If the quadratic polynomial x ^ 2 + 2kx-3k can be divided by X-1, try to find the value of K Note: it's 3k, not 3K ^ 2

If the quadratic polynomial x ^ 2 + 2kx-3k can be divided by X-1, try to find the value of K Note: it's 3k, not 3K ^ 2

The designer is a
Then x ^ 2 + 2kx-3k = a (x-1)
x=1,x-1=0
Right side = 0
So the left side is also equal to 0
x=1
x^2+2kx-3k=1+2k-3k=0
k=1

If the quadratic polynomial x2 + 2kx-3k2 can be divided by X-1, then the value of K is______ ．

∵ the polynomial x2 + 2kx-3k2 can be divided by X-1, ∵ X-1 is the factor of the polynomial x2 + 2kx-3k2, ∵ X-1 = 0, that is, x = 1 is the solution of the polynomial x2 + 2kx-3k2, ∵ 12 + 2k-3k2 = 0, solving the univariate quadratic equation about K, we get K1 = 1, K2 = - 13, so the answer is: k = 1 or K = - 13

If the quadratic polynomial x ^ 2 + 2kx-3k can be divided by X-1, then another factor of the polynomial is obtained_______

x+3
One factor is X-1, which means that when x = 1, the value of the algebraic formula is 0, and K = 1 is obtained by substituting it
Another factorization is x + 3

How to calculate trapezoidal sand cube

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How to calculate trapezoidal cube

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Verification: the cubic sum of three consecutive positive integers is a multiple of 9

Let me try to prove it
Suppose that the three consecutive positive integers are: A-1, a, a + 1 (a is an integer greater than 1)
So the cubic sum of the three positive integers is: (A-1) ^ 3 + A ^ 3 + (a + 1) ^ 3
=a^3-3a^2+3a-1+a^3+a^3+3a^2+3a+1
=3a^3+6a
=3a(a^2+2)
. back

M + n is a multiple of 6. It is proved that the cube of M minus the cube of n is also a multiple of 6
"m^3-n^3
=(m-n)(m^2+mn+n^2)
Because M-N is a multiple of 6, so“
I'm talking about M + n being a multiple of six
“6|m+n
And m ^ 3-N ^ 3 = (M + n) (n ^ 2-MN + m ^ 2) must be divisible by 6
But m ^ 3-N ^ 3 is not equal to (M + n) (n ^ 2-MN + m ^ 2)

This is a false proposition
If M = 4, n = 2, then M + n = 6 is qualified, but
M ^ 3-N ^ 3 = 64-8 = 56 is not a multiple of 6

Real explanation: when n is a positive integer, the cube of n minus n must be a multiple of 6

n^3-n
=n(n^2-1)
=n(n+1)(n-1)
It's (n-1) * n * (n + 1)
Do you see that? The result of multiplying three consecutive numbers must be a multiple of 6. Because at least one of the three numbers must be a multiple of 2 and one of them must be a multiple of 3. The result must be a multiple of 6

To prove that the cube of a minus a is a multiple of 6

It can be decomposed into a (a + 1) (A-1)
In these three consecutive integers,
One must be even (divisible by 2),
Every three cycles of a number divisible by three,
So there must be a number in it that can be divided by three,
2x3 = 6, so the cube of a minus a is a multiple of 6

Proof 1. When n is a positive integer, n ∧ 3-N must be a multiple of 6
2. The sum of the product of four continuous natural numbers and one must be a complete square number

1.
n∧3-n = n(n^2 -1) = n(n+1)(n-1)
-(1) If - n is a positive integer, then there must be a multiple of 3 in N, N + 1, n-1
-(2) If - n is a positive integer, then there must be a multiple of 2 in N, N + 1
So n (n + 1) (n-1) is a multiple of 6
two
n(n+1)(n+2)(n+3) +1
Open and close
n(n+1)(n+2)(n+3)+1
=n(n+3)(n+1)(n+2)+1
=(n^2+3n)(n^2+3n+2)+1
Let n ^ 2 + 3N = X
The above formula = x (x + 2) + 1
= X^2+2X+1
= (X+1)^2.
= (n^2+3n+1)^2.
So it must be a perfect square
In addition, 1 ^ 3 - 1 = 1 - 1 = 0, 0 is a multiple of 6