PEMDAS question: $F(x) = 3x^2 - x+2$. Find $[f(a)]^2$ How should I go about doing this? $(3a^2-a+2)^2$? 
Thus, $9a^4-a^2+4$
 A: To thrash an already-dead horse, I’ll remind you that just as $(a+b)^2=a^2+2ab+b^2$, so it’s the case that $(a+b+c)^2=a^2+2ab+b^2+2ac+2bc+c^2$.
A: Hint For any real numbers $x,y$ we have $$\color{green}{(x-y)^2 = x^2-2xy+y^2}, \quad \color{blue}{(x+y)^2 = x^2+2xy+y^2},$$
and note that with $\color{green}{x = (2-a)}$ and $y = 3a^2$ we get 
$$(3a^2-a+2)^2=\color{blue}{(x+y)^2} = \ldots$$
Can you continue from here?
A: $(3a^2-a+2)^2$ is correct but your expansion of this is incorrect. Remember that:$$(3a^2-a+2)^2=(3a^2-a+2)(3a^2-a+2)=3a^2(3a^2-a+2)-a(3a^2-a+2)+2(3a^2-a+2)$$
You might find this useful: expanding or removing brackets
Here is an example to illustrate the point:$$(3-1+2)^2=(4)^2=16$$If we did this using your result, we would get:$$(3-1+2)^2=3^2-1^2+2^2=9-1+4=12\ne16$$However:$$(3-1+2)^2=(3-1+2)(3-1+2)$$$$=3(3-1+2)-1(3-1+2)+2(3-1+2)$$$$=3(4)-1(4)+2(4)=12-4+8=16$$
More generally:$$(a+b+c)d=ad+bd+cd$$Now image $d=(e+f+g)$, then we would get:$$(a+b+c)d=(a+b+c)(e+f+g)$$$$=ad+bd+cd$$$$=a(e+f+g)+b(e+f+g)+c(e+f+g)$$
