New University Algebra

4. And in general; The coefficient of the term having ʼn terms before it, is equal to the algebraic sum of all the different products formed by multiplying the roots, with their signs changed if n is odd, n and n. Hence,

5.—The absolute term is the continued product of all the roots, with their signs changed when the number' denoting the degree of the equation is odd.

This principle will enable us to construct an equation, the roots of which are given, and the composition of eq. (1) shows that eq. (2) thus constructed can have no other than the assumed roots; for there is no value of x differing from one of these roots which can cause the first member of eq. (1) to disappear.

From this we might conclude that every equation involving but one unknown quantity, has as many roots as there are units in the exponent of its degree, and can have no more.

425. Admitting that every equation containing but one unknown quantity has at least one root, real or imaginary, it may be demonstrated that the first member of every equation of the mth degree, the second member being zero, may be regarded as the continued product of m binomial factors of the first degree with respect to the unknown quantity. We will first prove that,

If a is a root of an equation of the form

-1

x+Ax1+B.x2+.... Tx+U = 0,

its first member can be exactly divided by x-a.

(1)

For if we apply the rule for division, we shall finally arrive at a remainder which will not contain x; since for each quotient term obtained, the new dividend is at least one degree lower than that which precedes.

Calling the entire quotient Q and the remainder R, we shall have Tx+U= Q(x-a)+R, (2)

an identical equation. The substitution of a for x causes the first member, and also the first term in the second member of this equation, to vanish. Hence, R = 0. But by hypothesis R does not contain x; it is therefore equal to zero whatever value be attributed to x, and the division is exact.

426. The converse of the last principle is also true; that is, If the first member of the equation,

m-1

x+Ax1+Вx2+.... Tx+U=0,

can be exactly divided by x—a, then a is a root of the equation. For, suppose the division performed, and that the quotient is Q; then we shall have the identical equation,

m-1

m-2

x+Ax1+Bx2+.... Tx+U = Q(x—a).

But x = a causes the second member of this equation to vanish; it will therefore cause the first member to vanish, and consequently satisfy the given equation.

427. Every equation containing but one unknown quantity has a number of roots denoted by the exponent of its degree, and no

more.

Resuming the equation,

m-3

x+Ax1+BxTM-3+.... Tx+U=0,

and admitting that it has one root, a, x—a must be a factor of its first member; (425). The quotient which arises from the division of the polynomial,

m-1

x+4x+.... Tx+U,

by x-a, will be of the form

xm-1+А'xTM-2+....T′′x+U' ;

we shall therefore have the identical equation,

Now the second member of this equation will vanish for any value of x which reduces the second factor to zero.

If then the assumed root of the equation,

xm-1+A'xm2+.... T'x+U' = 0,

A third equation may be formed in the same way, and then a fourth, and so on, until the (m-1)th equation is finally reached, in which the second factor in the second member is of the first degree

with respect to x,

Taking this last equation, and substituting for its first member the second, in the next preceding equation, and thus continuing the process of substitution until the first equation of the series is arrived at, the result will be the following identical equation :

m-1

m-2

x" + Ax= 1 + Bu " " + } = { (x—a)(x—b)(x—c)....

....

Tx+U

(x-p)(x-q)

The second member of this equation vanishes for any one of the m values,

x = a, x = b, x = c,. . . . x = p, x = q,

and consequently these values are severally roots of the equation, x+4x-1+Bx-2.... Tx+U=0.

Moreover, no value of x that differs from some one of these values, can satisfy the equation; for no such value will cause any one of the factors in the second member of the identical equation to be zero, a condition requisite to make the product zero. equation therefore has m roots and no more.

428. From the foregoing principles we conclude,

The

1.—That in an equation in which the second term does not appear, that is, the term containing the next to the highest power of the unknown quantity,—the algebraic sum of the roots is 0.

2.-If an equation has no absolute term, at least one of its roots is 0. 3.—The absolute term being the continued product of all the roots of an equation, it must be exactly divisible by each of them. 4. An equation may be constructed, which shall have any assumed roots.

5. The degree of an equation may be reduced by 1 for each of its known roots.

EXAMPLES.

1. What is the equation having +2, -3 for its roots?

Ans. x+x-6=0.

2. What is the equation having the roots +1, -2, 4?

Ans. x+5x+2x-8= 0.

3. What is the equation having for its roots +3, -2, −1, +5 ? Ans. x-5x-7x2+29x+30 = 0.

4. What is the equation of which the roots are 1+√—5, 1-√-5, +√5,-√/5? Ans. x*—2x2+x2+10x-30 = 0.

5. What is the equation of which the roots are −1, −2, +3, 2+√—3, 2—√=3 ? Ans. x-4x+22x3-25x-42 = 0. 6. One root of the equation

are +2,-3; what is the depressed equation, and what are its roots? The depressed equation is

Ans.

x2-4x+6=0;

and its roots are 2+V-2, 2-V-2.

429. Any equation having fractional coefficients can be transformed into another in which the coefficients are entire, that of the first term being unity.

If the coefficient of the first term of the given equation is not unity, make it so by dividing through by this coefficient. Then the equation will be of the form

m-1

m-2

x+4x+Bx+.... Tx+U = 0,

in which it is supposed that some or all the coefficients, A, B, etc., are fractional.

Assume x =

2, a being entirely arbitrary, and substitute this

value of x in the equation; it then becomes