the equation, by dividing it by them (or a power of one of them equal to their number), and then proceed as above directed, with the new equation.

Example 1.-Examine and class the roots of the equation

Here, by Des Cartes' Rule, if all the roots are real, they are all positive, since there are three variations of sign. The criterion for equal roots being applied, give none.

To apply the criteria for imaginary roots :

Deguas' does not apply here in the outset, for no coefficient is 0; nor if we transform it to want the second term, does that rule give any result; for the transformed equation is (x 1)3 12 (x − 1)2 — 15 = 0.

Nor by the application of the change of sign of the alternate terms is any indication of the nature of the roots obtained; since there are not fewer permanencies than there were changes before.

Instead of trying other criteria, apply Horner's algorithm for reducing the equation by 1, 11, . . . then we have

1=0;

But as

and as the signs remain the same, there is no root between 0 and 1. the absolute term is so near vanishing, one of the roots is shown to be nearly 1, but a little more.

For if they have not both changed their signs in this interval, this is obvious: and if one of them should also have changed its sign, it will be possible, by lessening the interval, to ascertain which changes first. From this the nature of the roots can often be ascertained pretty readily.

2. If in approximating to the root a certain number of changes of sign are lost in the interval 1, reduce the roots of the equation whose roots are the reciprocals of these; then if as many changes are not gained as were lost before, there are as many imaginary roots as units in the difference. This is Budan's Criterion: but it cannot be considered a certain indication, except the search be for the greatest positive or negative roots of the equation.

3. Estimating all changes from left to right, number the variations as they arise in the two transformations between which we are to expect two roots; and at that step where one of these numbers first exceeds the other by two, note the coefficients themselves, and multiply each by the index of the power to which it belongs. Divide the right hand product in each case by the left hand one of those above-named; and if the sum of the fractions thus formed (disregarding their signs) be equal to, or greater than the interval between the numbers by which the transformations were effected, these are imaginary roots. But it does not follow that the said sum being less than the interval, indicates real roots, and it is necessary to diminish the limit, occasionally, several successive times, and apply the test each time. If, however, at each time the ratio of the said sum to the interval diminishes at each step, it is a satisfactory indication of real roots; but if it increase, then it is an equally satisfactory indication of imaginary roots, existing in the interval. This is Fourier's Criterion.

4. Change the signs of the alternate terms, and if there be not the same number of permanencies of sign now as there were of variations before, the difference of that number indicates so many imaginary roots.

there is one permanency gained; and therefore one, and only one root between

1 and 2.

By trying 10, we find all the roots +, and hence no root so great as 10.
By trying 4, 5, and 6 successively, we have the coefficients indicating

and as one variation is lost between 4 and 5, there is one root between those limits; and for a similar reason, another between 5 and 6.

Hence all the

roots are real, and their initial figures are found. The solution will be resumed. Ex. 2. Determine the character of the roots of the equation

1, x, and x + 1, denoted here by (− 1), (0), (1), are—

(0) + + + +

(1)

two roots in this interval.

++++++

three roots in this interval.

To ascertain whether these roots be real or not, take the intervals

5 between 0, 1, and 0 + 1; then we have the series, both coefficients and

Here are two variations lost in passing from a to z3, the transformations being -5 and 0. Hence multiplying these coefficients by the indices of the powers to which they belong, viz.

1 x 3

1 x 3

+

= +, which being

1.5 X 4

1 X 4

and taking the sum of the quotients greater than the interval, indicates that the two roots which should be sought for in that interval (—•5, 0), are imaginary, by Fourier's Criterion.

In a similar manner, by forming the equation in x + 5, they will find a pair of roots indicated in the interval between x and x + ·5, and one root between this latter and 1. Also, the second loss of a variation will make its appearance in this case under the first power of x, or at index (1); and the application of Fourier's Criterion will show them to be imaginary.

Ex. 3. What are the characters of the roots of the equation

=

in which intervals there are four variations lost, and hence four roots indicated. Try Budan's Criterion, that is, take the reciprocal equation, which gives all its signs, and hence no variations are gained. Hence var. lost — var. gained: 40 = 4 = number of these roots (all in this case), which are imaginary. This should, however, be subjected to Fourier's Criterion by the student himself. Farther, form the equation in x - 2, and its reciprocal, and Budan's Criterion again gives two roots imaginary for this interval. That is, all the roots of the equation are imaginary.

Ex. 4. Resolve the equation x*

Here the criterion of equal roots being applied, the result is found to be that the equation is equivalent to (x — 1)3 (x — 3) =0.

The student may resolve the following equations, so far as the character and initial figures of the roots are concerned :—

x-12x60x1 + 123x2 + 456789012= 0,

ენ · 8x3 + 32x1 — 74x3 + 104x2

x7-9x640x5

80x+25=0,

106x+178x3-184x2 + 105x = 25,

8902x6-4567x5 - 123x3 60x2+12x1,

THE principle of this method, viewed as an arithmetical operation, is the transformation of the equations by one or more figures of the root at a time, and a sure and regular method of discovering those figures; whilst the work itself is performed by means of the detached coefficients, without the introduction of one single symbol except those of operation.

RULE 1. Exclude the equal roots, if there be any, by VI.

2. Find the initial figures of the several terms, and apply where necessary the criteria for imaginary roots, by VII. and Notes.

3. Transform the equation so that its roots shall be diminished by the initial figure just found, if the root be positive, or increase it, if that root be negative.

4. Use the penultimate coefficient, as a trial divisor of the absolute term, by which to find the next figure of the root; observing that there is a small increase of the penultimate term to be made during the next operation, and which should in any case of doubt be allowed for t.

* Prior to the appearance of the method here given (which was in a paper published in the Philosophical Transactions of the Royal Society for 1819, by a most able mathematician, W. G. Horner, Esq. of Bath) no direct and certain method of finding the roots of equations beyond the fourth degree was known.

For let a, b, c, d, be the roots, of which d is that to which we are approaching, or which is the difference between one of the roots of the equation and the init: figure of that root by which all the roots were diminished at the last process.

Then the two terms, viz. the penultimate and absolute terms are respectively, by the composition of equations, abc +(ab + ab + be) d, and abed. The immediate or first figure of dis then, on account of the relative smallness of d, found from

abed

abc+(abac + be) d

by dropping the term (ab+ ac + be) d from the denominator. The same holds, evidently, however many roots there may be in the equation.

This becomes, as a method, so much more certain as we proceed further and further with the figures of d, as their local values give them upon the average a tenfold diminution at cach step of the process and it gives rise to a method of abbreviation to be noticed in a future page.

5. Diminish the roots again by this new figure, as before, and find in the same way the next figure of the root. Proceed thus till all the figures are found (which will be possible if the root itself be a finite number), or till as many figures are found as are required for the purpose in view.

6. The last line of coefficients are those of a new equation, whose roots are less than those of the given equation by the quantity (or root) already found; and if that root be completely determined, the last or absolute term is 0, and the equation is reduced one degree lower than the given one. The initial figures of this are now already known, and it may be resolved as the former was; but being of a lower degree, will involve less actual calculation. But if the root

term be taken

has been only approximately determined, then if the absolute 0, we may resolve the reduced equation as one of a lower

degree, and obtain values of the remaining roots nearly true.

It is, however, often, especially by the abbreviated process hereafter explained, p. 236, quite as convenient for the young and inexperienced calculator, to resume the original equation, and approximate to the other roots whose initial figures he has already found.

EXAMPLES.

Find the roots of x + x3 + x2 13089033x + 13089030 = 0. Here substitute successively for a the terms of the decimal scale, 1, 10, 100, 1000, &c., or, which evidently comes to the same thing, dividing the equation by x − 1, x — 10, x — 100, x — 1000, &c., and working with detached coefficients by means of the synthetic method of division.

Hence 1 is a root of the equation, the last term being 0, and only one root, since only one coefficient has vanished. Also from the law of signs there are

no other roots between 0 and 1; and the equation which contains the other roots is x3 + 2x2 + 3x 13089030 = 0.

Proceeding with the substitutions, it is found that 100 produces no change, but 1000 does produce one change, and hence that there is one, and only one, root between those limits.

In this way, the first digit is found to be 2 in the hundred's place, or 200. Proceed, therefore, according to the rule, and the work will stand thus :—

Hence, using the penultimate coefficient, 120803, as a trial divisor, and the absolute term as a dividend, the next figure of the root would be found to be 4 : but upon trial, on account of the increased division which this is to receive from

the preceding terms, the result will be too great, and hence 3 should be tried. In that case we have

Again, using the penultimate term as a divisor, and the absolute term as a dividend as before, the next figure is found to be 5. Hence,

Hence the last term being 0, 5 is a root of this equation, 3 + 697x2 + 163108x = 815540; 30+ 53 35 is a root of the equation ≈ + 602x2 + 120803x = 5008430; and 200 + 30+ 5, or 235, is a root of the equation x3 + 2x2 + 3x = 13089030. Also, the remaining roots of the given equation are those of the reduced equation whose coefficients have been just found above, viz. of x2 + 697x + 163108 = 0, which are imaginary. And hence the four roots of the given equation are 1, 235, and 398·5 ± √ · The actual work to be put down in the process is, however, much less than above exhibited, and stands as below.

- 165623.

+13089030 (1
-13089030

(235

When the root involves an interminable decimal, and it is desirable to compute several places of it, a considerable portion of the actual labour may be