# Generalising Cubic Roots in terms of a third variable

I have a quartic function: $$P(B)=-s\left(\frac{25}{3}×10^{-20}\right)B^4+s\left(\frac{5}{3}×10^{-13}\right)B^3-s\left(\frac{31}{30}×10^{-7}\right)B^2+s\left(\frac{1}{50}\right)B+\left(\frac{3}{50}\right)B-(6×10^4)$$ I want to write a generalised equation for the turning points of this function in terms of $s$. In order to do this, I need the derivative of the function, which I have determined to be: $$\frac{d}{dB}P(B)=-s\left(\frac{10}{3}×10^{-19}\right)B^3+s(5×10^{-13})B^2-s\left(\frac{31}{15}×10^{-7}\right)B+s\left(\frac{1}{50}\right)+\frac{3}{50}$$ I know that I can find the $B$ values of the turning points by solving $\frac{d}{dB}P(B)=0$. This is where I am having difficulty; I cannot find an effective method of finding the roots of this derivative. What is the best way of writing a generalised equation in terms of $s$ giving the $B$ values of the roots? Is it even possible to write an such an equation?

I intend to use this generalised equation in an Excel spreadsheet where I calculate the turning and inflection points for the function for different values of $s$. Because of this, factorising the derivative won't work.

There is another method which I have thought of, but I am not sure if it would work and I haven't been able to find anything online. I can determine the inflection points by finding the roots of the second derivative. Since this is a quadratic, I can use the quadratic formula to generalise this. If I could find the $B$ values of these two inflection point, could I use them to identify the turning points? Would the $B$ value of one turning point not be halfway between the two values, which is the average? Is there anyway I could use these two points to identify the other two turning points?

With the derivative you want to be zero, you are facing the problem of solving a cubic equation in $B$ $$sB^3 -15\times 10^5s B^2 +62\times 10^{10} sB -6\times 10^{16} s-18\times 10^{16}=0$$ If you use Cardano method for cubics, the discriminant evaluates to $$\Delta =4\times 10^{30} s^2 \left(2197 s^2-218700\right)$$ and, then, the number of real roots just depends on the sign of $\left(2197 s^2-218700\right)$. If this term is positive, there three distincts real roots; if it is zero, still three real roots but two are identical; if it is negative, there only one real root.
$$B_1=-\frac{1300000 s}{\sqrt[3]{3} \sqrt[3]{P}}-\frac{100000 \sqrt[3]{P}}{3^{2/3} s}+500000$$ $$B_2=\frac{650000 \left(1+i \sqrt{3}\right) s}{\sqrt[3]{3} \sqrt[3]{P}}+\frac{50000 \left(1-i \sqrt{3}\right) \sqrt[3]{P}}{3^{2/3} s}+500000$$ $$B_3=\frac{650000 \left(1-i \sqrt{3}\right) s}{\sqrt[3]{3} \sqrt[3]{P}}+\frac{50000 \left(1+i \sqrt{3}\right) \sqrt[3]{P}}{3^{2/3} s}+500000$$ in which $$P=\sqrt{3} \sqrt{218700 s^4-2197 s^6}-810 s^2$$