Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2
Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8
Now that 2(x-2)^2 is done lets move on to 3(x-2).
Use the distributive property and distribute the 3
3(x-2) = 3x - 6
All that is left is the -2
Now lets put it all together
2(x-2)^2 + 3(x-2)-2
2x^2 - 8x + 8 + 3x - 6 - 2
Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0
So all we have left is
2x^2 - 5x
Its false, standard notation would have an exponent.
When estimating, compatible numbers are numbers that are close in value to the actual numbers, and which make it easy to do mental arithmetic.
I think the answer is 384 in^3
