Algebra 2 help needed for simplifying?

(2i^5 - 2i^2)/(4i^3 - 1)

= [2(i) - 2(- 1)]/ [4 ( - i) - 1]

= [ 2i + 2]/ [ - 4i - 1]

Now you have to multiply top and bottom by 4i - 1

[(2i + 2)(4i - 1)]/[(- 1 - 4i)( -1 + 4i)]

= [8i^2 - 2i + 8i - 2]/[1 - 4i + 4i - 16i^2]

= [ - 8 + 6i - 2]/[ 1 - 16(-1)]

= [ -10 + 6i]/ 17

Will use j for i

2j^5 - 2j^2
---------------
4j^3 - 1

2j + 2
--------
-4j - 1

- 2 [ 1 + j ]
---------------
[ 1 + 4j ]

- 2 [ 1 + j ] [ 1 - 4j ]
------------------------------
17

- 2 [ 1 - 3j + 4 ]
-------------------
17

- 2 [ 5 - 3j ]
------------------
17

Assuming i is the imaginary unit ("square root of -1") then i² = -1 and i⁴ = 1. You can simplify all powers of i down to ±1 or ±i.

(2i⁵ - 2i²) / (4i³ - 1) = (2i + 2) / (-4i - 1)
= - (2 + 2i) / (1 + 4i)

Multiply by (1-4i)/(1-4i) to get rid of the complex denominator:

= - [(2 + 2i)(1 - 4i)] / [(1 + 4i)(1 - 4i)]
= - (2 - 8i + 2i + 8)/ [1 - (4i)²]
= - (10 - 6i) / 17
= (-10 + 6i) / 17