## The Remainder, Factor, and Rational Roots Theorems

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:

 The Remainder Theorem 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

 The Factor Theorem 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, $-a_0 q^n$.

Remembering that $p$ and $q$ share no common positive integer factors other than $1$, this must also be true for $p$ and $q^n$. Thus, if $p$ is to be a factor of $-a_0 q^n$, it must be a factor of $a_0$.

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

 The Rational Roots Theorem 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...

 Example Factor:  $2x^3 - 9x^2 - 11x + 30$ 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, and is rational, then it must take the form of a fraction whose numerator is a factor of the constant term $30$ and whose denominator is a factor of the leading coefficient $2$. Noting that the integer factors of $30$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15$ and $\pm 30$ and the integer factors of $2$ are simply $\pm 1$ and $\pm 2$, the possible values for a rational root $c$ are limited to $$\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \textstyle{\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \, \textrm{and} \, \pm\frac{15}{2}}$$ 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 completely. Nicely, under a supposition that $(2x^2 - 13x + 15)$ is the product of two binomials we find $$2x^2 - 13x + 15 = (2x-3)(x-5)$$ Thus, we arrive at our complete factorization of the given expression: $$2x^3 - 9x^2 - 11x + 30 = (x+2)(2x-3)(x-5)$$