Question

What is the standard form of (5+i)(6-5i)/2i(-3+7i)

Answer 1

Answer:

Since all parts of 2i(-3+7i) is in the  denominator, you have to use an  extra set of parentheses around the whole denominator. this is what you do

(5+i)(6−5i) / (2i(−3+7i))

= (30−25i+6i−5i^2) / (−6i+14i^2)

= (35−19i) / (−14−6i)

= (35−19i)(−7+3i) / (2(−7−3i)(−7+3i))

= (−245+105i+133i−57i^2) / (2(49−9i^2))

= (−188+238i) / (2(58))

= (−188+238i) / 116

= −47/29 + 119/58 iso that is how you do it