Factoring Cubic Equations I’ve been trying to figure out how to factor cubic equations by studying a few worksheets online such as the one here and was wondering is there any generalized way of factoring these types of equations or do we just need to remember a bunch of different cases. 

For example, how would you factor the following: 


  $x^3 – 7x^2 +7x + 15$


I'm have trouble factoring equations such as these without obvious factors.
Your example has integer coefficients. We deal mainly with that case, or the essentially equivalent case of rational coefficients, since we can always multiply through by the lcm of the denominators and get integer coefficients,

The following result, called the Rational Roots Theorem, is sometimes useful.

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ be a polynomial with integer coefficients $a_i$, and suppose $a_0\ne 0$. Then any rational root of $P(x)=0$  must have shape $\frac{c}{d}$, where $c$ divides $a_0$ and $d$  divides $a_n$, 

In our case, $c$ must divide $15$ and $d$ must divide $1$. That gives the candidates $\pm 1$, $\pm 3$, $\pm 5$, and $\pm 15$. 

Test them all. We get very lucky, $3$ works. To find the other roots, we can divide our polynomial by $x-3$, getting a quadratic, which we solve using the Quadratic Formula.

This procedure often works when we are dealing with a "made up" problem. In particular, if the cubic is in a school exercise, almost certainly it has a rational root.  But the procedure certainly does not always work. Most cubics with integer coefficients do not have a rational root.   And the procedure  only applies to polynomials with rational coefficients.

If there are no rational roots, we are in trouble. True, there is a formula for the roots of the general cubic, usually called the Cardano Formula. It dates back to the sixteenth century. One can also find formulas that use trigonometric or hyperbolic functions. But overall the formulas are not very useful.

There are good methods for quickly approximating the roots to high accuracy,
$x^3-7x^2+7x+15=(x+1)(x-3)(x-5)$