Question

2(4-i) + (2-i)(3+i

Answer 1

This works pretty much like it would if i were a normal variable (like x) in algebra. The only 'weirdness' is the fact that
\[\Large i^2 = -1\]

Answer 2

????????

Answer 3

Right...
\[\Large \color{green}2(4-i)+(2-i)(3+i)\]
See that 2 I marked in green Distribute it over on this group:\[\Large \color{green}2\color{red}{(4-i)}+(2-i)(3+i)\]

Answer 4

yes

Answer 5

Yes? Distribute it, what do you get?

Answer 6

\[(8-2i)+(2-i)(3+i)\]

Answer 7

Good. Now What about this pair?
You can use FOIL like you normally would if these were polynomials...
\[\Large (8-2i)+\color{blue}{(2-i)(3+i)}\]

Answer 8

\[ummm (8-2i)+(6-i-i)?????\]

Answer 9

I see... What if it were an x instead?
\[\Large (2-x)(3+x)\]
How would you FOIL this?

Answer 10

<shrugs> whichever works, I guess.
I can see that grouping is easier to apply. @jazzyfa30 ?
Check that out ^

Answer 11

Oh my goodness....\[\large \color{green}2(3+i)-\color{green}i(3+i)\]

Answer 12

good job

Answer 13

Oh, so you do remember your FOIL
Great :)
Now, apply it.
First (product) plus Outer (product) plus Inner (product) plus Last terms (product)

Answer 14

um ok