As all other examples of the same kind can be performed in a like way, we have the following rule:

Arrange the polynomials with reference to the same letter, and suppress all monomial factors in either polynomial. If any factor so suppressed is common to the two polynomials, set it aside as a factor of the greatest common divisor.

Multiply the polynomial containing the highest power of the leading letter thus prepared by such a factor as will make its first term exactly divisible by the first term of the other. Divide the first by the second, and continue the division until the greatest exponent of the leading letter, in the remainder, is at least one less than in the divisor. If the highest power of the leading letter in both polynomials is the same, either polynomial may be used as the divisor.

Take the divisor as a new dividend and this remainder as a divisor, and proceed as before; and continue until a remainder 0, or one independent of the leading letter, is found. In the first case, the last divisor, multiplied by common factors set aside, will be the greatest common divisor; in the second, the product of factors set aside, if any, will be the greatest common divisor. If there be no remainder 0, and no factors to set aside, the polynomials are prime with respect to each other, and have, of course, no common divisor.

(1) Find the G. C. D. of

3x-13x2+23x-21 and 12x+2x2-88x + 42.

Suppressing the factor 2 in the second polynomial, and introducing the factor 2 into the first, we have

Suppressing the factor-9 in the first remainder, and proceeding as before, we have

Suppressing the factor 4 in the second remainder, and proceeding as before, we have

Hence 3x7 is the G. C. D. sought.

(2) x 7x+8x2+28x-48 and -8x+19x - 14.

Ans. x 2.

(3) 4x+9x+2x2-2x-4 and 3x+5x2-x+2.
3x3 +5x2−x+2.

Ans. x+2.

(4) x+3x-8x 9x-3 and 25-2x-6x+4x2 Ans. (x+1)(x+1)(x + 1).

+13x+6.

(5) 625-4x-11 x3-3x2-3x-1 and 4x+2x3-18x2

+3x-5.

Ans. 2x4x2+x-1.

(6) 20x6-12x+16x-15x+14x2-15x+4

and 15x9x+47 x2-21x+28.

Ans. 5x-3x+4.

Ans. x+2x+3.

(7) 25 — x1 — 3 x3 — 2x2 + 14x + 21, x1+x2-2x+6.

To find the G. C. D. of three or more quantities,

Find the G. C. D. of the first and second, then the G. C. D. of this result and the third quantity, and so on to the last.

(9) 3x2-6x, 2x3 — 4x2, and x2y — 2xy.

(10) 3-9x2+26x-24, x3- 10x2 + 31x — 30,

and x11x2+38x-40.

Ans. x-2.

(11) x* — 10x2 +9, x2+10x3 +20 x2 — 10 x − 21, and x+4x-22x2-4x+21.

(12) x-a, x3+a3, and x2- a2.

(13) a+b3, a-b', and a + b3.

Ans. x2-1.

Ans. x+a.

Ans. a+b.

Ans. 3x-y.

(14) 3x3-7x2y +5xy2 -- y3, x2y + 3xy2 — 3 x3 — y3, and 3x+5y + xy3 — y3.

LEAST COMMON MULTIPLE.

72. A multiple of a number is any quantity that can be divided by that number without a remainder. Thus, 8a2b is a multiple of 8, also of a2 and of b.

73. A common multiple of two or more quantities is a quantity that can be divided by each separately without a remainder. Thus, 24a31⁄23 is a common multiple of 6ax and 4a2x.

74. The least common multiple of two or more quantities is the simplest quantity that can be divided by each without a remainder. Thus, 12 ab2x2 is the least common multiple of 2a'x, 4ab', and 6a2b2x2.

75. Since the common multiple is a dividend of each of the quantities, and since the division is exact, the common multiple must contain every prime factor in all the quantities; and, if the same factor enters more than once, it must enter an equal number of times into the common multiple.

When the given quantities can be factored by any of the methods already given, the least common multiple may be found by the following rule:

Resolve each of the quantities into its prime factors.

Take each factor as many times as it enters any one of the quantities, and form the continued product of these factors. It will be the least common multiple.

Exercises.

NOTE. For convenience, L. C. M. will be used to denote least common multiple.

1. Find the L. C. M. of 12a3b2c2 and 8a2b3.

Now, since 2 enters three times as a factor, it must enter three times in the common multiple; 3 must enter once; a, three times; b, three times; and c, twice: hence 2.2.2.3 aaabbbcc, or 24 a3b3c2, is the L. C. M.

Find the L. C. M. of

2. 6a, 5ab, and 25 abc2.

Ans. 150 a2bc2.

3. 3a2b, 9abc, and 27 a2x2.

Ans. 27 a2bcx2.

4. 4a2x2y2, 8a3xy, 16 a1y3, and 24a3y*x.

Ans. 48 a3x2y.

6. a+b, a2—b2, and a2+2ab+b2. Ans. (a+b)2(a — b).

7. 3 a3b2, 9a2x2, 18a1y3, 3a2y2.

Ans. 18 a'b'x'y3.

8. 8a (ab), 15 a5(a - b), and 12a (a2-b2).

Ans. 120a3(a - b)2(a+b).

75 a. When the given quantities cannot readily be factored, another method is used.

The L. C. M. of two polynomials contains all the factors of each polynomial; the G. C. D. contains all the factors common to the two polynomials. Hence, to find the L. C. M. of two polynomials, we have the following rule:

Find the G. C. D. of the two quantities. Divide one of the quantities by it, then multiply the other quantity by the

quotient.

Exercises.

Find the L. C. M. of the following:1. 2x2-xy-6y' and 3x2-8xy + 4y3. Their G. C. D. is x-2y: hence their L. C. M. is 2x2-xy-6 y2 × (3x2 — 8 xy +4y2) = (2x + 3y)(3x2 — 8 xy + 4y3).

X- 2y

2. 3x2-5x+2 and 4x3-4 x2 − x + 1.

Ans. (3 x − 2)(4 x3 − 4 x2 − x + 1).

3. 6x2-x-1 and 2x2+3x-2.

Ans. (3x+1)(2x2 + 3 x − 2).

To find the L. C. M. of several quantities,

Find the L. C. M. of the first and second, then of that result and the third, and so on to the last.

Find the L. C. M. of the following:

4. x-6x+11x-6, x3-9x+26x-24,

and x3 — 8x2 + 19x – 12.

Ans. (x-1)(x-2)(x − 3)(x — 4).

5. x10x+9, x+10x+20x2-10x-21,

and x+4x-22x-4x+21.

Ans. (x2-1)(x2 - 9)(x+7).