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 $x=\frac{-b}{a}=-\frac{\mathrm{second} \; \mathrm{coefficient}}{\mathrm{first} \; \mathrm{coefficient}}$.
As you've mentioned, you know that
$$y_1 + y_2 + y_3 = 1$$
$$y_1y_2 + y_1y_3 + y_2y_3 = -4$$
$$y_1y_2y_3 = -3$$
To find a polynomial $x^3 - ax^2 + bx - c$ whose roots are $y_1 + y_2, y_1 + y_3$ and $y_2 + y_3$, you can use Vieta's formulas in terms of elementary symmetric polynomials of these roots. That is, you want to calculate:
$$a = (y_1 + y_2) + (y_1 + y_3) + (y_2 + y_3)$$
$$b = (y_1 + y_2)(y_1 + y_3) + (y_1 + y_2)(y_2 + y_3) + (y_1 + y_3)(y_2 + y_3)$$
$$c = (y_1 + y_2)(y_1 + y_3)(y_2 + y_3)$$
The first one is easy to find, it is equal to $a = 2(y_1 + y_2 + y_3) = 2$.
The computations for $b$ and $c$ are more tedious, but they are symmetric polynomials in $y_1, y_2$ and $y_3$ so they can be expressed in terms of elementary symmetric polynomials (and hence in terms of known values) using something like Gauss' algorithm.
In our case we have $b = (y_1+y_2+y_3)^2 + (y_1y_2 + y_1y_3 + y_2y_3) = -3$ and we have $c = (y_1 + y_2 + y_3)(y_1y_2 + y_1y_3 + y_2y_3) - y_1y_2y_3 = -1$.
So the polynomial $x^3 - 2x^2 - 3x + 1$ should have the right roots, assuming my calculations are correct.
Best Answer
Observe that $2x^3+6x^2 = 2(x^3+3x^2) = 2(x+1)^3 - 6x-2$. Now let $X=x+1$, and divising by 3, your equation is now something like $X^3+pX+t=0$, which can be solved by Cardano method.