How do you evaluate #4(8-(4*2))+1#?

1 Answer
May 13, 2017

#4(8-(4xx2))+1 = 1#

Explanation:

In a calculation with different operations, they have to be done in a specific order, given by the acronym BODMAS, PEDMAS, or similar.

There are 2 terms. Each term must be simplified to a single answer which are added in the last step.

#color(green)(4(8-(4xx2)))color(blue)(+1)#

Within each term, brackets must be done first, then the powers and roots, and then multiplication and division.

#color(green)(4(8-(4xx2)))color(blue)(+1)#

There are 2 sets of brackets - start with the inner one.

#4(8-color(red)((4xx2))color(blue)(+1)#
#color(white)(.............)darr#
#=4(color(red)(8-8))color(blue)(+1)#
#color(white)(......l..)darr#
#=color(red)(4xx0)color(blue)(+1)#
#color(white)(.......)darr#
#=" "color(red)(0)color(blue)(+1)#

#=1#

Notice that there is nothing to be done with the last term #(+1)#.
It is just carried down with each line until the last step where it is added.