Superior Limit of Positive Roots.

highest negative term; the sum of the negative terms, neglecting their signs, must evidently be less than that of the series

S+SxSx2 + &c. ... + Sxm,

for each term of this series is greater than the corresponding negative term of the equation.

But this series is a geometrical progression of which S is the first term, Sam the last term, and x the ratio; so that its sum is, by example 3, of art. 186,

and must be greater than any positive term, as x", or

x - 1 <x and (x — 1)n — m −1 < xn−m −1,

we must have

(x − 1)n — m < (x — 1) xn — m − 1 <S; ·

If we, then, denote by L this superior limit of the positive

Limits of Negative Roots.

that is, a superior limit of the positive roots is unity, increased by that root of the greatest negative coefficient, whose index is equal to the number of terms which would precede the first negative term, if no terms were wanting.

209. Problem. To find an inferior limit of the positive roots.

Solution. Substitute in the given equation for r, the value

x=

1

and find, by the preceding article, a superior limit of the positive values of y, after the equation is reduced to the form of art. 168; and denote this limit by L'.

1 x >

is an inferior limit of the positive roots of the

given equation.

210. Problem. To find the limits of the negative roots of an equation.

Solution. Substitute for x

x=- y,

and the positive roots of the equation thus formed are the

Integral Root.

negative roots of the given equation; and, therefore, the limits of its positive roots become, by changing their signs, the required limits.

SECTION III.

Commensurable Roots.

211. A Commensurable Root is a real root which can be exactly expressed by whole numbers or fractions.

212. Problem. To find the commensurable roots of the equation

xn+axn−1+bxn−2+ &c. + 1x + m = 0,

in which a, b, &c. are all integers, either positive or negative.

Solution. Let one of the commensurable roots be, when reduced to its lowest terms,

As this root must verify the given equation, we have

whence, multiplying by q-1, and transposing, we obtain

and, therefore, as the second member is integral, the first member must also be integral, or we must have

Method of finding Integral Roots.

that is, every commensurable root of the given equa

tion must be an integer.

Again, the substitution of

x= P

in the given equation, produces

p2 + a pr−1 + &c. . . + k p2 + lp+ m = 0; whence, dividing by p, and transposing, we obtain

and, therefore, as the second member is integral, the first member must be so likewise; that is, every integral root must be a divisor of m.

so that this integral root must likewise be a divisor of m'.

Method of finding Integral Roots.

this integral root must be a divisor of m", m'", m1ˇ, &c.; and the last condition to be satisfied is

Hence to find all the commensurate roots of the given equation, write in the same horizontal line all the integral divisors of m, which are contained between the extreme limits of the roots.

Write below these divisors all the corresponding values of m', m", &c. which are integral, remembering that a divisor cannot be a root, when the values which it gives for either m', m", m'", &c. is fractional.

Proceed in this way till the values of m[n-1] are obtained, and those divisors only are roots which give value of this quantity.

·P for the

Solution. The extreme limits of the real roots are +7,4,

and

and, therefore, 2, - - 1,

5 are roots of the given

equation, and its first member, divided by the factor

(x − 2) (x + 1). (x + 5) = x3 + 4 x2

-7x

10,