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How to factorize x ^ 3 + 2x ^ 2 + 1 = 0
The formula for finding the root of the univariate cubic equation x ^ 3 + a1x ^ 2 + a2x + a3 = 0 (also known as kardan formula) is: let P = - (A1 ^ 2 / 3) + A2, q = 2A ^ 3 / 27-a1a2 / 3 + a3 have
x1=[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)+[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)=u+v
x2=-1/2(u+v)+i/3^(1/2)*(u-v)
x3=-1/2(u+v)-i/3^(1/2)*(u-v)
So the result is (x-x1) (x-x2) (x-x3)
In the range of complex number, decompose factor X ^ 3 + 2x ^ 2 + 3x + 6
More detailed process ~ thank you~
x^3+2x^2+3x+6
=x^2（x+2）+3(x+2)
=(x+2)(x^2+3)
=(x + 2) (x + radical 3I) (x-radical 3I)
Factorization in complex number: x ^ 3-x ^ 2 + 2x=
The solution of x ^ 2-x + 2 = 0 is (1 ± I √ 7) / 2
So the original formula = x (x-1 / 2 + I √ 7 / 2) (x-1 / 2-I √ 7 / 2)
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