How do you simplify #(8i)(-4i)#?

2 Answers
Mar 17, 2018

#32#

Explanation:

When performing the basic mathematical functions on terms that contain #i# within them, it is easiest to recognize it as a variable when starting off.

We can begin to treat this as a normal multiplication problem:

#(8i)(-4i)#

#=-32i^2#

Now that we have this, we need to understand how setting #i# to a power other than #1# changes the value.

#i^0=1#
#i^1=sqrt(-1)#
#i^2=-1#
#i^3=-i#
#i^4=1#
#i^5=sqrt(-1)#
#...#

(Click here for more information regarding imaginary numbers)

In this instance, however, the #i^2# would then become #-1#.

Knowing this we can write our expression as:

#-32(-1)#

#=32#

Mar 17, 2018

#32#

Explanation:

#color(orange)"Reminder "color(white)(x)i^2=(sqrt(-1))^2=-1#

#rArr(8i)(-4i)#

#=-32i^2=-32xx-1=32#