## Appendix

### Rational root test

We saw in the section Solving polynomial equations that, to find the roots of a polynomial (with integer coefficients), a good place to look is the factors of the constant term. In fact, we proved that any integer root must be a factor of the constant term.

As long as we're prepared to try a few more possibilities, we can determine whether there are any roots which are rational numbers (i.e., fractions).

###### Theorem (Rational root test)

Let $$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$ be a polynomial with integer coefficients. If $$\dfrac{r}{s}$$ is a rational root of $$f(x)$$, and $$\dfrac{r}{s}$$ is in simplest form, then $$r$$ is a factor of $$a_0$$ and $$s$$ is a factor of $$a_n$$.

###### Proof

Since $$f\big(\dfrac{r}{s}\big) = 0$$, we have $a_n \left( \frac{r}{s} \right)^n + a_{n-1} \left( \frac{r}{s} \right)^{n-1} + a_{n-2} \left( \frac{r}{s} \right)^{n-2} + \dots + a_1 \left(\frac{r}{s}\right) + a_0 = 0.$

Multiplying through by $$s^n$$ gives $a_n r^n + a_{n-1} r^{n-1} s + a_{n-2} r^{n-2} s^2 + \dots + a_2 r^2 s^{n-2} + a_1 r s^{n-1} + a_0 s^n = 0.$

Examining this equation, we see that every term on the left-hand side except the last is divisible by $$r$$; so we have a multiple of $$r$$, plus $$a_0 s^n$$, equal to 0. Hence $$a_0 s^n$$ is a multiple of $$r$$; equivalently, $$r$$ is a factor of $$a_0 s^n$$. But, as $$\frac{r}{s}$$ is in simplest form, $$r$$~and $$s$$ have no factor in common; hence $$r$$ is a factor of $$a_0$$.

Similarly, every term on the left-hand side except the first is divisible by $$s$$, so $$a_n r^n$$ is also divisible by $$s$$; equivalently, $$s$$ is a factor of $$a_n r^n$$. Again, since $$r$$ and $$s$$ have no common factor, $$s$$ is a factor of~$$a_n$$.

$$\Box$$

Next page - Appendix - Fundamental theorem of algebra