How do you multiply #3i(4+2i)(2+5i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Narad T. Feb 9, 2017 The answer is #=-72-6i# Explanation: Remember that #i^2=-1# Therefore, #(4+2i)(2+5i)=8+20i+4i+10i^2# #=8-10+24i# #=-2+24i# so, #3i(4+2i)(2+5i)=3i(-2+24i)# #=-6i+72i^2# #=-72-6i# Answer link Related questions How do I multiply complex numbers? How do I multiply complex numbers in polar form? What is the formula for multiplying complex numbers in trigonometric form? How do I use the modulus and argument to square #(1+i)#? What is the geometric interpretation of multiplying two complex numbers? What is the product of #3+2i# and #1+7i#? How do I use DeMoivre's theorem to solve #z^3-1=0#? How do I find the product of two imaginary numbers? How do you simplify #(2+4i)(2-4i)#? How do you multiply #(-2-8i)(6+7i)#? See all questions in Multiplication of Complex Numbers Impact of this question 2707 views around the world You can reuse this answer Creative Commons License