How do you simplify #(3-i)(1+2i)(2+3i)#?
1 Answer
Mar 20, 2016
Explanation:
Multiply out one pair of numbers as a product of binomials, simplify using
You can use FOIL if you find it helpful.
#(3-i)(1+2i)#
#=overbrace((3*1))^"First" + overbrace((3*2i))^"Outside" + overbrace((-i*1))^"Inside" + overbrace((-i * 2i))^"Last"#
#=3+6i-i-2i^2#
#=3+5i+2#
#=5+5i#
Similarly:
#(5+5i)(2+3i)=10+15i+10i-15=-5+25i#
So:
#(3-i)(1+2i)(2+3i) = -5+25i#