here's the expression: (4j−2)2−(2+4j)2
here's what i did:
(4j−2)(4j−2)−(2+4j)(2+4j)
(16j2−8j−8j+4)−(4+8j+8j+16j2)
(16j2−16j+4)−(16j2+16j+4)
16j2−16j+4−16j2−16j−4 (combine like terms)
−32j
i know i can factor out more, but 32 has many factors (2, 4, 8, 16). can i use any factor or is there a rule that i must use a specific set (like the lowest)? i just wanted to make sure i was factoring this last term correctly!
It all looks correct and there isn't much else you can do with it!
You can write it in its simplest form such as: -32j =-2^5j
+9488 Maybe your teacher wants you to factor the given expression as a difference of squares? Like this....
a2−b2=(a+b)(a−b) so.....
(4j−2)2−(2+4j)2=((4j−2)+(2+4j))((4j−2)−(2+4j)) (4j−2)2−(2+4j)2=((4j−2)+(2+4j))((4j−2)−(2+4j))
Simplifying further...
(4j−2)2−(2+4j)2=(4j−2+2+4j)(4j−2−2−4j) (4j−2)2−(2+4j)2=(4j+4j)(−2−2) (4j−2)2−(2+4j)2=(8j)(−4) (4j−2)2−(2+4j)2=−32j
If you want to factor -32j , then you want to get it into its prime factors.
( For example, if you chose to factor it into -1 * 4 * 8 * j then you could still factor out more,
because 4 = 2 * 2 and 8 = 4 * 2 
)
-32j = -1 * 2 * 2 * 2 * 2 * 2 = -25j in its prime factored form.
