How do you evaluate #(- 3x + 1) ^ { 2} + ( 2x - 3) ^ { 2} - 2( 3x + 2) ( 2- 3x )#?

1 Answer
Ben
Mar 23, 2018

Expand and combine, and you get
#31x^2-18x+2#

Explanation:

There will be a lot of FOILing going on here first, let's write out all of the parentheses:

#(-3x+1)^2+(2x-3)^2-2(3x+2)(2-3x)#

#(-3x+1)(-3x+1)+(2x-3)(2x-3)-2(3x+2)(2-3x)#

Remember, FOIL stands for
F irst
O uter
I nner
L ast

So let's FOIL the first set, #(-3x+1)(-3x+1)#:

First: #(-3x)(-3x)=9x^2#
Outer: #(-3x)(1)=-3x#
Inner: #(1)(-3x)=-3x#
Last: #(1)(1)=1#

Putting this all together:

#9x^2-3x-3x+1=9x^2-6x+1#

Let's put that back into the original equation:

#(9x^2-6x+1)+(2x-3)(2x-3)-2(3x+2)(2-3x)#

FOIL-ing the other two sets:

#(2x-3)(2x-3)=4x^2-12x+9#

#2(3x+2)(2-3x)=2(-9x^2+4)=-18x^2+8#

Again, replace in the original statement:

#(9x^2-6x+1)+(4x^2-6x+9)-(-18x^2+8)#

Now, take the terms out of the parentheses. For the third term, this means applying the minus to both the #-18x^2#, making it positive, and the 8, making that negative:

#9x^2-6x+1+4x^2-6x+9+18x^2-8#

Finally, combine like terms and you're done!

#x^2(9+4+18)+x(-6-12)+(1+9-8)#

#31x^2-18x+2#

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