<s>[INST] Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$ 
How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$

I simplified it to:
$$x(x^2+3x-5)=4$$
But I don't know where to go from here.
 [/INST] You are given $x^3+3x^2-5x-4=0$
First, find one of the roots of the polynomial. By way of the rational root theorem, we find that -4 is one of the roots, therefore x + 4 is a factor. Next, split the polynomial in accordance with x + 4 as follows:
$x^3 + 4x^2 - x^2 -4x -x - 4 = 0$
Next factor each pair of terms from left to right as follows:
$x^2(x + 4) -x(x + 4) - 1(x + 4) = 0$
Since x + 4 is common to each factorization as expected, factor it out:
$(x + 4)(x^2 - x - 1)= 0$
By zero product property:
$x + 4 = 0$
$x^2 - x - 1 = 0$
Obviously $x = -4$ is one of the solutions. The other solution can be found by completing the square:
$x^2 - x = 1$
$x^2 - x + \left(\dfrac{1}{2}\right)^2 = 1 + \left(\dfrac{1}{2}\right)^2$
$\left(x - \dfrac{1}{2}\right)^2 = 1 + \dfrac{1}{4}$
$\left(x - \dfrac{1}{2}\right)^2 = \dfrac{5}{4} $
$x = \dfrac{\pm\sqrt{5} + 1}{2}$
</s>