The equation

x=8

has apparently but one root, viz., 2; but by the method of the preceding article we can discover two other roots. Dividing x3-8 by x-2, we obtain

x2+2x+4=0.

Solving this equation, we find

x=−1±√−3..

Thus, the three roots of the equation x'=8 are

2; −1+√ −3; −1−√ −3.

These last two values may be verified by multiplication as follows:

If the last term of an equation vanishes, as in the example

x+2x+3x2+6x=0,

the equation is divisible by x-0, and, consequently, O is one of its roots.

If the last two terms vanish, then two of its roots are equal to 0.

PROPOSITION III.

To discover the law of the coefficients of every equation. (309.) In order to discover the law of the coefficients, let us form the equation whose roots are

a, b, c, d, . . . . . l.

This equation will contain the factors (x-a), (x—b), (x−c). &c.; that is, we shall have

(x-a) (x-b) (x−c) (x-d)......(x-1)=0.

If we perform the multiplication as in Art. 264, we shall

1. The coefficient of the second term of any equation is equal to the sum of all the roots with their signs changed.

2. The coefficient of the third term is equal to the sum of the products of all the roots taken two and two.

3. The coefficient of the fourth term is equal to the sum of the products of all the roots taken three and three, with their signs changed.

4. The last term is the product of all the roots with their signs changed.

It will be perceived that these properties include those of quadratic equations mentioned on pages 163 and 164.

If the roots are all negative, the signs of all the terms of the equation will be positive, because the factors of which the equation is composed are all positive.

If the roots are all positive, the signs of the terms will be alternately and —.

Ex. 1. Form the equation whose roots are 1, 2, and 3.

For this purpose, we must multiply together the factors x-1, x-2, x-3, and we obtain

x2-6x2+11x-6=0.

This example conforms to the rules above given for the coefficients. Thus, the coefficient of the second term is equal to the sum of all the roots (1+2+3) with their signs changed.

The coefficient of the third term is the sum of the products of the roots taken two and two; thus,

1×2+1×3+2×3=11.

The last term is the product of all the roots (1×2×3) with their signs changed.

Ex. 2. Form the equation whose roots are 2, 3, 5, and 6. Ans. x-4x3-29x2+156x-180=0. Show how these coefficients conform to the laws above given.

Ex. 3. Form the equation whose roots are 1, 3, 5, −2, −4, -6.

Ans. x+3x3-41x1-87x3 +400x2-444x-720=0. (310.) Every rational root of an equation is a divisor of the last term; for, since this term is the product of all the roots, it must be divisible by each of them. If, then, we wish to find a root by trial, we know at once what numbers we must employ. For example, take the equation

x3-x-6=0.

If this equation has a rational root, it must be a divisor of the last term, 6; hence we must try the numbers 1, 2, 3, 6. either positive or negative.

Hence we see that 2 is one of the roots of the given equation, and by the method of Art. 307, we shall find the remaining roots to be

PROPOSITION IV.

(311.) No equation whose coefficients are all integers, and that of the first term unity, can have a root equal to a rational fraction.

For, take the general equation of the third degree,

x2+Ax2+Bx+C=0,

a

and suppose, if possible, that the fraction is one value of x.

b

this fraction being reduced to its lowest terms. If we substitute this value for z in the given equation, we shall have

Multiplying each term by b2, and transposing, we obtain

Now, by supposition, A, B, C, a and b are whole numbers. Hence the entire right-hand member of the equation is a whole number.

a

But by hypothesis, is an irreducible fraction; that is, a

b

and b contain no common factor. Consequently, a3 and b will

contain no common factor, that is,

a

a3

is a fraction in its lowest

b

terms. Hence, if were a root of the proposed equation, we

b

should have a fraction in its lowest terms equal to a whole number, which is absurd.

The same mode of demonstration is applicable to the general equation of the mth degree.

This proposition only asserts that in an equation such as is here described, the real roots must be integers, or they can not be exactly expressed in numbers. They may often be expressed approximately by fractions, as is seen in the examples on pages 288-301. A real root which can not be exactly expressed in numbers is called incommensurable.

PROPOSITION V.

(312.) If the signs of the alternate terms in an equation are changed, the signs of all the roots will be changed.

If we take the general equation of the mth degree, and change the signs of the alternate terms, we shall have

...

=

0 (1): or, changing the sign of every term of the last equation,

Now, substituting +a for x in equation (m) will give the same result as substituting -a in equation (1), if m be an even number; or, substituting -a in equation (2), if m be an odd number. If, then, a is a root of equation (m), —a will be a root of equation (1), and, of course, a root of equation (2), which is identical with it.

Hence we see that the positive roots may be changed into negative roots, and the reverse, by simply changing the signs of the alternate terms; so that the finding the real roots of any equation is reduced to finding positive roots only.

Ex. 1. The roots of the equation

are 1, 3, and -2.

x3-2x2-5x+6=0

What are the roots of the equation

x2+2x2-5x-6=0?

Ex. 2. The roots of the equation

x2-6x2+11x-6=0

are 1, 2, and 3. What are the roots of the equation

x3+6x2+11x+6=0?

PROPOSITION VI.

(313.) If an equation whose coefficients are all real, contains imaginary roots, the number of these roots must be even. If an equation whose coefficients are all real, has a root of the form

For, let a+b√-1 be substituted for x in the equation, the result will consist of a series of terms, of which those involving only the powers of a, and the even powers of b√-1 will be real, and those which involve the odd powers of b√ −1 will be imaginary. If we denote the sum of the real terms by P, and the sum of the imaginary terms by Q✔-1, then we must have

P+Q√−1=0,

which relation can only exist when P=0 and Q=0.

Again, let a−b√ −1 be substituted for x in the proposed equation, the only difference in the result will be in the signs of the odd powers of b√-1, so that the result will be PQV-1. But we have found that P=0 and Q=0; hence

P-Q√=I=0.