GUYS i have to multiply these bionomials and write the result in standard form its (3-2i)(4+3i) and i got 18+i is that it or did i do something wrong?

Question

DevotedHeart · Answer

yeah you mess up do it again

DevotedHeart · Answer

use the FOIL method

lolboooohruuu · Answer

like this? (3-2i)(4+3i) = (3×4) + (3×3i) + (-2i×4) + (-2i×3i) and i get   12 + 9i - 8i - 6i²  and then  i²  -1 so it would be  12 + 9i - 8i - 6(-1)
12 + 9i - 8i + 6
then i Combine like terms
and i got 18 + i

DevotedHeart · Answer

@maccinkylie wrote:

I got 7i but I don't know what binomials are

wrong mac and cheese

maccinkylie · Answer

@devotedheart wrote:

@maccinkylie wrote:

I got 7i but I don't know what binomials are

wrong mac and cheese

I'm in 8th

DevotedHeart · Answer

@lolboooohruuu wrote:

like this? (3-2i)(4+3i) = (3×4) + (3×3i) + (-2i×4) + (-2i×3i) and i get 12 + 9i - 8i - 6i² and then i² -1 so it would be 12 + 9i - 8i - 6(-1) 12 + 9i - 8i + 6 then i Combine like terms and i got 18 + i

bro wthhh use chatgpt

DevotedHeart · Answer

@maccinkylie wrote:

@devotedheart wrote:

@maccinkylie wrote:

I got 7i but I don't know what binomials are

wrong mac and cheese

I'm in 8th

yeah but still wrong

maccinkylie · Answer

ok I also don't know how to do it I just did 3-2 = 1 and 4+3 = 7 and multiplied

DevotedHeart · Answer

@maccinkylie wrote:

ok I also don't know how to do it I just did 3-2 = 1 and 4+3 = 7 and multiplied

wellll that is not how multiplying binomials works you gotta use the FOIL distributive property

lolboooohruuu · Answer

oh its pre calc, im in 12th but its collage math 😭

Luigi0210 · Answer

You are correct:

$\large (3-2i)(4+3i) = 12+9i-8i-6i^2 $

where $\large i^2 = -1 $

So you end up with $\large (12-6(-1))+(9i-8i) = 18+i $

CecePitbull · Answer

@lolboooohruuu wrote:

oh its pre calc, im in 12th but its collage math 😭

college* how can u not spell in 12?

sllo · Answer

yes, you did it correctly! The result is 18 + 𝑖 18+i.

tf1w5f5613ef16 · Answer

I got 1i

TJH · Answer

standerd form 6i2+1i+12