How do you write #5i(3+2i)(8+3i)# in standard form?
1 Answer
Sep 17, 2016
Explanation:
Before multiplying by 5i, distribute the brackets.
#(3+2i)(8+3i)=24+9i+16i+6i^2=24+25i+6i^2#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#
#rArr24+25i+6i^2=24+25i-6=18+25i# We now multiply this result by 5i
#5i(18+25i)=90i+125i^2#
#=-125+90ilarr" in standard form"#
