here's the expression: \({(4j - 2)^{2} - (2+4j)^2}\)
here's what i did:
\({(4j - 2)(4j - 2) - (2 + 4j)(2 + 4j)}\)
\({(16j^{2}-8j-8j+4)-(4+8j+8j+16j^{2})}\)
\({(16j^{2} - 16j +4)-(16j^{2}+16j+4)}\)
\({16j^{2} - 16j+4-16j^{2}-16j-4}\) (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
+9466 Maybe your teacher wants you to factor the given expression as a difference of squares? Like this....
\({\color{RedOrange}a}^2-{\color{JungleGreen}b}^2= ({\color{RedOrange}a}+{\color{JungleGreen}b})({\color{RedOrange}a}-{\color{JungleGreen}b}) \) so.....
\({\color{RedOrange}(4j-2)}^2-{\color{JungleGreen}(2+4j)}^2\,=\, ({\color{RedOrange}(4j-2)}+{\color{JungleGreen}(2+4j)})({\color{RedOrange}(4j-2)}-{\color{JungleGreen}(2+4j)}) \\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,((4j-2)+(2+4j))((4j-2)-(2+4j))\)
Simplifying further...
\( \phantom{(4j-2)^2-(2+4j)^2}\,=\,(4j-2+2+4j)(4j-2-2-4j)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,(4j+4j)(-2-2)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,(8j)(-4)\\~\\ \phantom{(4j-2)^2-(2+4j)^2}\,=\,-32j\)
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.
