Find radical expressions for all three roots of $t^3-15t-4=0$.

Question

I understand that I can use the trigonometric identity $4\cos^3(\theta)-3\cos(\theta)-\cos(3\theta)=0 \tag{2}$ to find the three real roots, and Cardano's formula gives the radical expression for t=4; but how exactly do I find the radical expressions for the other two real roots? Any help would be appreciated!

Answer 1

If $t=4$ is a solution, then it will be in this form

$t^3-15t-4=(t-4)(t^2+bt+1)$

By putting a suitable number for $t$

For example $t=1$

$-18=(-3)(2+b)$

$b=4$

$t^3-15t-4=(t-4)(t^2+4t+1)$

That radical part can be find in different ways: Horner's method, Polynomial division,...