Least number of roots for $f(x) = g(x)^2+4g(x)-320.$ 
Let $g(x)$ be a quadratic polynomial s.t $$f(x)=g(x)^2+4g(x)-320.$$ If $g(2016)=0,$ what is the least number of real roots that $f(x)$ can have?

Differentiating $f(x)$ results in $f’(x)= 2g(x)+4g’(x)$ so if $2g(x)+4g’(x) \geqslant 0$, then $f(x)$ would have at least one root. Is there a way I could confirm this statement since I don’t know anything about $g’(x)$?
 A: Certainly since $g$ is quadratic it goes to $\infty$ as $x$ goes to $\pm \infty$, and thus the same holds for $f$ since the leading term is $g(x)^2$. You can factor $f$: $$f(x) = (g(x) - 16)(g(x)+20).$$ Now $f(2016) = -320  < 0$ Since $f(-\infty) = f(\infty) = \infty$, this gives at least one root in $(-\infty,2016)$ and one root in $(2016,\infty)$ by the intermediate value theorem. Thus $f$ has at least two real roots. You can prove that this is the least guaranteed with an example where $f$ has exactly two real roots. 
One such example: let $g(x) = (x-2016)^2$. Then $(g(x) + 20) > 0$ while $(g(x) - 16)$ has exactly two roots at $x = 2012,2020$. Thus by the factorization above, the real roots of $f$ are $x = 2012,2020$. 
EDIT: as pointed out in the comment, I assumed that $g$ is a positive quadratic. Of course, it could be the case that $g$ goes to $-\infty$ as $x \to \pm \infty$. However, since the leading term is $g(x)^2$, this does not change the form of $f(x)$. Anything I said about $f$ is true as stated. 
