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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:

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 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

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