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.
 A: The manipulation that you performed does not help. Use the rational root test: since the leading coefficient is $1$ and the constant term is $4$, the only possible rational roots of the cubic are fractions of the form $a/b$, where $a$ is a divisor of $4$ and $b$ is a divisor of $1$. In other words, if the cubic has any rational root at all, it must be one of the numbers $\pm1,\pm2$, and $\pm4$. Actually calculation shows that 
$$(-4)^3+3(-4)^2-5(-4)-4=-64+48+20-4=0\;,$$
so $-4$ is a root of the cubic $x^3+3x^2-5x-4$. Now use the fact that $r$ is a root of a polynomial if and only if $x-r$ is a factor of that polynomial to conclude that $x^3+3x^2-5x-4$ is divisible by $x-(-4)=x+4$. When you perform this division $-$ either by polynomial long division or by synthetic division $-$ you’ll get a quadratic as the quotient. Let’s say that you get $x^2+bx+c$ as the quotient; then you know that 
$$x^3+3x^2-5x-4=(x+4)(x^2+bx+c)\;.$$
This tells you that $x=-4$ is one solution to your original equation, and the others are the solutions to the quadratic equation $x^2+bx+c=0$, which you can find using the quadratic formula.
A: Hint: One might hope that there is a rational solution. There needn't be, but if this is a school exercise there probably is.
By the Rational Root Theorem, such a solution must be a divisor of $4$. Try all of them. We get lucky, there is an integer solution.  (Don't forget that the divisors of $4$ include some negative numbers.)
After you have found a solution $x=a$, divide the polynomial by $x-a$. You will end up with a quadratic, which you solve in the usual way to find the remaining roots.
A: All of the answers are fine, but I feel that it should be expanded upon why your intended approach does not work. It is a mistake that I have seen many, many times, and it is probably worth addressing.
Why do we factor expressions? If we are trying to solve
$$
f(x) = 0
$$
and we can factor $f(x) = h(x)g(x)$, then this is the same as
$$
h(x)g(x) = 0
$$
The key fact now is that if the product of two numbers/functions is identically zero, then one of the two numbers/functions must be zero themselves. That is, we must have either $h(x) = 0$ or $g(x) = 0$. This is why we factor in this setting.
Now, by contrast, if we are trying to solve, say,
$$
f(x) = 4
$$
and we factor $f(x) = h(x)g(x)$, then this yields
$$
h(x)g(x) = 4
$$
which tells us nothing whatsoever! Maybe $h(x) = 1/2$ and $g(x) = 8$. Maybe $h(x) = 2/x$ and $g(x) = 2x$. We can say absolutely nothing about this case, and so the factorization of $f(x)$ in this case is not helpful. We could, however, look at
$$
f(x) - 4 = 0
$$
and then try to factor $f(x) - 4 = r(x)s(x)$, which would then be reduced to solving $r(x) = 0$ and $s(x) = 0$.
A: 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} $
$x = \dfrac{\pm\sqrt{5} + 1}{2}$
