okay help with this problem and then one more and I wont need anymore help tonight.
(2-3i(3-i)-(4-i)(4+i)

- anonymous

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- anonymous

\[(2-3i)(3-i)-(4-i)(4+i)\]

- anonymous

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

- anonymous

theres supposed to be one more parenthases after 3i

Looking for something else?

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## More answers

- anonymous

am i foiling this or is that not the right way to go about this problem?

- jim_thompson5910

you can FOIL or use the table like last time. Which option works best for you?

- anonymous

foil the table is a little confusing.

- jim_thompson5910

ok what do you get when you FOIL (2-3i)(3-i)

- bradely

(2 -3i)(3-i)-(4-i)(4+i)
(2)(3-i)-(3i)(3-i)-(4^2-i^2)
Multiply find the value
Note i^2 = -1
Source: mathskey.com

- anonymous

3-11i

- jim_thompson5910

correct

- jim_thompson5910

how about `(4-i)(4+i)` ?

- anonymous

that is 15

- anonymous

brb

- jim_thompson5910

not 15, but you are close

- jim_thompson5910

try again

- anonymous

okay

- anonymous

where did i go wrong with my problem i will show you my work.

- bradely

(4-i)(4+i)
(a-b)(a+b) =a^2-b^2
4^2 -i^2
16-(-1)=17

- anonymous

could the answer be 187i+51?

- anonymous

|dw:1441072256502:dw|

- jim_thompson5910

i^2 is -1
so \(\Large 16-\color{red}{i^2}\) turns into \(\Large 16-\color{red}{(-1)}\)

- anonymous

im not sure @sayomi

- jim_thompson5910

Do you see what I mean @huntergirl1 ?

- anonymous

yes now i do @jim_thompson5910

- jim_thompson5910

ok great, so
16-(-1) = 16+1 = 17
---------------------
\[\large \color{red}{(2-3i)(3-i)}-\color{blue}{(4-i)(4+i)} =\color{red}{3-11i}-\color{blue}{17} = -14-11i\]

- anonymous

@sayomi you were correct and yes @jim_thompson5910 i do understand

- jim_thompson5910

I'm glad it's making sense

- anonymous

so the ending result is 51-187i

- anonymous

okay do you wnt me to start a new window for the next problem? or to place it here?

- jim_thompson5910

how are you getting `51-187i` ?

- jim_thompson5910

I wrote the answer above

- anonymous

bc you put 17 back into the problem

- jim_thompson5910

are you saying that they wanted you to multiply `(2-3i)(3-i)` and `(4-i)(4+i)` ??

- anonymous

actually i put those answers both into the website they were both incorrect

- anonymous

hwy was the answer supposed to be -14-11i?

- anonymous

why

- jim_thompson5910

because the 3 and -17 combine to -14

- jim_thompson5910

(3-11i) - (17) = 3-11i - 17 = (3-17) + (-11i) = -14 - 11i

- anonymous

oh okay.

- anonymous

will you be on to check a few more problems before I submit them?

- jim_thompson5910

sure

- anonymous

thank you!!!!!!!!!!!!

- jim_thompson5910

no problem

- anonymous

okay i have one to check its similar to the last problem

- anonymous

|dw:1441074082197:dw|

- anonymous

what did i do wrong?

- jim_thompson5910

you didn't foil correctly

- anonymous

okay i guess the sqrt part mixed me up

- anonymous

i also tried the box method too

- jim_thompson5910

|dw:1441074547608:dw|

- jim_thompson5910

what we do is make a 2x2 box
2 rows and 2 columns
the terms of each factor will go along the outside like this
|dw:1441074595637:dw|

- jim_thompson5910

agreed so far?

- anonymous

okay thats how i started it on my paper too

- jim_thompson5910

ok go ahead and fill it out

- anonymous

|dw:1441074610390:dw|

- jim_thompson5910

-4 times \(\large -i\sqrt{2}\) should be \(\large +4i\sqrt{2}\)
the two negatives cancel
|dw:1441074917203:dw|

- anonymous

right?

- jim_thompson5910

same with the -i and -i being multiplied in the lower right box
|dw:1441074956036:dw|

- anonymous

so instead of a negative is a possitive

- jim_thompson5910

so after filling the boxes and simplifying the expressions in each box, we end up with
|dw:1441075014939:dw|
since \[\Large i^2*\sqrt{4} = -1*2 = -2\]

- anonymous

okay

- anonymous

so is the answer 18 instead?

- anonymous

no :( dang it please help me i have until 11:50 to submit these

- jim_thompson5910

now add up the terms inside the boxes
\[\Large 16 + 4i\sqrt{2} + 4i\sqrt{2}+(-2) = 14+8i\sqrt{2}\]

- jim_thompson5910

|dw:1441075415932:dw|

- anonymous

okay that makes since

- jim_thompson5910

|dw:1441075435745:dw|