The tool of polynomial long division -- especially as applied to dividing polynomials $f(x)$ by a polynomial of the form $(x-c)$ where $c$ is some constant value -- turns out to have some very important consequences should we be trying to factor that polynomial $f(x)$.

To see this, suppose we polynomial long division to divide $f(x)$ by $(x-c)$ to discover polynomials $q(x)$ and $r(x)$ so that $$\frac{f(x)}{x-c} = q(x) + \frac{r(x)}{x-c}$$ Importantly, since the fraction on the right will always be a "proper" rational expression (where the degree of the numerator is strictly less than the degree of the denominator) and the denominator is $(x-c)$, a polynomial of degree $0$, it must be the case that $r(x)$ is just a constant. Renaming $r(x)$ as just $r$ because of this, and multiplying both sides by $(x-c)$, we then have $$f(x) = q(x) \cdot (x-c) + r$$ Consider what this means for the value of $f(c)$: $$f(c) = q(c) \cdot (c-c) + r = q(c) \cdot 0 + r = r$$ As such, just by evaluating $f(c)$ we can discover the remainder of $\displaystyle{\frac{f(x)}{x-c}}$.

Let us provide a name for this wonderful result:

For any polynomial $f(x)$, the remainder of $\displaystyle{\frac{f(x)}{x-c}}$ is simply $f(c)$. |

Of course, the theorem above implies that if $f(c)$ is zero, the remainder $\displaystyle{\frac{f(x)}{x-c} = 0}$

That is to say, when $f(c) = 0$, it must be the case that $(x-c)$ goes into $f(x)$ evenly. Equivalently, when $f(x)=0$, at least one factor of $f(x)$ must be $(x-c)$.

Thus, one way to try to factor a polynomial $f(x)$ is to hunt for a value of $x$ that solves $f(x)=0$. We codify this result in its own theorem, known as

If for some polynomial $f(x)$, we have $f(c) = 0$, then $(x-c)$ is a factor of $f(x)$. |

The question remains however -- how do we determine a value of $c$ for a given polynomial $f(x)$ where $f(c) = 0$?

Interestingly, if the polynomial in question, $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$, lies in $\mathbb{Z}[x]$ (i.e., has only integer coefficients, $a_i$) and there is a *rational* value $c$ that solves $f(x)=0$, there (nicely) are only a small number of possibilities!

To see this, suppose $f(c)=0$ and $c = \frac{p}{q}$ for integers $p$ and $q$ and where the fraction is in lowest terms (i.e., $p$ and $q$ share no common positive integer factors other than $1$). Then $f(c)$ equals the following:
$$a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right) + \cdots + a_1 \left(\frac{p}{q}\right) + a_0 = 0$$
Multiplying both sides by $q^n$ to clear the denominators, we have
$$a_n p^n + a_{n-1} p^{n-1} q + \cdots a_1 p q^{n-1} + a_0 q^n = 0$$
Subtracting $a_0 q^n$ from both sides and factoring out a $p$ from the terms that remain on the left tells us
$$p(a_n p^{n-1} + a_{n-1} p^{n-2} q + \cdots a_1 q^{n-1}) = -a_0 q^n$$
Having expressed the left side as the product of an integer $p$ and another integer $(a_n p^{n-1} + a_{n-1} p^{n-2} q + \cdots a_1 q^{n-1})$ (*Recall we know this long expression must be an integer by the closure of integers under addition and multiplication*), we can conclude $p$ must be an integer factor of the expression on the right side, $-a_0 q^n$.

Likewise, if we start over with $$a_n p^n + a_{n-1} p^{n-1} q + \cdots a_1 p q^{n-1} + a_0 q^n = 0$$ but this time subtract the leading term, $a_n p^n$ from both sides and factor out the common $q$ on what remains, we have $$q(a_{n-1} p^{n-1} + a_{n-2} qp^{n-2} + \cdots + a_0 q^{n-1}) = -a_n p^n$$ The same reasoning can be applied again to see that $q$ must divide $-a_n p^n$, can't divide $p^n$, and so must be a factor of $a_n$.

Putting these two results together, we have

If $c$ is a rational solution/root to $f(x) = 0$, where $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ is a polynomial in $\mathbb{Z}[x]$ (i.e., one with only integer coefficients), then $c = \frac{p}{q}$ where $p$ evenly divides the constant term, $a_0$, and $q$ evenly divides the leading coefficient, $a_n$. |

Let's see this last theorem in action as it helps us factor something that would otherwise prove very difficult to factor...

The usual factoring tricks and techniques don't seem to help here -- there is no common factor on all the terms; this doesn't fit the form of a square (or cube) of a binomial; it is not a difference of squares, nor a sum/difference of cubes; assuming it is a product of two binomials doesn't lead us anywhere; and factoring by grouping (preumably in pairs) doesn't seem to work either -- even after rearranging the terms. As such, we hope that we can find some value of $c$ such that $f(c)=0$. If we can do so, then the Factor Theorem will guarantee that $(x-c)$ will be one of the factors of $f(x) = 2x^3 - 9x^2 - 11x + 30$. We can then "divide out" this $(x-c)$ factor from $f(x)$, to find the other factor (which itself might factor further). This leaves us with the question, though: "How do we find a $c$ value where $f(c) = 0$?" The Rational Root Theorem says that if such a $c$ exists, Granted, this is a long list of possibilities -- but we may not need to check them all. With that in mind, let us check the easiest ones to check first.
$$\begin{array}{rcl}
f(1) &=& 2(1)^3 - 9(1)^2 - 11(1)+30 = 2-9-11+30 = 8 \neq 0 \\
f(-1) &=& 2(-1)^3 - 9(-1)^2 - 11(-1) + 30 = -2 - 9 + 11 + 30 = 30 \neq 0\\
f(2) &=& 2(2)^3 - 9(2)^2 - 11(2) + 30 = 16 - 36 - 22 + 30 = -12 \neq 0\\
f(-2) &=& 2(-2)^3 - 9(-2)^2 - 11(-2) + 30 = -16 - 36 + 22 + 30 = 0 \quad \checkmark
\end{array}$$
We found one! As $f(-2)=0$, it must be that $(x-(-2)) = (x+2)$ is a factor of $f(x) = 2x^3 - 9x^2 - 11x + 30$. As such, we can do the long division to find
$$\frac{2x^3 - 9x^2 - 11x + 30}{x+2} = 2x^2 - 13x + 15$$
which means
$$2x^3 - 9x^2 - 11x + 30 = (x+2)(2x^2 - 13x + 15)$$
Now, all that remains is to make sure we have factored things |