Question

Write the product in standard form.
( 9 + 5i)( 9 + 8i)

Answer 1

You do this pretty much "as if" you were FOIL'ing a binomial.

The difference will be: in the final term, you will have a factor of \(i^2\), so that will simplify to a (-1).

Then just combine the constant (non-i) terms, and you'll get i's in the 2 middle terms so you combine those also.

Then just put it all in the form: a + bi

Answer 2

(a+b)(c+d)= a*c + a*d + b*c + b*d

Answer 3

I'll do a slightly different one as an example:

\[\large ( 3 + 2i)( 6 - 7i)
\\\large=18-21i+12i-14i^2\\\large
\\\large=18-9i-14(-1)
\\\large=18+14-9i\\\large=32-9i\]

Answer 4

thank you both! I got 41+117i

Answer 5

I think you should look at it again

Answer 6

the question was (9+5i)(9+8i)
I distributed and combined like terms and got 81+117i+40i^2
then I plugged in -1 for i^2 and got -40.
-40 + 81 gave me 41 + 117i.

Answer 7

41 is correct, the problem is in your middle terms (the "O" and "I" of the FOIL). Check those out again. :)

Answer 8

The 81 and the 40i^2 are correct, it's the 117i that is the issue. :)