how to factor this cubic polynomial Let $f(t)=36t^3-19t+5$ be a cubic polynomial.
How we can factor $f$ to its roots?
Mathematica says that $f(t)=(-1+2 t) (-1+3 t) (5+6 t)$. How?
 A: The Rational root theorem can be used here to find all roots of this polynomial
A: Hint $\ 6f(t) = (6t)^3\! -19(6t) + 30 = x^3\! - 19x + 30 =: g(x)\,$ for $\,x = 6t.$ By the Rational Root Test the only possible rational roots of $g$ are integer factors of $30.\,$
Remark $\ $ This is a generalization of the high-school "AC method" and it works generally to reduce finding rational roots to finding integer roots. See here for much further discussion, including more general ring-theoretic viewpoints
A: Looks like using the rational root theorem is a good bet for this problem. If you feel it is tedious to use the rational root theorem to find all three roots, then use it to find just one root (or just plain guess and check until you find one root, like $t=\frac{1}{2}$) and then use polynomial division. Since $t=\frac{1}{2}$ is a root, you can divide your polynomial by $\left(t-\frac{1}{2}\right)$ and will get $$36t^3-19t+5 =\left(t-\frac{1}{2}\right)(36t^2+18t-10)$$ From here you should be able to factor the quadratic portion $(36t^2+18t-10)$ by hand. The rest of the algebra should follow as: $$\left(t-\frac{1}{2}\right)(36t^2+18t-10) = \left(t-\frac{1}{2}\right)2\cdot(18t^2+9t-5) \\ = \left(2\left[t-\frac{1}{2}\right]\right)(6t+5)(3t-1) \\ = (2t-1)(6t+5)(3t-1)$$ This technique will work for whichever of the three roots you begin with.
A: And without using Rational root theorem:
$$36t^3-10t+5=36t^3-30t^2+30t^2-25t+6t+5$$
$$36t^3-10t+5=6t(6t^2-5t+1)+5(6t^2-5t+1)$$
$$\,\,\,\,\,\,\,=(6t^2-5t+1)(6t+5)$$
$$\,\,\,\,\,\,\,\,\,\,=(2t-1)(3t-1)(6t+5)$$
