Therefore we infer that there are four distinct and separate roots; one is -1, for X vanishes for this value of x; another between 0 and 1; a third is 2, and a fourth is between 3 and 4. The common measure -2 indicates that the polynomial X is divisible by (x-2)2; and hence there are two roots equal to 2 (Prop. VII, Cor. 1.)

HORNER'S METhod of resolviNG NUMERICAL EQUATIONS OF ALL ORDERS.

176. The method of approximating to the roots of numerical equations of all orders, discovered by W. G. Horner, Esq., of Bath, is a process of very remarkable simplicity and elegance, consisting simply in a succession of transformations of one equation to another, each transformed equation as it arises having its roots less or greater than those of the preceding by the corresponding figure in the root of the proposed equation. We have shown how to discover the initial figures of the roots, by the theorem of STURM; and by making the penultimate coefficient in each transformation available as a trial divisor of the absolute term, we are enabled to discover the succeeding figure of the root; and thus proceeding from one transformation to another, we are enabled to evolve, one by one, the figures of the root of the given equation, and push it to any degree of accuracy required.

GENERAL RULES.

1. Find the number and situation of the roots by Sturm's or Budan's theorem, and let the root required to be found be positive.

2. Transform the equation into another, whose roots shall be less than those of the proposed equation, by the initial figure of the root.

3. Divide the absolute term of the transformed equation by the trial divisor, or penultimate coefficient, and the next figure of the root will be obtained, by which diminish the root of the transformed equation as before, and proceed in this manner till the root be found to the required accuracy.

Note 1. When a negative root is to be found, change the signs of the alternate terms of the equation, and proceed as for a positive root.

Note 2. When three or four decimal places in the root are obtained, the operation may be contracted, and much labour saved, as will be seen in the following examples:

EXAMPLES.

(1.) Find all the roots of the cubic equation

x3-7x+7=0.

By Sturm's theorem, the several functions are (Note, p. 297.)

X = = x3-7x+7
X1=3x2-7
X2=2x-3
X= +

Hence, for x=+∞ the signs are + + + + no variation

+three variations;

therefore the equation has three real roots, one negative, and two positive. To determine the initial figures of these roots, we have

for a= O signs +

hence there are two roots between 1 and 2, and one between

But in order to ascertain the first figures in the decimal parts of the two roots situated between 1 and 2, we shall transform the preceding functions into others, in which the value of x is diminished by unity. Thus for the function X, we have this operation:

And transforming the others in the same way, we obtain the functions
Y=y3+3y2-4y+1; Y1=3y2+6y-4; Y-2y-1; Y,=+.

Therefore the initial figures of the three roots are 1·3, 1·6, and —3.

1/4/7/615

We have thus found one root x=1.356895867......., and the coefficients of the successive transformed equations are indicated by the asterisks in each column.

Another root is x=1.692021471

For the negative root, change the signs of the second and fourth terms.

20....

3616

203616

4

3632

18791228

Hence the three roots of the proposed cubic equation are

Note.-Since each successive figure in the decimal part of the root extends the right hand column three places of decimals, the middle column two places, and the left hand column one, therefore, in the contractions we must cut off two figures from the left hand column, one from the middle column, and none from the right hand column; and thus we cut off in effect three decimal places from each column.

(2.) Find the roots of the equation x3+11x2−102x+181=0.

We have already found the roots to be nearly 3·21, 3·22, and —17. (See Example 4, page 300.)

In a similar manner, the two remaining roots will be found to be 3-22952121

and -17-44264896.

+3x-100=0, to find the number and situation of

hence two roots are real, and two imaginary; and the real roots must have contrary signs, for the last term of the equation is negative. To find the situation of the roots.

In this example, the function X, vanishes for x=-1, and for the same value of x, the functions X and X, have contrary signs, agreeably to Lemma 2, and writing + or for 0 gives the same number of variations. The initial figures of the root are, therefore, 2 and -3.

The same otherwise, by Budan's method.

If the roots of this equation be all real, the permanencies and variation indicate three negative roots and one positive root.

In the transformation by 2, one variation is left, and in transforming by 3, there is no variation left; therefore, the positive root is between 2 and 3. (2.) For the negative roots.

Here two variations are lost in the direct transformation, and no variations are left in the reciprocal transformation; therefore, the two roots in the interval 0 and 1 are imaginary.

Hence the negative root is obviously situated between −3 and -4.