<s>[INST] Find the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$ According to my textbook the $f$ of $g$ where $f(x) =x^2+2x+1$ and $g(x)=2x-3$  is $4x^2-8x+2$, but I get $4x^2-8x+4$. 
What am I doing wrong?
My Steps:
$(2x-3)^2 + 2(2x-3)+1$
$(2x-3)(2x-3)+4x-6+1$
$4x^2-12x+9+4x-6+1$
$4x^2-8x+4$
 [/INST] A simpler approach to your calculation is to factor $f$ first:
$$f(x) = x^2 + 2x + 1 = (x+1)^2.$$
Then, $$f(g(x)) = (g(x)+1)^2 = ((2x-3)+1)^2 = (2x-2)^2 = 2^2(x-1)^2 = 4x^2 - 8x + 4.$$
To check that the book's solution cannot be correct, it suffices to choose $x = 0$, which leads to $g(0) = -3$, and $$f(g(0)) = f(-3) = (-3)^2 + 2(-3) + 1 = 9 - 6 + 1 = 4.$$  Your computation is consistent with this result, since $$4(0)^2 - 8(0) + 4 = 4,$$ but the book's solution is not, since it gives $2$ when $x = 0$.
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