How do you simplify #2 ^ 2 + (13 - 8/4) * 2- (6b - 4) ^ 2#?

2 Answers

Answer:

#-(6b-4)^2+26# or
#2(-18b^2+24b+5)#

Explanation:

Whether you use BODMAS, BIDMAS OR PEDMAS, brackets and parentheses always come first in an expression, followed by exponents and roots, after which comes division and multiplication, and finally addition and subtraction.

For both answers, the first thing to do is to simplify the question. The first thing I do is rearrange the #*2# to be in front of the first brackets, #(13-8/4)#, so that the expression now looks like this:

#2^2+2(13-8/4)-(6b-4)^2#.

The second thing I do is simplify #8/4=2#, and then solve the first brackets, which becomes #(13-2)=11#. From this, I can solve the first bit of the expression, like so:

#2^2+2(11)-(6b-4)^2#

#=4+22-(6b-4)^2#

#=26-(6b-4)^2#

For the first answer, I simply rearrange the equation just so:

#-(6b-4)^2+26#

and I am done.

Now for the second answer. We have to be careful here because there is a minus sign in front of the second brackets. #-(6b-4)^2# does mean to take away what is in the brackets, but when variables are involved and it is not as simple as that, #-# also means to multiply everything inside the brackets by #-1#. With that in mind, I expand the brackets like so

#26-(6b-4)(6b-4)" "# (since #(a-b)^2=(a-b)(a-b)#)

and from there I multiply each term to get #26-(36b^2-48b+16)#.

This is where I multiply each term inside the brackets by #-1#, so that the signs of each term changes. Doing this I get

#26-36b^2+48b-16#

From here I simplify by adding #26+(-16)# to get

#-36b^2+48b+10#.

Although you could stop here, a last step that can be done involves taking out the common factor of 2 from this whole expression, to end up with the final answer of

#2(-18b^2+24b+5)#

I hope that helped!

Jun 6, 2017

Answer:

# -36b^2 +48b +10#

Explanation:

In an expression with different operations there is a well-defined order in which they have to be done.

First count how many TERMS there are.

Each arithmetic term will simplify to one value while an algebraic term might simplify to several unlike terms.

Within each term:
do BRACKETS first,
then POWERS AND ROOTS,
then MULTIPLY AND DIVIDE.

The simplified terms are only ADDED ANDSUBTRACTED in the LAST step.

We are working with #3# terms and will simplify each separately:
Take note of what happens within each term from one line to the next.

#" "color(blue)(2^2)color(green)( +(13-8/4) xx 2)" "color(red)( -(6b-4)^2)#

#=color(blue)(4)color(green)(" + "(13-2) xx 2)" "color(red)( -(6b-4)(6b-4))#

#=color(blue)(4)color(green)(" + "(11) xx 2)color(red)(" " -(36b^2-24b-24b+16))#

#=color(blue)(4)color(green)(" + "22color(red)(" " -(36b^2-48b+16))#

#=color(blue)(4)color(green)(+22color(red)(-36b^2+48b-16))" "larr# note that the signs change

Re-arrange the terms with the algebraic terms first, and the subtraction at the end:

#=color(red)(-36b^2+48b) color(blue)(" + "4)color(green)(" + "22)color(red)(" - "16)#

#= -36b^2 +48b +26-16#

#= -36b^2 +48b +10#