Guessing one root of a cubic equation for Hit and Trial Suppose I have a cubic equation as $$15x^3-4x^2-25x+14=0$$
By Hit and Trial method I know that one of the roots is $x=1$.
And hence I can solve the cubic equation wit ease as it will take the form of $$(x-1)(ax^2+bx+c)=0$$

But what if the cubic equation is $$15 x^3-64 x^2-69 x+70 = 0$$
One of the roots is $x=5$

How do I guess that? Like...I have to try for $x=0,1,-1,2,-2,3,4,5$. It takes time and huge calculations...
I am only talking about simple roots like $x \in \mathbb{Z}$ (integers) and maybe $x \in [-5,5]$ or maybe $x=\frac{1}{2}$ or $x=\frac{-1}{2}$


The methods I am aware of is drawing an approx graph of the cubic equation using maxima/minima or using Newton's method with maybe 2 iterations and check for integers near that.

So are there any easy yet reliable methods of guessing one root.
You can use the rational root theorem to guess some roots.

Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.

For example, in the integer case, one can take $q=1$, and one only has to test $p = \pm 1$, $p = \pm 2$, $p= \pm 5$, and then you have it. (you of course could test $\pm 7$, $\pm 10$, $\pm 14$, $\pm 35$ and $\pm 70$ as well, but it would be less work to just divide by the $x-5$)
Another technique:

Descartes rule of Signs. The number of positive roots of $P(x)$ is equal to the number of sign changes is the sequence formed by the coefficients of $P(x)$ or it is less by an even number.
The number of negative roots of $P(x)$ is equal to the number of sign changes is the sequence formed by the coefficients of $P(-x)$ or it is less by an even number.
Roots with multiplicity $n$ are counted $n$ times.

In this case it is useless. But if it is a polynomial like $x^3+3x^2+4x+2$, you now know that there are no positive roots.
Last technique:

Estimating roots. If you've tested all eligible roots that are smaller than, say, 10, then you can use this. If the first two coefficients are not extremely small compared to the last two, then its a good idea to look to you eligible roots near to $$-\frac{\mathrm{second} \; \mathrm{coefficient}}{\mathrm{first} \; \mathrm{coefficient}}$$

If the root is larger than 10, the $x^3$ and $x^2$ term are large enough to be able to ignore the smaller terms, thus this is a good estimation. Write $a$ for the first coefficient and $b$ for the second, $c$ for the third and $d$ for the fourth.
Now, if $x$ is large, then $ax^3+bx^2+cx+d \sim ax^3+bx^2$. Therefore the roots will be comparable. But $ax^3+bx^2=0$ gives $x^2=0$ or $ax+b=0$. In the first case $x$ isn't large enough. In the second case we have $ax+b=0$, thus $ax=-b$, thus $x=\frac{-b}{a}=-\frac{\mathrm{second} \; \mathrm{coefficient}}{\mathrm{first} \; \mathrm{coefficient}}$.
Considering the equation to be $ax^3+bx^2+cx+d=0$ check for $d$ is product of roots. so you can take the divisors of $d$ for trial
Example : $15x^3-64x^2-69x+70=0$
Here the product of the roots is $70$ divisors of $70$ are $1,2,5,7,10,14,35,70$ so you can try these numbers instead of $1,2,3,4,5,6,7,8,9,10$.
Make sure you try the negative numbers too.