II., Cor.), which is infinite; consequently, √P2+Q must be infinite. But when p+q√-1 is a root of the equation f(x)=0, √P+Q is zero. Hence, in this case, neither p nor q can be infinite.

243. An objection may be brought against the preceding reasoning that ought not to be concealed. It may be denied that the modulus of the product above referred to is simply the modulus of (p+q√ −1)" in the case of p or q infinite; for it may be maintained that although in this case all the quantities within the parenthesis after the 1 become zero, yet the combination of these with (p+q-1)", which involves infinite quantities, may produce quantities also infinite; and thus the modulus of the product may differ from the modulus of (p+q√ −1)" by a quantity infinitely great. It is not to be denied that there is weight in this objection. But it is not difficult to see that although the true modulus may thus differ from the modulus of (p+q√ −1)" by an infinite quantity, yet the modulus of (p+q√−1)", involving higher powers than enter into the part neglected, is infinitely greater than that part. This part, therefore, is justly regarded as nothing in comparison to the part preserved, the former standing in relation to the latter as a finite quantity to infinity.

But the proposition may be established somewhat differently, as follows: Substituting (p+q√ —1) for x in f(x), we have

P+Q√−1=

(p+q√ −1)"+An−1(p+9 √ −1)"~1+....A1(p+q√−1)+N.

Call the aggregate of all these terms after the first P'+Q'√-1; then it is plain that the modulus of the first term, that is, (√p2+q2)", must infinitely exceed the modulus √P+Q of the remaining terms whenever por q is infinite, because in this latter modulus so high a power of the infinite quantity por q can not enter as enters into the former. Now the modulus of the whole expression, that is, of the sum of (p+q√−1)" and P'+Q'√ −1, is not less than the difference of the moduli of these quantities themselves (Lemma I.), which difference is infinite. Hence, as before, √P+Q' must be infinite when p or q is infinite.

PROPOSITION IV.

244. Every equation containing but one unknown quantity has as many roots as there are units in the highest power of the unknown quantity.

Let f(x)=0 be an equation of the nth degree; then if a, be a root of this equation, we have, by last proposition,

(x-a,)f(x)= f(x)=0,

where f(x) represents the quotient arising from the division of f(x) by x—ɑ1, and will be a polynomial, arranged according to the powers of x, one degree lower than the given polynomial f(x). Now, if a, is also a root of the equation f(x)=0, it is obvious that f(x) must be divisible by x- -a2, for x-a1 is not divisible by x-a, (see Art. 84, Note); hence, if f(x), a polynomial of a degree one lower than f1 (r), or of a degree two lower than f(x), represent the quotient of f(x) divided by x-a2, we have

(x—a,)(x—a2)ƒ2(x)=f(x)=0.

Proceeding in this manner, if a3, A4, A5, ••

.....

a are roots of the equation,

the degree of the quotient reducing by one each time, the equation will assume the form

(x-a,)(x-a)(x-a3)..... (x—a1)=0;

and, consequently, there are as many roots as factors, that is, as units in the highest power of x, the unknown quantity; for the last equation will be verified by any one of the n conditions,

and since the equation, being of the nth degree, contains n of these factors of the 1st degree, (r-a,), &c., there are n roots.

Corollary 1. When one root of an equation is known, the depressed equation containing the remaining roots is readily found by synthetic division. Corollary 2. The number of factors of the 2° degree in an equation is n(n-1) ÷1.2; of the 3o, n(n−1)(n—2)÷1.2.3, and so on (see Art. 203).

EXAMPLES.

(1) One root of the equation 1-25x2+60x—36=0 is 3; find the equation containing the remaining roots.

is the equation containing the remaining roots.

(2) Two roots of the equation 1—12x3+48x2—68x+15=0 are 3 and 5; find the quadratic containing the remaining roots.

is the equation containing the two remaining roots.

(3) One root of the cubic equation x3-6x2+11x-6=0 is 1; find the quadratic containing the other roots.

Ans. x2-5x+6=0

(4) Two roots of the biquadratic equation 4x^—14x3-5x2+31x+6=0 are 2 and 3; find the reduced equation.

Ans. 4x+6x+1=0.

(5) One root of the cubic equation r3+3x2-16x+12=0 is 1; find the remaining roots.

Ans. 2 and -6.

(6) Two roots of the biquadratic equation x-6x+24x-16=0 are 2 and -2; find the other two roots.

PROPOSITION V.

Ans. 3√5.

245. To form the equation whose roots are a1, A2, A3, A4, ••••• an

The polynomial, f(x), which constitutes the first member of the equation required, being equal to the continued product of x—A1, x—A2, x—ɑ3, ••• r-a, by the last proposition, we have

(x-a,)(r-a)(x—az) ...... (x—a1)=0;

and by performing the multiplication here indicated, we have, when

By continuing the multiplication to the last, the equation will be found whose roots are those proposed; and from what has been done we learn that (1) The coefficient of the second term in the resulting polynomial will be the sum of all the roots with their signs changed.

(2) The coefficient of the third term will be the sum of the products of every two roots with their signs changed.

(3) The coefficient of the fourth term will be the sum of the products of every three roots with their signs changed.

(4) The coefficient of the fifth terin will be the sum of the products of every four roots with their signs changed, and so on; the last or absolute term being the product of all the roots with their signs changed.*

* I. The generality of this law may be proved as follows: Let us suppose it to hold good for the product of n binomial factors, we shall prove that it will for the product of n+1 of these. Let

represent the product of n binomial factors, in which A, represents the sum a1+a+a ̧ +, &c., +a, of the n second terms of the binomials, A, the sum of their products two and two, A, the sum of their products three and three, and so on, and A, the product of all the n second terms aα, &c., a

Introduce now a new factor (x+1). Performing the multiplication of the above polynomial by this new factor,

—AL
is composed of A,, the sum of all the
-ant!

second terms of the n binomials (x—a1), (x—a2), &c., and a+1, the second term of the (n+1)th binomial, and is, therefore, equal to the sum of the second terms of the n+1 bino

mials. The coefficient of the third term

+A2
is composed of A,, the sum of the prod
+Ajan+1

ucts of the n second terms two and two, and A1+ the sum of the n second terms, each

Corollary 1.-If the coefficient of the second term in any equation be 0, that is, if the second term be absent, the sum of the positive roots is equal to the sum of the negative roots.

Corollary 2.-If the signs of the terms of the equation be all positive, the roots will be all negative, and if the signs be alternately positive and negative, the roots will be all positive.

Corollary 3.-Every root of an equation is a divisor of the last or absolute

term.

+A2

multiplied by the new second term an+1; hence

of the n-1 second terms two and two.

will be the sum of the products

The last term A4+1 is the product of A, which is the product of all the n second terms multiplied by the new second term an+1, so that A1 is the product of all the n+1 sec

ond terms.

We have thus proved that if the law for the formation of the coefficients above stated hold good for a certain number of binomial factors n, it will hold good for one more, or n+1. We have seen, by experiment, that it holds good for four, it therefore holds good for five; if for five, it must for six, and so on ad infinitum.

II. One might imagine, at first view, that the above relations would make known the roots. They give at once equations into which these roots enter, and which are equal in number to the coefficients of the equation (excepting the coefficient of the first term, which unity). The number of these coefficients is equal to the number of the roots of the equation. Unfortunately, when we seek to resolve these secondary equations, we are led to the very equation proposed, so that no progress is made.

For simplicity, I will take the equation of the 3° degree.

Designating the three roots by a, b, c, we have, to determine the roots, the three re

lations

P--a-b-c
Q=ab+ac+bc..
R--abc

(2)

To deduce from them an equation which contains but the unknown a, the most simple mode of proceeding is, to multiply the 10 by a2, the 20 by a, and add them to the 3°. There results

Pa2+Qa+R=—a3—a2b—a2c

+ab+a2c+abc

-abc.

Reducing, and transposing the term-a3, we have

a+Pa+Qa+R=0.

The unknown quantities b and c are thus eliminated, but the equation resulting is of the same degree with the proposed. From the symmetrical form of the relations (2) we perceive that the elimination of a and b, or a and c, would have been attended with similar consequences.

III. To find the sum of the squares of the roots of any equation.

A1=a+b+c... +l;

.. A ̧2=a2+b2+c2...+12+2(ab+ac+be+....)

=sum of the squares +2A,;

.. sum of squares A12-2A.

To find the sum of the reciprocals of the roots.

(—1)°—1A ̧—1—bc......l+ac...l+ab..i+..

Corollary 4.-In any equation, when the roots are all real, and the last or absolute term very small compared with the coefficients of the other terms, then will the roots of such an equation be also very small.

EXAMPLES.

(1) Form the equation whose roots are 2, 3, 5, and -6

Here we have simply to perform the multiplication indicated in the equa

tion

(x-2)(x-3)(x-5)(x+6)=0,

and this is best done by detached coefficients in the following manner:

•*. x^—4x3—29x2+156x—180=0 is the equation sought.

(2) Form the equation whose roots are 1, 2, and -3.

(3) Form the equation whose roots are 3, 4, 2+ √3, and 2— √3. (4) Form the equation whose roots are 3+ √5, 3— √5, and −6.

(2) -7x+6=0.

ANSWERS.

(3) xa—3x3—15x2+49x—12=0.

(4) -32x+24=0.

PROPOSITION VI.

246. No equation whose coefficients are all integers, and that of the highest power of the unknown quantity unity, can have a fractional root. If possible, let the equation

+A-1+...+A323+A ̧2+A,x+N=0,

whose coefficients are all integral, have a fractional root, expressed in its low

a

est terms by If we substitute this for x, and multiply the resulting equation by b1, we shall have

ō+A2-1α11+...+Aza3b»¬3+Aab”—2+Nba¬1=0.

In this polynomial, every term after the first is integral; hence the first term

must be integral also. But being a fraction in its lowest terms, b

απ

b

must also

be a fraction in its lowest terms, and can not be an integral. (See Note to Art. 84.) Therefore the proposed equation can not have a fractional root.

PROPOSITION VII.

247. If the signs of the alternate terms in an equation be changed, the signs of all the roots will be changed.

Let

1x+An=0

....

(1)

be an equation; then, changing the signs of the alternate terms, we have