Question

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

Answer 1

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`