I want to solve the following equation: $$144s^3-408s^2+349s-85 = 0$$ I know that the solution is: $$(s-1)(12s-17)(12s-5)=0$$ which implies gives, $s=1$ or $s=\dfrac{17}{12} $ or $s=\dfrac{5}{12}$.
However, I don't understand how to factor the left hand side into a product of three terms.
Thanks in advance!
If you know that s=1 is solution, you might try to rewrite $144s^3-144s^2-264s^2+264s+85s-85=0$, then you factor with s-1