The function $f\left(x\right)=a{x}^3+b{x}^2+cx+d=0$ has three positive roots. If the sum of the roots of $f\left(x\right)=0$ is 4, the largest possible integral values of $\frac{\mathrm{c}}{\mathrm{a}}$, is

Let $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ be the three positive roots of f(x)=0
We know that $(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }{)}^2={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2+2(\mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha })$
$\Rightarrow 2\sum \mathrm{\alpha \beta }={\left(\sum \mathrm{\alpha }\right)}^2-\sum {\mathrm{\alpha }}^2\Rightarrow 2\left(\frac{\mathrm{c}}{\mathrm{a}}\right)=4^2-\sum {\mathrm{\alpha }}^2\dots \dots .\left(1\right)$
Again we have $(\mathrm{\alpha }-\mathrm{\beta }{)}^2+(\mathrm{\beta }-\mathrm{\gamma }{)}^2+(\mathrm{\gamma }-\mathrm{\alpha }{)}^2=2\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2\right)-2(\mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha })\ge 0$
$$\begin{gathered} {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2\ge \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }\Rightarrow \sum {\mathrm{\alpha }}^2\ge \sum \mathrm{\alpha \beta }\Rightarrow -\sum {\mathrm{\alpha }}^2\le -\sum \mathrm{\alpha \beta } \\ \text{ From (1) }\stackrel{2}{min} \left(\frac{\mathrm{c}}{\mathrm{a}}\right)=16-\sum {\mathrm{\alpha }}^2\le 16-\sum \mathrm{\alpha \beta }=16-\frac{\mathrm{c}}{\mathrm{a}}\Rightarrow 3\left(\frac{\mathrm{c}}{\mathrm{a}}\right)\le 16\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}\le \frac{16}{3}\text{. } \end{gathered}$$
Largest possible integral value for $\frac{\mathrm{c}}{\mathrm{a}}$, is 5.