As the signs are alternately + and —, there will never be any uncertainty as regards this last term. Let us view, then, the operations which must be performed.

1o. We calculate the sums S1, S2, S3.. up to Sea by means of the known relations S+P=0, S2+PS1+2Q=0, &c.

2o. In the formula which expresses fa we make successively a=1, 2, 3, ...n, and we thus have, to determine the n sums S1, S2, S3, ...fa,

f=mS-SIS1, f=mS,-4S,S1+3S,S2, &c.

3°. Finally, the relations existing between these n sums and the n coefficients p, q, r, will give the values of these coefficients, viz.,

...

p=Si, q={(S2+psi), r=} (S+p£1⁄2 +9/i), &c.

364. A method entirely analogous to that which has been employed in find ing the equation of the squares of the differences can be employed in a great number of cases, and particularly in those where the roots of the transformed equation are similar, and entire powers of the difference, of the sum, of the product, or of the quotient of any two roots whatsoever of the given equation. For example, suppose that each new root is to be the power k of the sum a+b of two roots of equation [A]. Taking n=m(m-1), the transformed equation ought to have the form

Sa=(a+b)ka+(a+c)ka+...+(b+c)ku+, &c.,

the calculation will reduce itself to expressing sa by a general formula. To do this, we take the function

4(x)=(x+a)ka+(x+b)ka+(x+c)ka+, &c.,

But if, before the development, we substitute in (r) successively a, b, c, ..., instead of x, the sum of the resultants will be equal to 2a+2ka Skai hence it is easy to perceive that by making the same substitutions in the development, we shall have

2a+2ka SkamSka+kaS,Ska-1... +mSka

Finally, we derive from this equation the required formula,

Ja=(m—2ka-1) Ska+kaS,Ska−1+

ka(ka-1),

1.2

S2Ska-2+, &c.

When ka is even, we stop at the term which contains S with two equal indices, and we take only the half of it; but when ka is uneven, we stop at the term in which the two indices are (ka-1) and (ka+1), and we take the entire term.

QUADRATIC FACTORS OF EQUATIONS.

365. Every equation of an even degree has at least one real quadratic factor. Let the proposed equation be

EE

2

...

+P1=0, having roots a, b, c, &c., and let n=2μ, μ being an odd number. Let it be transformed (Art. 362) into an equation whose roots are the combinations of every two of its roots, of the form y=a +b+mab, m being any number; also, let the transformed equation be m(y)=0; then its coefficients will be symmetrical functions of a, b, c, &c., and, therefore, rational and known functions of P1, P2, &c.; and its degree will 2μ (2μ-1) be which is odd; therefore, pm (y)=0 will have at least one real root, whatever be the value of m. Hence, making m=1, 2, 3, ... {μ(2μ−1)+1}, successively, each of the equations 1(y)=0, 2(y)=0, &c., will have at least one real root; that is, we shall have μ(2μ—1)+1 real values for combinations of two roots of the proposed equation, of the form a+b+mab; but there are only μ(2μ-1) such combinations which are differently composed of the roots a, b, c, &c.; therefore, two of these combinations, for which we have obtained real values, must involve the same pair of the quantities a, b, c, &c.; let this pair of roots be a, b, and a, a', the real roots of the corresponding equations m(y)=0, m'(y)=0, so that

a+b+mab=a, a+b+m'ab=a';

therefore, a+b and ab are real, and the proposed equation has at least one real quadratic factor, and two roots, either real, or of the form aß √ −1. Hence every equation whose degree is only once divisible by 2 has at least one real quadratic factor.

We shall now prove that if it be true that every equation has at least one real quadratic factor when its degree is r times divisible by 2, or when n=2'μ, where is odd, the same is true when the degree of the equation is r+1 times divisible by 2. For, let n=2+1μ; then the degree of the transformed equation will be 2μμ(2+1μ-1), which is only r times divisible by 2; therefore, by supposition, the transformed equation, „(y)=0, will have two roots, either real or imaginary. If they are real, then, exactly in the same way as for the preceding case of the index being only once divisible by 2, it may be shown that the proposed equation has at least one real quadratic factor. If they are imaginary, we shall have y=a+ẞ√−1, each of which expresses the value of some one of the combinations a+b+mab, a+c+mac, &c. Suppose, therefore, that we have a+b+mab=a+ß √−1; then, as shown above, we can give m such a value m', that m(y)=0 shall have a root corresponding to the combination of the same letters, so that a+b+m'ab=a'+ß' √ ~1; from which equations we can obtain values of ab and a+b under the forms

a+b=x+8 √ −1,
aby'+'-1,

..x2-(y+d√-1)x+r'+d' √ −1 is a factor of f(x);

but if any real expression have a factor of the form M+N-1, it must also have one of the form M-N-1;

••• x2 —(y—d √√ −1).r‍+y' —d′ √ −1 is a factor of f(x);

if, therefore, these two expressions have no simple factor in common, their product will be a biquadratic factor of ƒ (r),

(x2—yx+y')2+(dx—d')2,

which can always be resolved into two real quadratic factors. (See solution of Biquadratics.) If they have a factor in common, since they may be written

x2-yx+y-√—1(d.x-d′), x2-yx+y′+ √ −1(dx-8'),

it can only be of the form r-ɛ; and the factors themselves become (x−k+2 √ −1)(x—e), (x—k—λ √ —1)(x —e) ;

and, therefore, the proposed equation admits the real quadratic factor

Hence an equation whose degree 2+1μ will have a real quadratic factor, provided an equation whose degree 2μ has one; but we have proved this to be the case when r=1; therefore it is universally true that every equation of an even degree has at least one real quadratic factor. If now this factor be expelled, the depressed equation will have its coefficients real and its degree even, and will, therefore, as before, have one real quadratic factor. Hence the first member of every equation of an even degree may be resolved into real quadratic factors.

366. Hence if we divide the first member of any equation

by x2+ax+b, admitting no terms into the quotient that have r in the denominator, we shall at last obtain a remainder of the form Ar+B, A and B being rational functions of a and b; and in order that x2+ax+b may be a quadratic factor of the proposed equation, it is necessary and sufficient that this remainder should equal zero for all values of x, which requires that we separately have A=0, B=0. The different pairs of values, real or imaginary, of a and b which satisfy these equations will give all the quadratic factors of the proposed; and as the number of these factors is n(n-1) (Art. 244, Cor. 2), the final equation for determining one of the quantities a, b, obtained by eliminating the other between the two preceding equations, will be of the degree in(n-1), which exceeds n, if n>3; therefore, the determination of the quadratic factors of an equation will generally present greater difficulties than the solution of the equation.

As the proposed equation has necessarily in or (n-1) real quadratic factors, according as n is even or odd, there will always exist the same number of pairs of real values of a and b, satisfying the equations A=0, B=0; and if any of these pairs of real values be commensurable, they may be easily found; and the commensurable quadratic factors being known, the equation may be depressed.

EXAMPLES.

(1) To resolve x-6x2+nx—3—0 into its factors. Dividing by x2+ax+b, we find a remainder,

(n+2ab+6a—a3)x — (a3b—b2—6b+3);

therefore, to determine a and b, we have

n+2ab+6a-a3—0,

a2b―b2―6b+3=0.

Solving the former with respect to b, and substituting in the latter, we find

(a2-4)3=n2—64, or a=√4+Vn2-64; from whence b, and the other quadratic factor,

may be determined.

x2—ax+a2—b—6,

(2) The resolution of x^+pr3+qx2+rx+s into its two quadratic factors, x2+mx+n, x2+m'x+n, may be effected by the following formulæ :

23-(3p2-8q)2+(3p1-16p2q+16q+16pr-64s)z-(8r-4pq+p3)2=0,

which has necessarily a real root.

ELIMINATION BY SYMMETRIC FUNCTIONS.

367. Symmetric functions furnish a method of elimination which has the advantage of making known the degree of the final equation.

...

in which P, Q..., P', Q'....... are functions of y. If we could resolve (1) with respect to x, we would derive from it m values, a, b, c..., of x, which would be functions of y; and, by substituting these values of x in equation (2), we would have, for determining the values of y, m equations free from x, viz.,

But, in general, the resolution of equation (1) is impossible, and the problem is to obtain a final equation which embraces all the values of y without distinction.

We shall have an equation which will fulfill this condition by multiplying together the m equations (3), for the resulting equation will be satisfied by each value of y derived from any one of them, and it can not be satisfied in any other way. But the factors of this resultant can only change places, whatever permutations we may make between the quantities a, b, c...; the product, then, will only contain entire and rational symmetric functions of these quantities; hence we shall be able to express these factors by means of the coefficients of equation (1), and in this way we shall have the final equation in y.

This method of elimination leads, in general, to very tedious calculations; but it has the advantage of giving a final equation containing all the roots that it ought to embrace, without any complication of foreign roots.

368. This method has also the advantage of leading to a general theorem with respect to the degree of the final equation. In the preceding article the first equation is of the degree m, the second of the degree n, and P, Q..., P', Q'... are any functions whatsoever of y; but, for the theorem in question, these functions must evidently be polynomes, such that the sum of the exponents of x and y shall be, at most, equal to m in each term of equation (1), and, at most, equal to n in each term of equation (2). We have, then, to determine to what degree y can be raised in the symmetric functions which compose the product of equations (3).

Each term of this product is the product of m terms taken respectively from

the m equations (3); hence, designating these terms by Yaa, Y'b3, Y", the term of the product will be YY'Y"...abc... But the product of these m equations being symmetric with respect to the quantities a, b, c..., all the terms should have the same form with the one that we have given above; consequently, we know that the product embraces all the terms represented by

YY'Y"... ×S(aab3c¥.....)

(4)

We have now to determine the degree of y in this expression. Observing that the degree of y in Y is, at most, equal to n-a, in Y' to n-ß, in Y" to n―y, &c., we shall readily see that in YY'Y"... its degree will be, at most, equal to mn—a-ẞ-y.... On the other hand, if we refer back to the relations (Art. 356) from which the sums S1, S2, S3, &c., are derived, we shall see that, P being, at most, of the first degree in y, Q of the second, R of the third, and so on, the degree of y in these sums can not surpass the subscript number of S; and, in like manner, if we refer (Art. 359) to the formulas which express double, triple, &c., functions, we shall perceive that in S(abc...) the degree of y can not surpass a++... Hence in expression (4) the degree of y will be, at most, equal to mn.

The same remark will apply to all the symmetric functions whose sum composes the product of the m equations (3); therefore, lastly, the final equation can not be of a degree superior to mn.

But we

The demonstration seems to require that equation (1) contain m. can suppose that at first am had a coefficient, A, independent of y, and that we have divided the whole equation by A. The final equation ought to subsist, whatever may be the value of A; we can make A=0, and it is evident that this supposition will not raise the degree of the final equation. Finally, the theorem is to be thus understood: that the elimination between two general equations, the one of the degree m, the other of the degree n, ought to give a final equation of the degree mn; but that, in particular cases, the degree of the final equation can be less than mn.

EXAMPLES.

The two equations, x-y"=0, x"+ay"+by+c=0, although very simple, will give a final equation fully of the degree mn; for, by substituting in the second the value of r derived from the first, it becomes y+ay"+by+c=0. On the other hand, in eliminating x between the equations x-y=0, x"+ay"+by+c=0, we obtain a final equation of a degree less than mn, viz., y+ay"+by+c=0.

369. For extending the theorem to any number whatsoever of equations, we have the general theorem given by Bezout, viz., that If, between equations equal in number to that of the unknowns, we eliminate all the unknowns, except one, the degree of the final equation will be, at most, equal to the product of the degrees of these equations.

Before Bezout, the theorem had been known for the case of two equations; and Cramer, in the appendix to his Introduction to the Analysis of Right Lines, has given a very simple demonstration, which, in reality, does not differ from that which we have stated. It has been a desideratum that the same demonstration should be capable of being applied to all other cases; this has been accomplished by Poisson, in a memoir which appeared in the eleventh volume of the Journal de l'Ecole Polytechnique.