SOLUTION: Multiply (2+5i)(2+3i) (3-4i)^2

Algebra ->  Radicals -> SOLUTION: Multiply (2+5i)(2+3i) (3-4i)^2      Log On


   

Question 145134: Multiply
(2+5i)(2+3i)
(3-4i)^2
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
For these problems you just need to use foil.
RECALL: i^2 = -1

(2+5i)(2+3i)
F: 2 * 2 = 4
O: 3i * 2 = 6i
I: 5i * 2 = 10i
L: 5i * 3i = 15i^2 = -15
(2+5i)(2+3i) = 4 + 6i + 10i -15 = -11 +16i

(3-4i)^2 = (3-4i)(3-4i)
F: 3 * 3 = 9
O: -4i * 2 = -8i
I: -4i * 2 = -8i
L: -4i * -4i = 16i^2 = -16
(3-4i)(3-4i) = 9 + -8i + -8i -16 = -7 -16i