Greatest Common Divisor. 2, give the first lines of the following form of the process, which is similar to that in art. 62. 14 x3-5 a x2-39 a2x-14 a x2 + 5 a2 x +39 a3 x3 54 a x2+39 a2x-63 a3-14 x2 +5 a x +39 a2 a2 9 7x Column 3, in this form, is the line of quotients. The 1st line of col. 1 is first divided by that of col. 2, and the remainder is the 3d line of col. 2; this remainder, simplified by the suppression of the factor a, is the 4th line of col. 2, and is used to divide the 1st line of col. 1. The 2d line of col. 1 is the 1st line multiplied by 7 in order to render its first term divisible by the first term of the new divisor; the remainder of the division is the 4th line of col. 1, which is simplified in the 5th line by the suppression of the factor 3 a. The 6th line of col. 1 is the 5th line, multiplied by 7 in order to render its first term divisible by the first term of the divisor already used; for it is to be observed, that a divisor should continue to be used until a remainder is obtained in which the first term ceases to be divisible by the first term of the divisor, that is, until the exponent of its leading letter is smaller than that in the first term of the divisor. The remainder arising from the division of the 5th line of col. 1 by the 4th line of col. 2 is the 8th line of col. 1, which, reduced by the suppression of the factor 68 a is the last line of col. 1. The remainder of the division of the 4th line of col. 2 and the last line of col. 1 is the 6th line of col. 2, which reduced by the suppression of the factor 13 x, is the Greatest Common Divisor. 7th line of col. 2, being the same with the last line of col. 1. The remainder of the last division is therefore zero, and the last divisor 2x + 3 a is the greatest common divisor. 2. Find the greatest common divisor of 21 a3 b x7 21 a4 b x6 - 168 a7 b x3 and 14 a2 b3 c x4 14 a3 b3 c x3 + 28 a4 b3 c x2. -42 a5 b3 c x 140 a6 b3 c. Solution. Since 7 a2 b is a monomial factor of the two given quantities, suppress it, and they become 3 a x7 3 a2 x6 24 a5 x3. 2 b2 c x4-2 a b2 c x3+4 a2 b2 c x2-6 a3 b2 c x-20 a1 b2 c. The greatest common divisor of these two quantities, found as in the preceding example, is x-2 a, which, multiplied by the common monomial factor 7a2b, gives 7 a2 b (x-2 a) for the required greatest common divisor. - a3 and Ans. x-a. 3. Find the greatest common divisor of x3 x2-a2. V 4. Find the greatest common divisor of 15 63 and 3 a3 + 6 a2 b + 6 a b2 +363. 5 a3 — 10 a2 b+ Ans. a+b. 5. Find the greatest common divisor of 24+x3+x2+x —4 and x2+2x3 + 3 x2 + 4 x— 10. Ans. X- 1. 6. Find the greatest common divisor of 7 a x5+21 a xa† 14 ax and 3 r6+3x5+ 3 x4 −3x2. Ans. x2+x. 7. Find the greatest common divisor of 81 a4 x4. 24 a7 x and 3 a x7-2 a2 x6 +3 a3 x3—2 a4 x4. Ans. 3 a x2-2 a2x. 8. Find the greatest common divisor of 23+x— 10 and x4-16. Ans. x-2. 65. When there are several terms in the given polynomials, which contain the same power of the letter according to which the terms are arranged, these terms are to be united in one, as in art. 53, and the compound terms thus formed are to be treated as monomials. and +32/34 +12 x2 x2 y | y Solution. The factor (x2+3x) y is a common factor of all the terms, and is therefore to be suppressed, in order to be multiplied by the greatest common divisor of the remaining polynomials. The polynomials thus become The suppression of the factor (x-3) y in the first of these polynomials reduces it to by which the second is to be divided, and the rest of the process is as follows: y3-3x|y2 · 4 x y + 4x|y3. - 2 x | y2 —3 x y +2x | 1 y3-2xy2—3 xy+2x|y3+ +2 +2 y2 2 y -y +2 y+2 1 Ans. (y + 2) (x2 + 3 x) y = (x2+3x) (y2+2y). Greatest Common Divisor. The third line of col. 1 is the remainder of the division of the 1st line of col. 1 by the 1st line of col. 2; and this remainder, reduced by the suppression of the factor x is the 4th line of col. 1. The 5th line of col. 2 is the remainder of the division of the 1st line of col. 2 by the 4th line of col. 1, and this remainder, reduced by the suppression of the factors x + 1 is the last line of col. 2. The 4th line of col. 1 is exactly divisible by the last line of col. 2, and therefore the greatest common divisor is the product of (x2+3x) y by y + 2. Ans. (x2+3x) (y2+2y). 2. Find the greatest common divisor of the polynomials a2+b2 + c2+2ab+2ac+2b c and a2- b2 — c2 — 2b c. Ans. a+b+c. 3. Find the greatest common divisor of the polynomials a1— 2 b2 | a2 + b4 and a3 +36a2 + 3b2 | a + b3 -2 c2 - 2 b2 c2 + c4 c2bc2. Ans. a2+2ab+ b2 — c2. 4. Find the greatest common divisor of the polynomials 4 xy5—3 x y1— x2 | y3 + x3 | y2 — x3 | y +3 and -1 x2 | y1-3x2 | y3 + x3 + 2 x2 x 2 Ans. y (y-1)(x — 1). 67. Problem. To reduce fractions to a common denominator. Solution. Multiply both terms of each fraction by the product of all the other denominators. Common Denominator. For the value of each fraction is, from art. 55, not changed by this process; and as each of the denominators thus obtained is the product of all the denominators, the fractions are all reduced to the same denominator. 68. But fractions can be reduced to a common denominator which is smaller than their continued product, whenever their denominators have a common multiple less than this product. For, by art. 55, Fractions may be reduced to a common denominator, which is a common multiple of their denominators, by multiplying both their terms by the quotients, respectively obtained from the division of the common denominator by their denominators. 69. Corollary. An entire quantity may, by the preceding article, be reduced to an equivalent fractional expression having any required denominator, by regarding it as a fraction, the denominator of which is unity. |