Yeah, I've read all the comments already. The selected answer assumes that $x, 4, 6, 5$ are given, and we are given a temporary variable $T$ to work with. Because after doing $x(x) +4$ the result will be in temporary variable and now to get $x(x(x) + 4)$ we definitely need to know $x$. So, there's no way that $x$ IS the temporary variable.
Now, as pointed out in the comments, particularly one which describes the three address code for the selected answer, directly references $x$ in the 4th line.
This answer also, doesn't need to store $x^2$ in the variable cause it is being used only once, and $x$ is readily available, this answer seems correct.
Correct me if I'm wrong. Sorry for ping.