What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di)
choices: a)c = 2, d = 3 b)c = –2, d = 3 c)c = 3, d = –2 d)c = –3, d = –2

The correct answer is:

D)c = –3, d = –2

Explanation:

To simplify this, we first distribute the i through the first set of parentheses:

i(2+3i) = i(2) + i(3i) = 2i + 3i²

Since i² = -1, this means we have

2i + 3(-1) = 2i - 3

Now we multiply this by the second set of parentheses:

(2i-3)(c+di) = 2i(c) + 2i(di) - 3(c) - 3(di)

= 2ci + 2di² - 3c - 3di

= 2ci + 2d(-1) - 3c - 3di

= 2ci - 2d - 3c - 3di

We need 2ci and -3di to cancel. This means we want the product 2c and the product 3d to be the same; this happens if c = -3 (2*-3 = -6) and d = -2 (3*-2 = -6).

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative