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1.

1 a, a, 1a is an arithmetical progression, whose terms increase (numerically) by the addition of -1; while if a is negative, the progression becomes 1+ a, a, and 1+a, whose terms decrease by the addition of To find the positive roots, we put 1, 0, and 1, successively, for x, and get 64, 105, and 144 for the corresponding results; then taking 3 (the least required divisor of 105) for, we have 2, 3, 4 for the representatives of 1-a; and dividing 64, 105, and 144, 2, 3, and 4, we get 32, -35, -36 for the quotients, which we shall use to find the remaining roots. Taking 5, the least divisor of 35 (as before), for -a, we get 4, 5, and 6 for the corresponding values of 1-a, -1-a. Dividing 32, 35, and -36, severally, by 4, 5, and 6, we get 8, 7, and 6 for the quotients, which may be taken for the series 1 + a, a, and 1+ a.

1 a, a, and 1-a, severally, by

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a, and

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Hence, putting 3, 5, 7, separately for x in the equation, it will be found to be satisfied, and of course 3, 5, and 7 are its roots.

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Remark. The elegant process used in the solution of this example is called the Method of Divisors, and is due to Newton. (See p. 38 of his Universal Arithmetic, second edition, published in 1728.)

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6. To find the integral roots of x3 +9x2 + 14x — 24 = 0. Here we shall have 24+ 1 for a superior limit to the positive roots, and (9 + 1) = 10 for an inferior limit to the negative roots; and it is easy to perceive that 1 is a positive root, and the only positive root.

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To find the negative roots, we shall suppose that a is a negative root; then, since the first member of the equation must be exactly divisible by a+a, if we put 2, 0, and 2 for a in the equation, the results, 48, 24, and - 24, must severally have 2+ a, a, and 2+ a for corresponding divisors.

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Hence, we get 8, 6, and 4 for one set of corresponding divi24, and 24 severally by 8, 6, and sors, and dividing 48,

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4, we get 6, 4, and 6, which give 6, 4, and 2 for the

remaining set of corresponding divisors.

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Substituting 1, 4, and 6 for x in the equation, we find that it is satisfied, and of course 1, — 4, 6 are its roots.

Remark. If we had found 6 of the first set of divisors not to be a root of the equation, we should have used 48, - 24, and 24 to get the remaining divisors; and we must proceed in like manner in every case, since such divisors do not correspond to a root of the proposed equation.

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7. To find the rational roots of the equation 6x3 5x

22x240.

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If the equation has rational roots, it is clear that it must have rational factors, which may be represented by ax-b, such that a and b are integers, a being a divisor of 6, and b of 24; consequently, either a or b, or each of them (neglecting the sign of b), may be 1, if required.

Since the first member of the equation must be exactly divisible by axb, if we put 1, 0, 1 successively for x, the results must have a-b, b, and a-b for corresponding divisors when the root is positive, and a + b, b, and +b will be the corresponding divisors when the root is negative.

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Hence, putting 1, 0, and 1, successively, for x in the equation, we get 3, 24, and 35 for the corresponding results, which have 3, 2, and 1 for divisors, and putting a + b = 3, a+b=1, we get a = 1, and we find that

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b = 2,
a root of the equation.

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Dividing 3, 24, and 35 by 3, 2, and 1, we get 1, 12, and 35, from which we are to find the remaining roots; and it is clear that they can not give any negative root.

To get the positive roots, we take 2 (a factor of 6) for a; and adding — b to it, we have 2 - b for a divisor of 1; consequently, if we put 3 (a factor of 12) for b, the divisor 2-b becomes

1, and thence we have sponding divisors of 1, 12, and 35.

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Hence, since the equation is exactly divisible by 2x

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Dividing 1, 12, and 35 by 1, 3, 5, we get 1, 4, and -7; consequently, we have a-b1, b = −4, and ab7, which give a = 3; consequently, 3x-4

4

is an exact divisor of the equation, and x=

is its remaining

3

root.

Hence, the equation is reduced to (x + 2)(2x − 3)(3x — 4) 0, which is satisfied by putting either of its factors equal to 0.

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Remarks.-1st. If we multiply the terms of the equation by 62, and put y for 6x, then it will be reduced to (6x) — 5(6x)2 — 132(6x) + 864 = 0, or y3 — 5y2 — 132y + 864=0; consequently, if we find the rational roots of this equation, as in the 5th and 6th examples, and divide each of them by 6, we shall get the rational roots of the proposed equation; and it is plain that we may proceed in the same way in all analogous cases.

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2d. When an equation is reduced to the form y" + A1y′′−1 + A2y-2 + .... + An-13 + A2 = 0, such that n is a positive integer, and A1, A2, A3.... An positive or negative integers, according to the nature of the case, then the rational roots of the equation shall be integers.

For since the rational roots must clearly be either integers

b

or rational fractions, if possible, let a stand for a rational

x=

a

root; then it is clear that we may suppose b and a to be prime to each other, or to have 1 for their greatest common divisor.

b

Because is a root of the equation, it must be exactly

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and a are prime to each other, the equation must be exactly divisible by ay — b.

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Since in this division y is arbitrary, and the term y" different from any other term of the equation, it is clear that ay -b can not exactly divide the equation, unless ay is an exact divisor of y".

Hence, since it is clear that y" is not exactly divisible by ay, when a is an integer different from 1, it follows that we must have a = 1; consequently, the rational roots of the equation must be of the form y=b an integer.

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8. To find the rational roots of "64x4 -328x+574x2 393x+90= 0.”

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Because the equation is complete, and has no permanence of signs, it follows from the rule of Descartes that it can not have any negative roots.

Let, then, ac-b (as in the preceding example) stand for any simple rational divisor of the first member of the equation, such that a is a factor of 64, while b is a factor of 90.

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Putting 1, 0, and 1 successively for x in the equation and in the divisor, we have the results 7, 90, and 1449, whose corresponding divisors are ab, b, and a-b, which increase numerically by the addition of a.

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Putting 1, 2, and 3 for the corresponding divisors (which are the simplest), we have a b1, − b = −2, and a- - b = - 3, which give a = 1 and b = 2; conse quently, the divisor axb becomes

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2; and since this is an exact divisor of the first member of the equation, we have x2 = one of the roots.

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Dividing 7, 90, and 1449 by 1, 2, and 3, we have the quotients - 7, — 45, — 483, from which we are to find the remaining roots.

Putting 2 (the simplest divisor of 64, excepting 1, already used) for a, and taking 1, the simplest divisor of 45, for b,

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Because 2-1 exactly divides the first member of the equation, it follows that, by putting 2x-1=0, we shall for another root of the equation; consequently,

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dividing 7, 45, and 483 by 1, -1, and 3, we have -7, 45, and 161 for the quotients, which will enable us to find the remaining roots.

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Putting a = 4 and b=3, we have ab=1, − b = — 3, - ab 7 for the divisors of -7, 45, and 161; consequently, since the equation is divisible by 4x-3, we have x = for another root of the equation.

3

4

Dividing -7, 45, and 161 by 1, 3, and -7, we get the quotients7, 15, and 23; consequently, since all the roots but one have been found, we must have a—b—— 7,

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b = − 15, — a b 23, which give a 8, b = 15, and 8x15 must be a divisor of the equation, and thence x = 15 the remaining root.

8

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Remark. If we multiply the equation by 64 82, and

put Say, or x=, we have (8x) — 41(8x)3 + 574(8x)2 —

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8'

3144(8x) + 57600, or y-41y + 574y2 — 3144ỷ + 5760 = 0.

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Solving the equation in y (as in the 5th and 6th examples), we get 4, 6, 15, and 16 for the roots; and dividing each. of these by 8, we get the roots of the proposed equation, as before.

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9. To resolve x. 17- 20x60 into rational factors, and thence to find its roots.

Since it is clear that the equation has no rational factors of the first degree, we shall try to resolve it into rational factors of the second degree; and it is clear, since the coefficient of a is 1, that + ax + b is the proper form for a factor of the second degree.

If we put 1, 0, and 1 successively for a in the equation and in the factor, we shall get the results - 42, -6, and — 2, which have 1 + a + b, b, and 1-a+b for corresponding divisors, noticing that a + b, b, and a + b are in arithmetical progression, being formed by the successive additions of a to the first and second terms.

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After a few trials, we find that it will answer our purpose to put 1+ a + b = - 6, b 3, and 1 - a + b = 2, which a4; consequently, putting

give a =

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and b, we have x2 + ax + b = x2 - 4. 3.

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3 for a

Because the division of the equation by a 4x-3 gives the quotient + 4x + 2, it is reduced to (x2-4 −- 3) (x2 + 4x+2)=0; consequently, since the equation is satisfied by putting either of its factors equal to 0, we get, from x2 - 4x -30 and a + 4x + 2 = 0, 2 + √7, 2 − √7, − 2 + √2, and 2-1/2 for the roots of the proposed equation.

10. To resolve 6x4

5x3- 2x2-29x+10=0 (if possible)

into two rational quadratic factors.

It is clear that we may represent either of the quadratic

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