This is a proof, please give detailed process
Thank you!

Mathematical induction (1) when n = 1, 1 ^ 3-1 = 0 can be divided by 6, when n = 2, 2 ^ 3 / 2 = 6 can be divided by 6. (2) suppose that when n = K (k is a positive integer) k ^ 3-K can be divided by 6, then when n = K + 1, (K + 1) ^ 3 - (K + 1) = (K + 1) [(K + 1) ^ 2-1] = (K + 1) (K + 2) KK (K + 1) (K + 2) is the product of three consecutive positive integers

Is the cube of polynomial 1-A + the square of a quadratic trinomial, cubic trinomial, cubic trinomial or quintic binomial

The cube of 1-A + the square of a is a cubic trinomial
There are three terms in this polynomial, and the highest degree is degree 3

If the polynomial (K's Square-1) x's square + (k-1) about X is a binomial of first degree, find K

Wrong, there should be x outside k-1, and the constant term is missing
It's a simple formula
So there is no quadratic term
So the coefficient of X & # 178 is 0
k²-1=0
k=±1
The coefficient of the first term is not equal to 0
k-1≠0
So k = - 1

Square of 12, square of - 3, cube of - 2, square of - 9, cube of 3

Square of 12, square of - 3, cube of - 2, square of - 9, cube of 3
=144÷9×（-8）-81÷27
=-128-3
=-131

Proof: in any five natural numbers, there must be three numbers, and their sum is a multiple of 3
Why each drawer has at least three numbers? Hee hee, I don't understand!

Is this a question for me?
The first floor quoted my answer to this question in 2009
I graduated in 2009 for one year, and now I have worked for more than three years
But if you look at my answer at that time carefully, you can try your best to explain it again
First of all, divide all natural numbers into three drawers, which should be understandable
Secondly, any five natural numbers must be taken from the three drawers above
1. If one or two of the three drawers do not take out, there must be at least three numbers in the same drawer
Divide three numbers in the same drawer by 3 to get the same remainder R, so the sum of them must be a multiple of 3
2. Each of the three drawers has to take data, then the method must be 1, 2, 2
In this way, take one number from each drawer, then the remainder of the three numbers divided by 3 are 0, 1 and 2 respectively. Therefore, their sum must be divisible by 3 (0 + 1 + 2 is divisible by 3)
From the above, no matter how five numbers are distributed, the sum of three must be a multiple of three
I don't know if the landlord can understand after I explain this?

Why is the sum of 3 of any 5 natural numbers a multiple of 3
To prove

If at least three of the five random natural numbers are in the same drawer, then the three numbers are divided by 3 to get the same remainder R, so their sum must be a multiple of 3

If the quadratic polynomial x2 + 2x + K can be divided by X-1, then the value of K is?