How to factor x^2 -8x +17 with imaginary i (more) how to... please

16x^4 -81
x^4 -10x^2 + 25


i need to find the zeros, this kind of factoring includes imaginary number "i"

x^2 -8x +17
= (x^2 -8x +16) + 17-16
= (x-4)^2 + 1
= (x-4)^2 - i^2
= (x -4 -i) (x -4 +i)
so x = 4+i or 4-i

16x^4 -81
= (4x^2)^2 - 9^2
= (4x^2 -9)(4x^2 +9)
= ( [2x]^2 -3^2 ) (4x^2 +9)
= (2x -3)(2x+3)(4x^2 -9i^2)
= (2x-3)(2x +3)( 2x -3i)(2x+3i)
so x = 3/2 , -3/2 , 3i/2 or -3i/2

x^4 -10x^2 + 25
= ( x^2 -5)(x^2 -5)
= (x^2 - [ sqrt 5]^2 ) (x^2 - [ sqrt 5]^2 )
= (x - sqrt5) (x + sqrt 5)(x - sqrt5) (x + sqrt 5)
so x = sqrt 5 or - sqrt 5

2 roots, both contain imaginary numbers (which always occur in pairs)
just use the quadratic equation and you will get a negative square root, and use i for the negative number.
4+i and 4-i

x^2 - 8x + 17 = 0
x^2 - 8x = -17
x^2 - 8x + 16 = -17 + 16
(x - 4)(x - 4) = -1
(x - 4)^2 = -1
x - 4 = ± √-1
x - 4 = ± i
x = 4 ± i


16x^4 - 81 = 0
(4x^2 + 9)(4x^2 - 9) = 0
(4x^2 + 9)(2x + 3)(2x - 3) = 0

4x^2 + 9 = 0
4x^2 = -9
x^2 = -9/4
x = ± √(-9/4)

x = ± 3i/2
x = -3/2
x = 3/2


x^4 - 10x^2 + 25 = 0
(x^2 - 5)(x^2 - 5) = 0

x^2 - 5 = 0
x^2 = 5
x = ± √5

Quadratic Formula always works, X = -B + or - the square root of B squared - 4(a)(c) divided by 2(a)