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By continuing this process, it may be shown that the first member will ultimately be resolved into n binomial factors of the form &-2, 2-6, 2-c, etc. Hence equation (1) may be written under the form

(x-a)(x-7)(x-c)(x-d).... (x-k)(x-1)=0. (5.) This equation may be satisfied by any one of the n values, x=a, x=b, x=c, etc., and, consequently, these values are the roots of the equation.

The equation has no more than n roots, because if we ascribe to x a value which is not one of the n values a, b, c, etc., this value will not cause any one of the factors of Eq. (5) to be zero, and the product of several factors can not be zero when neither of the factors is zero.

If both members of Eq. (5) be divided by either of the factors x—a, b, etc., it will be reduced to an equation of the next inferior degree; and if we can depress any equation to a quadratic, its roots can be determined by methods already explained. Ex. 1. One root of the equation

3 + 3x - 16x+12=0 is 1; what are the other roots ? Ex. 2. Two roots of the equation

+ -100 + 35x250x+24=0 are 1 and 3; what are the other roots ? Ex. 3. Two roots of the equation

2c4 - 12.23 +48x2 - 68x+15=0 are 3 and 5; what are the other roots ? Ans. 2+V8. Ex. 4. Two roots of the equation

4x4 – 1423-52% +31x+6=0 are 2 and 3; what are the other roots ?

- 3+15 Ans.

4 Ex. 5. Two roots of the equation

24 — 623 +24x—16=0 are 2 and -2; what are the other roots ? Ans. 3+V5.

437. The n roots of an equation of the nth degree are not necessarily all different from each other. Any number, and, indeed, all of them, may be equal. When we say that an equation of the nth degree bas n roots, we simply mean that its first member can be resolved into n binomial factors, equal or unequal, and each factor contains one root.

Thus the equation x3 — 6x2 + 12x–8=0 can be resolved into the factors (x-2)(x-2)(x-2)=0, or (x-2)3=0; whence it appears that the three roots of this equation are 2, 2, 2. But, in general, the several roots of an equation differ from each other numerically.

The equation x3=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 a2 +23+4=0. Solving this equation, we find x=-1=V-3. Thus, the three roots of the equation x3=8 are

2; -1+V-3; -1-V-3. The student should verify the last two values by actual multiplication.

Ex. 1. Find the four roots of the equation 24—81=0.
Ex. 2. Find the six roots of the equation 26—64=0.

of

438. The coefficient of the second term in the equation of the nth degree is equal to the algebraic sum of the roots with their signs changed.

The coefficient of the third term is equal to the algebraic sum of the products of all the roots, taken in sets of two.

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

The last term is equal to the continued product of all the roots with their signs changed.

Let a, b, c, d, .... I, represent the roots of an equation of the nth degree. This equation will accordingly contain the factors 2-a, x-b, etc.; that is, we shall have

(x—a)(x—b)(x-c)(x-d).... (x-1)=0. If we perform the multiplication as in Art. 351, we shall have

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tad +bd

+cd

ac" — alan-1+ab an-3- abc 2n-3+ .... (abcd .... 1)=0; -6 tac - abd

- acd -d

-bcd etc.

etc. etc. which results are seen to conform to the laws above stated. By the method employed in Art. 352 it may be proved that if these laws hold true for the product of n binomial factors, they will also hold true for the product of n+1 binomial factors. But we have found by actual multiplication that these laws are true for the product of four factors, hence they are true for the product of five factors. Being true for five, they must be true for six, and so on for any number of factors.

It will be perceived that these properties include those of quadratic equations mentioned on pages 203-5.

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

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

If the sum of the positive roots is numerically equal to the sum of the negative roots, their algebraic sum will be zero; consequently the coefficient of the second term of the equation will be zero, and that term will disappear from the equation. Conversely, if the second term of the equation is wanting, the sum of the positive roots is numerically equal to the sum of the negative roots.

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

For this purpose we must multiply together the factors 2—1, 2—2, 2—3, and we obtain 23 — 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,

1x2+1x3+2x3=11.

The last term is the product of all the roots, 1 x 2 x 3, with their signs changed. Ex. 2. Form the equation whose roots are 2, 3, 5, and -6.

Ans. Oct - 4x3 – 29x2 +156–180=0. Show how these coefficients conform to the laws above given. Ex. 3. Form the equation whose roots are

1, 1, 1, -1, and -2. Ex. 4. Form the equation whose roots are

1, 3, 5, -2, -4, and -6. Ans. 2 + 3x4 - 41c* -8703 +400m-+4443-720=0. Ex. 5. Form the equation whose roots are 1V-2 and 2+1

-3. Ex. 6. Form the equation whose roots are

1=V-1 and 2 + V3.

439. Since the last term is the continued product of all the roots of an equation, it must be exactly divisible by each of them.

For example, take the equation 23 —X—6=0. Its roots must all be divisors of the last term, 6; hence, if the equation has a rational root, it must be one of the numbers 1, 2, 3, or 6, either positive or negative; and, by trial, we can easily ascertain whether either of these numbers will satisfy the equation. We thus find that +2 is one of the roots, and, by the method of Art. 436, we find the remaining roots to be –10V-2.

If the last term of an equation vanishes, as in the example *4 +2x3 + 3x2 +6x=0, the equation is divisible by x—0, and consequently 0 is one of its roots. If the last two terms van. ish, then two of its roots are equal to zero.

440. If the coefficients of an equation are whole numbers, and the coefficient of its first term unity, the equation can not have a root which is a rational fraction.

Suppose, if possible, that is a root of the general equation of the nth degree, where s represents a rational fraction expressed in its lowest terms. Substitute this value for x in the given equation, and we have

6-T Bang

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a
ont A an-1

it +7 +V=0. Multiplying each term by bn-1, and transposing, we obtain

q=-(Aan-1+Ban=26+ ... +Tabn-s+Vbn-1). Now, by supposition, a, b, A, B, C, etc., are whole numbers; hence the right-hand member of the equation is a whole number.

.

a

But, by hypothesis, õ is an irreducible fraction ; that is, a and b contain no common factor. Consequently, an and b will contain no common factor; that is, 7" is a fraction in its lowest terms. Hence the supposition that the irreducible fraction

is a root of the equation leads to this absurdity, that an irreducible fraction is equal to a whole number.

This proposition only asserts that every commensurable root must be an integer. The roots can not be of the form of 3, 4, $, etc. The equation may have other roots which are incommensurable or imaginary, as 2=V3, 1=V_2.

b

441. Any equation having fractional coefficients can be trans. formed into another which has all its coefficients integers, and the coefficient of its first term unity. Reduce the equation to the form

A3" +B.x*-1+Con-+.... +Tx+V=0, in which A, B, C, etc., are all integers, either positive or nego ative. Substitute for x the value x=

y

A' and the equation becomes

Byn-1, Cyn-2

+ An-1

+ An-1 AN-2

=

A which, multiplied by An-1, becomes ya +Byn-1+ ACyn-8+ ... + An-Ty+An-1V=0,

o

+

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