An introduction to the modern theory of equations, by Florian Cajori.

THEORY OF EQUATIONS

Writing - for x, we get
(7jn +, +a -1 + a, n +.. + ca,, = 0.
Multiplying by kc"- and transposing all integral terms,
- a h= -1- a -- aJ-k..- a.,,k-1
hin
This equation is impossible, since the fraction -, which is in
its lowest terms, cannot be equal to an integral number.
Hence, - cannot be a root of the given equation.
Hc
55. Integral Roots. Since the equation with integral
coefficients, n + alxl'- -+ ' + a, = 0,
cannot have rational fractional roots, and since a,, is numerically
equal to the product of all the roots (~ 13), it is evident that all
commensurable roots are exact divisors of an and may be found
by testing the factors of a,,. By ~ 4 a factor c is a root, if
f(x) is divisible by x - c without a remainder.
If the coefficient of xn is not unity, but ao, then we may
divide through by a0 and transform the equation into another
whose roots are those of the given equation multiplied by a0
(~ 29). In the new equation the coefficient of xn is unity and
all the other coefficients are integral. Hence, all its commensurable roots are integral.
Ex. 1. Find the commensurable roots of x3 - 7 x - 6 = 0.
The commensurable roots must be found among the values ~ 1, ~ 2,
+ 3, ~ 6, which are all factors of - 6. By Descartes' Rule of Signs we
see that there is only one positive root. By substitution or by synthetic
division we find that +- 1 is not a root, that - 1 is a root. We may now
either depress the degree of the equation by dividing by x + 1 and then
solve the resulting quadratic, or we may try the other factors. We obtain
- 2 and + 3 as the values of the other roots.
Ex. 2. Find the commensurable roots of
2 -3 - 2 -3 = 0.