Factoring x ^ 4 + 4 in complex number

Factoring x ^ 4 + 4 in complex number

The factorization of the upper floor can be √ (2I) I X & # 8308; + 4 = (X & # 178; + 2I) * (X & # 178; - 2I) suppose (a + bi) &# 178; = 2I, then a & # 178; - B & # 178; + 2abi = 2I, then a & # 178; - B & # 178; = 0, ab = 1A = 1, B = 1 or a = - 1, B = - 1, so (1 + I) &# 178; = 2I, or (- 1-I) & # 178; = 2I

Factorization: x ^ 2 + 5Y ^ 2 (in complex number)

x²+5y²
=x²-5i²y²
=(x-√5iy)(x+√5iy)

Factoring factor 2x & # 178; - 4x + 5 in complex set C is equal to?

2X "- 4x + 5 = 2x" - 4x + 2 + 3 = 2 [(x - 1) "+ 3 / 2] = 2 [(x - 1)" - (- 6 / 4)] = 2 {(x - 1) "- [(√ 6 / 2) I]} = 2 [x - 1 - (√ 6 / 2) I] [x - 1 + (√ 6 / 2) I] or = 2 [2x / 2 - 2 / 2 - (√ 6 / 2) I] [2x /