Find the Critical Points: $f(x) =(x^2-1)^3$ This question probably has more to do with my Algebra skills than Calculus. Nonetheless, can someone explain why the factored "term" is not set to zero (0) [second picture]. Thanks in advance. 


 A: $f(x)=(x^2-1)^3$ is a sixth-degree polynomial with $x=\pm 1$ being triple zeroes.
Moreover, $f(x)$ is an even function, hence $f'(x)$ (that is a fifth-degree polynomial) has double zeroes in $x=\pm 1$ and a simple zero in $x=0$. The critical points of $f(x)$ may occur only at the zeroes of the derivative or at the endpoints of the given interval $I=[-1,5]$, hence to compute $\max_{x\in I}f(x)$ and $\min_{x\in I}f(x)$ it is enough to compute $f(-1),f(0),f(1),f(5)$.
The same approach works for the second exercise, too, by considering $F(t)^2$ in place of $F(t)$, getting a third-degree polynomial.
A: Converted comments into answer as requested by OP.
If $a×b=0$ then either $a=0$ or $b=0$ or both a and b are zero. In your case you had $2(12.5−t^2)=0$ so either $2=0$ (clearly false) or $12.5−t^2=0$. Both cannot be zero since we know that clearly $2\ne0$.
In general you can safely divide both sides of an equation by something that you know can never be zero and not lose any solutions. So, in your case, where you had $2(12.5−t^2)=0$ you could simply divide both sides by $2$ leaving you with:$$12.5−t^2=0$$
If you had $2t(12.5−t^2)=0$ then again you can first divide both sides by $2$ leaving you with:$$t(12.5−t^2)=0$$
Leading to:$$t=0$$or:$$12.5−t^2=0$$
