Page:Elementary algebra (1896).djvu/502

484 Rh By substitution we have

a" + pa + pra"? +--+, 10 + Pn = O.

Transposing and dividing throughout by a, we obtain

in which it is evident that = must be an integer. Denoting = by Q and transposing — p,_1,

Q+ Pri=—-— poa" — pia? — a)

Dividing again by a gives

Q+ Dai

a =e — pa" — pa" — a.

Again, as before, the first member of the equation must be an integer. Denoting it by Q, and proceeding as before, we must after n divisions obtain a result

Q,-1 + Pi

=—1. a

Hence if a represents one of the integral divisors of the last term we have the following rule:

Divide the last term by a and add the coefficient of x to the quotient.

Divide this sum by a, and -if the quotient is an integer add to it the coefficient of x^2.

Proceed in this manner, and if a is a root of the equation each quotient will be an integer and the last quotient will be -1.

The advantage of Newton’s method is that the obtaining of a fractional quotient at any point of the division shows at once that the divisor is not a root of the equation.

Ex. Find the integral roots of «+423 «2—16%—12=0. By Descartes’ Rule the equation cannot have more than one positive root, nor more than three negative roots.

The integral divisors of —12 are +1, +2, +3, +4, +6. Substitution shows that — 1 is a root, and that + 1 is not a root.