I would like help with cubic polynomials. I have done research on the Web as I do not have access to the library but come up empty. I tried using grouping and free term method but the examples are way too easy compared to the problem I have. Can someone help please? Here is the problem:
$$3x^3 - 36x^2 + 135x + 3 = 0$$
The quickest way to find the roots of a cubic polynomial is to use the cubic formula:
Cardano's Formula:
Given a depressed cubic $x^3+qx+r=0$, a root is given by$$\begin{align*} & x=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{1/3}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{1/3}\end{align*}$$ To get a cubic $x^3+ax^2+bx+c=0$, substitute $x=\frac {y-a}3$
Your cubic, doesn't have nice rational roots. Which is why grouping or rational root theorem doesn't work. The roots are $$\begin{align*} & x=4-\sqrt[3]{\frac 2{53-\sqrt{2805}}}-\sqrt[3]{\frac12\left(53-\sqrt{2805}\right)}\\ & x=4+\frac {1\pm i\sqrt3}2\sqrt[3]{\frac {53-\sqrt{2805}}2}+\frac {1\mp i\sqrt3}{\sqrt[3]{4(53-\sqrt{2805})}}\end{align*}$$
Hint -
Let $3x^3 - 36x^2 + 135x + 3 = 0$
Taking 3 common and discarded we have,
$x^3 - 12x^2 + 45x + 1 = 0$
Use hit and trial to find one root and to proceed further.