Show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$. Given $f=x^3+9x^2+24x-40n^3+40xn^2+94n^2-12x^2n-62nx-74n+20$ has real roots, show that the largest root of $f$ is greater than $5n$  where $n(\ge 3)\in \mathbb N$.
I tried to do it by directly finding the roots in wolfram alpha for any $n\ge 3$, however I am getting the roots as complex which is against the hypothesis of the problem.
Also the roots obtained in Wolfram Alpha are very nasty or bad which is making my life my difficult.
Kindly help me out on how to find all the real roots of $f$ and show that they are $>5n$
 A: Your polynomial equation is
$$\begin{equation}\begin{aligned}
f(x)&=x^3+9x^2+24x-40n^3+40xn^2\\
& \; \; \; +94n^2-12x^2n-62nx-74n+20
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note you have
$$\begin{equation}\begin{aligned}
f(5n) & = (5n)^3 + 9(5n)^2 + 24(5n) - 40n^3 \\
& \; \; \; \; \; + 40(5n)n^2 + 94n^2 - 12(5n)^2n - 62n(5n) - 74n + 20 \\
& = 125n^3 + 225n^2 + 120n - 40n^3 + 200n^3 \\
& \; \; \; \; \; + 94n^2 - 300n^3 - 310n^2 - 74n + 20 \\
& = -15n^3 + 9n^2 + 46n + 20
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
At $n = 3$, you thus get
$$\begin{equation}\begin{aligned}
f(15) & = -15(3)^3 + 9(3)^2 + 46(3) + 20 \\
& = -166
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
However, since the coefficient of the highest power of $f(x)$ is $1$ in $x^3$, this means that $\lim_{x \to \infty}f(x) = \infty$. Since $f(x)$ is continuous, there must be a root larger than $5n$ for when $n = 3$.
To confirm this is also true for all $n \gt 3$, one way is take the derivative of \eqref{eq2A}, as shown below
$$\frac{df(5n)}{dn} = -45n^2 + 18n + 46 \tag{4}\label{eq4A}$$
Using the quadratic formula to get the roots gives
$$\begin{equation}\begin{aligned}
n & = \frac{-18 \pm \sqrt{18^2 - 4(-45)(46)}}{2(-45)} \\
& = \frac{3 \mp \sqrt{3^2 + 3(46)}}{15} \\
& = \frac{3 \mp \sqrt{3(3 + 46)}}{15} \\
& = \frac{3 \mp 7\sqrt{3}}{15} \\
& \approx -0.61, 1.01
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
The quadratic polynomial in \eqref{eq4A} being a concave-down parabola means its values are only positive with $n$ in the approximate range of $(-0.61,1.01)$, with it being negative everywhere else. Thus, for $n \ge 3$, the derivative is negative, so the value of \eqref{eq2A} would keep decreasing, confirming there's always a real root $\gt 5n$ for \eqref{eq1A}.
A: Hint:  As the cubic is eventually positive, it is enough to show that $f(5n)<0$, i.e. $9n^2+46n+20 < 15n^3$.  Can you show that, say using induction?
