Factoring the polynomial x2-4x + 4, the result is () A. （x+2）2B. （x-2）2C. x（x-4）+4D. （x+2）（x-2）

X2-4x + 4 = (X-2) 2

Factoring factor 4x ^ 2-3 in the range of real numbers

4x^2-3
=(2x)^2-(√3）^2
=(2x-√3)(2x+√3)
☆⌒_ - [hope to help you~

Factoring the factor 4x ^ 2-4x-1 in real numbers

Let 4x ^ 2-4x-1 = 0
The solution is x = (4 ± 4 √ 2) / 8 = (1 ± √ 2) / 2
so
4X^2-4X-1＝4[x-(1-√2)/2][x-(1+√2)/2]

In the complex number field, the rational number field decomposes f (x) = x ^ 9 + x ^ 8. X ^ 2 + x ^ 1 + 1 into the product of irreducible factors!

Let XK = cos [2K π / 10] + isin [2K π / 10] (k = 1,2,9)
Then f (x) = (x-x1) (x-x2), (x-x9) (decomposed in complex field)

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