Im looking to for the steps to solve this: 3(x^2 -2)² (2x-3)^2
apparent answer: 12x^6 - 36x^5 - 21x^4 +144x^3 - 60x^2 - 144x + 108
So you are looking to expand this? The word solve implies there is an equals sign somewhere in the problem.
$\displaystyle \displaystyle 3(x^2-2)^2(2x-3)^2$
This is the same as
$\displaystyle \displaystyle 3(x^2-2)(x^2-2)(2x-3)(2x-3)$
Now use the distributive law in pairs $\displaystyle \displaystyle (a+b)(c+d) = ac+ad+bc+bd$
If it's not equal to anything, then there is nothing to solve. Perhaps you are being asked to expand the expression. In this case you just need to do some multiplication using the distributive property several times.
To start you off, $\displaystyle (2x-3)^2=(2x-3)(2x-3)=4x^2-6x-6x+9=4x^2-12x+9$.
Leave the 3 out for now. It just makes the coefficients ugly. Okay, so you've got $\displaystyle (x^2 - 2)^2 = x^4 - 4x^2 + 4$ and $\displaystyle (2x - 3)^2 = 4x^2 - 12x + 9$. Now you need to do
$\displaystyle (x^2 - 2)^2(2x - 3)^2 = (x^4 - 4x^2 + 4)(4x^2 - 12x + 9)$.
-Dan