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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!

Guest Jul 12, 2018

#1**+1 **

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

Guest Jul 12, 2018

#2**+1 **

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 = -2^{5}j in its prime factored form.

hectictar Jul 12, 2018