Please imitate the above method and try to factorize the polynomial (x? - 2x) (x? - 2x + 2) + 1

Please imitate the above method and try to factorize the polynomial (x? - 2x) (x? - 2x + 2) + 1

(x²-2x）（x²-2x+2）+1
=(x²-2x)²+2(x²-2x)+1
=[(x²-2x)+1]²
=(x²-2x+1)²
=(x-1)^4

Factorization: 2A 2x-2ax + 1 / 2x (to be decomposed!)

2a²x-2ax+1/2x
=1/2x(4a²-4a+1)
=1/2x(2a-1)²
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Factorization factor 2A (x? 2 + 1) - 2aX? 2x? 2x + 2x + 1 / 2

2a(x²+1）²-2ax²
=2a[(x²+1)²-x²]
=2a(x²+1+x)(x²+1-x)
2x²+2x+1/2
=1/2(4x²+4x+1)
=1/2(2x+1)²
If there is anything you don't understand, you can ask,

If there is a factor (X-2) in 2x? - x? - 5x + K, then the value of K is

2x³-x²-5x+k
=2x³-4x²+3x²-6x+x+k
=2x²(x-2)+3x(x-2)+(x+k)
If there is a factor (X-2) in 2x? - x? - 5x + K, then
x+k=x-2
k=-2

Xiao Ming thinks that there must be a factor (x-3) in the polynomial 2x? - 5x-3. He thinks that 2x? - 5x-3 = 2x? - 6x + x-3 = 2x (x-3) + (x-3) = (2x + 1) (x-3) Do you think it is reasonable for him to do so? If you think so, is there a factor (x-3) in the polynomial x? - 2x-3? If you think it is unreasonable, please point out the mistakes

Right
x²-2x-3
=x²-3x+x-3
=x(x-3)+(x-3)
=(x-3)(x+1)
So there are x-3's

We know that one factor of the polynomial 4x ⁴ + 8x? - 3x? - 7x is 2x + 3, find another factor? Urgent!

Since a factor of the polynomial 4x ⁴ + 8x cubic - 3x? - 7x is 2x + 3, the factorization of 4x ⁴ + 8x? - 3x? - 7x must include the factor of 2x + 3, so we can get 4x ⁴ + 6x? + 2x? + 3x? - 6x? - 9x + 2X, and the last 2x should have a pair of 3. I don't know why it is missing
According to the normal situation, 4x ⁴⁴ + 8x? - 3x? - 7x + 3 will get 4x ⁴ + 6x? + 2x? + 3x? - 6x? - 9x + 2x + 3,
In this paper, we obtain 2x 3 (2x + 3) + x 2 (2x + 3) - 3x (2x + 3) + (2x + 3). If a constant term 3 is missing, then the final factorization result should be (2x + 3) (2x 3 + X? - 3x + 1), then this factor is 2x? + X? - 3x + 1
Little advice, little power

Given that x2 + X-6 is a factor of the polynomial 2x4 + x3-ax2 + BX + A + B-1, then a=______ ；b=______ .

Let another factor be 2x2 + MX + N, then (x2 + X-6) (2x2 + MX + n)
=2x4+（m+2）x3+（m+n-12）x2+（n-6m）x-6n
Then:
m+2＝1
m+n−12＝−a
n−6m＝b
a+b−1＝−6n
The solution is as follows:
m＝−1
n＝−3
a＝16
b＝3
So the answer is: 16, 3

On the polynomial of X, after factorizing the factor 6x? - 11x + m, there is a factor (2x-3). Try to find the value of M (solve it in two ways)

Idea: because only 2x * 3x = 6x
So there must be 3x in the other factor
Let the other factor be (3x + a)
Then 6x? - 11x + M = (2x-3) (3x + a)
That is, 6x? - 11x + M = 6x? + 2ax-9x-3a = 6x? + (2a-9) x-3a
So - 11 = 2a-9, M = - 3a
So a = - 1. M = 3
Note: according to the undetermined coefficient method, similar to the determination of the analytic formula y = KX + B in the first order function, we find y = 2x + 3, that is, k = 2, B = 3, the same as that of k = 2, B = 3, where 6x? - 11x + M = 6x? + (2a-9) x-3a
So - 11 = 2a-9, M = - 3a

On the polynomial of X, 6x ^ 2-11x + m has a factor after factorization and a factor is (2x-3). Try to find the value of M

There is a factor of 2x-3
Then the solution of equation 2x-3 = 0 is a root of equation 6x? - 11x + M = 0
x=1.5
By substituting, we can get:
6*1.5²-11*1.5+m=0
13.5-16.5+m=0
M=3
verification:
6x²-11x+3
=(2x-3)(3x-1)

On the polynomial of X 2x ^ 2-11x + m, there is a factor x-3 after decomposition. Try to find the value of M

Let (x-3) (2x + k) = 2x ^ 2-11x + M
2X^2+kx-6x-3k=2X^2-11X+M
(k-6)x-3k=-11x+m
So K-6 = - 11
-3k=m
So k = - 5, M = 15