Factoring is a process by which an expression can be rewritten in the form of product of a number of expressions called as factors. Evidently the degree of the factors will be of lower compared to the original polynomial. Hence it provides an easier method of finding the roots of the polynomial by using the zero product property. A root means the value of the variable which makes the entire value of the polynomial expression as $0$.

A polynomial can be expressed in different forms. One of such forms is called as factored form. As a corollary of fundamental theorem of algebra, a polynomial of degree ‘$n$’ will be having ‘$n$’ number of factors.  Therefore a polynomial $p(x)$ of degree ‘$n$’ can be expressed in factored form as $p(x)$ = $a(x – x_1)(x – x_2) ….(x – x_n)$, where ‘$a$’ is a constant, which is also the leading coefficient of the polynomial. If the value of the variable equals any one the constants $x_1, x_2, … x_n$, then the value of the entire polynomial is $0$. Hence $x_1, x_2, … x_n$ are nothing but the roots of the polynomial. Some or all of the roots can also be imaginary. Also the real roots of a polynomial function are nothing but its $x$-intercepts.

A cubic polynomial is a polynomial in degree $3$. Hence will have three roots and also can be expressed in the factored form as $p(x)$ = $a(x – x_1)(x – x_2)(x – x_3)$.  The roots of a cubic polynomial have an interesting property that at least one of its roots will be real. This fact can be derived from the study of end behavior of a cubic polynomial. Since its degree is $3$ which is an odd number, the two ends of the graph of the polynomial will be in the opposite directions and obviously hence, the graph has to cross the x-axis at least one point at one point. 

There are many techniques of factoring a cubic polynomial and the correct method is selected by closely studying the given polynomial.  As we all know a quadratic can certainly be factored in different ways.  Hence it can be realized that if one factor is found by even by trial and error, the rest of the process becomes simple as the remaining factor has to be a quadratic.
This method of factoring a cubic polynomial is the easiest and quickest. Since there are usually four terms in a cubic polynomial regrouping into two binomials is simple. However, the scope of this method is limited because only very few cubic polynomials could possibly be suitably grouped. In other words, only in rare cases this method can be applied. We will understand this method and also its limitations when we work out some examples.
An immense help is extended by Rational Root theorem in finding the roots of a polynomial. It states that if a polynomial is in the form of
$p(x)$ = $a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + …… a_1 x + a_0$, then the possible rational roots are from all possible combinations of,

$\pm$ $\frac{a\ factor\ of\ a_0}{a\ factor\ of\ a_n}$ 

The theorem carefully uses the term ‘possible rational roots’. Hence it is not necessary that all the combinations of this ratio will work. Also if, none of the combinations works, then the conclusion is either the roots are irrational or imaginary. 

Therefore, for a cubic polynomial we can find at least one possible rational root. Having done that, we come to know one factor of the polynomial. Then either by long division or synthetic division of the given polynomial by the known factor, we get the quotient in quadratic form. And the same can now be factored by regrouping or by using quadratic formula or by the method of completing square. 
Example 1: 

Factor the polynomial $2x^3 + 4x^2 – 8x - 16$

Solution:

We can group the given cubic polynomial as $(2x^3 + 4x^2)\ –\ (8x + 16)$

Factoring each group $2x^2(x + 4)\ –\ 8(x + 2)$

Now using $(x + 4)$ as common factor, the factors are 

$(x + 2)(2x^2 – 8)$ = $2(x + 2)(x^2 – 4)$

                             = $2(x + 2)(x + 2)(x – 2)$

                             = $2(x+2)^2(x – 2)$

In this case the terms are such that we can easily figure out the common factors and group accordingly. 
Example 2:

Find the roots of the polynomial $x^3 - 2x^2 + 5x + 26$ and hence factor completely.

Solution:

The roots of the given polynomial are the solutions of the polynomial equation $x^3 - 2x^2 + 5x + 26$ = $0$

We find there is no scope of grouping the terms to get common factors.  Hence let us use the rational root theorem. The possible rational roots are from all possible 
combinations of,

$\pm$ $\frac{a\ factor\ of\ 26}{a\ factor\ of\ 1}$

In other words the possible rational roots can be listed as, $\pm 1,\ \pm 2$ and $\pm 13$. Out of these, by trial and error we can see $x$ = $-2$ satisfies the equation $x^3 - 2x^2 + 5x + 26$ = $0$. Hence $-2$ is one of the roots of and hence $(x + 2)$ is a factor of the given polynomial. 

The quotient of  $\frac{2x^3 - 2x^2 + 5x + 26}{x + 2}$ is $(x^2 -4x + 13)$. Therefore, the remaining roots of the given polynomial are the solutions of the equation $x^2 -4x + 13$ = $0$. 

This is a quadratic which cannot be solved by factoring. Hence, let us use the method of completing square.

$x^2 - 4x + 13$ = $0$

$x^2 - 4x$ = $-13$

$x^2 - 4x + 4$ = $-13 + 4$ (Adding $4$ on both sides to complete the square on the left side)

$(x – 2)^2$ = $-9$

So, $x – 2$ = $\pm$ $\sqrt{(-9)}$ = $\pm\ (\sqrt9)(\sqrt{(-1)}$ = $\pm 3i$.

Therefore, $x$ = $2 + 3i$ and $x$ = $2 – 3i$.

Thus, the remaining two roots of the given polynomial are $(2 + 3i)$ and $(2 – 3i)$ and the remaining factors are $(x – 2 – 3i)$ and $(x – 2 +3i)$

So, the given polynomial $x^3 - 2x^2 + 5x + 26$  is completely factored as $(x + 2)(x – 2 – 3i)(x – 2 +3i)$