it may at first be observed, that if they admit of a common divisor, which should be independent of the letter a, it must divide separately each of the quantities by which the different powers of a are multiplied, (Art. 143), as well as the quantities b4c2b2c4 and 63-b2c, which comprehend not at all this letter. The question is therefore reduced to finding the common divisors of the quantities baca and b-c, and, to verify af terward, if, among these divisors, there be found some that would also divide b3-bc2 and ba-bc, b4c2-b2c4 and 63b2c. Dividing b2c2 by b-c, we find an exact quotient b+c: b-c is therefore a common divisor of the quantities b2-c2 and b―c, and it appears that they cannot have any other divisor, because the quantity b-c is divisible but by itself and unity. We must therefore try if it would divide the other quantities referred to above, or, which is equally as well, if it would divide the two proposed quantities; but it will be found to succeed, the quotients coming out exactly, (b+c)a2+(b2+bc)a3+h3c2+b2c3 ; and In order to bring these last expressions to the greatest possible degree of simplicity, it is expedient to try if the first be not divisible by b+c; this division being effected, it succeeds, and we have now only to seek the greatest common divisor of these very simple quantities; a1+ba3+b2 c2, and a2+ba+b2. Operating on these, according to the Rule, (Art. 141), we will arrive, after the second division, at a remainder containing the letter a in the first power only; and as this remainder is not the common divisor, hence we may conclude that the letter a does not make a part of the common divisor sought, which is consequently composed but of the factor b-c. Ex. 7. Required the greatest common divisor of (d2 —c2) Xa+c-daca and 4da3-(2c+4cd)a+2c3. Arranging these quantities according to d, we have it is evident, by inspection only, that ac2 is a divisor of the first, and a-c of the second. But a-c2 is divisible by ac; therefore a-c is a divisor of the two proposed quantities: Dividing both the one and the other by a-c, the quotients will be (a+c)×(d2-c2), and 4ad-2c2; which, by inspection, are found to have no common divisor, consequently a-c is the greatest common divisor of the proposed quantities. Ex. 8. Required the greatest common divisor of y1-x1 and y3—y2x—yx2+x3. Ans. y2-x2. Ex. 9. Required the greatest common divisor of 2a—ba and ab®. Ans. a-b2. Ex. 10. Required the greatest common divisor of a1+a3b— ab3-b4 and aa+a2b2+ba. Ans. a2 +ab+b2. Ex. 11. Required the greatest common divisor of a2 — 2ax +x2 and a3a2x — ax2 +x3. Ans. a2-2ax+x2. Ex. 12. Find the greatest common divisor of 6x3—8yx2+ 2y2x and 12r2-15yx+3y2. Ans. x-y. Ex. 13. Find the greatest common divisor of 36b2a6-1862 a5-27b2a962a3 and 2762a5-18b2 a* - 9b2a3. Ans. 9b2a4-9b2a3. Ex. 14. Find the greatest common divisor of (c-d)a2+ (2bc2bd)a+(b2 c—b3d) and (bc-bd+c2—cd)a+(b2d+bc2 -b2c-bcd). Ans. c-d. Ex. 15. Find the greatest common divisor of qnp3+3np2 q2 -2npq3 — 2nq1 and 2mp2 q2 - 4mp*—mp3q+3mpq3. Ans. q-p. 2 Ex. 16. Find the greatest common divisor of x3+9x2+ 27x-98 and x2+12x-28. Aps. x- -2. § III. METHOD OF FINDING THE LEAST COMMON MULTIPLE OF TWO OR MORE QUANTITIES. 145. The least common multiple of two or more quantities is the least quantity in which each of them is contained without a remainder. Thus, 20abc is the least common multiple of 5a, 4ac, and 2b. 146. The least common multiple of any number of quantities, literal or numeral, monomial or polynomial, may be easily found thus : Resolve each quantity into its simplest factors, putting the product of equal factors when there are any in the form of powers, then multiply all together the highest powers of every root concerned, and th eproduct will be the least common multiple required. Ex. 1. Required the least common multiple of a3b3x, acbx2, abc2d. Here the quantities are already exhibited in the form required. Therefore the least common multiple is a3b3c2 dx2. Ex. 2. Required the least common multiple of 2a3 x, 4ax 2, and 6x3. Here the literal quantities are already in the form required. The coefficients resolved into their simplest factors become 2, 22, 2X3. The least common multiple is therefore Ex. 3. Required the least common multiple of 12a2y (a+b), 6a3y2+12a3by3+6ab2 y3, and 4a2 y2. These quantities resolved into their simplest factors become 22 X3 Xay(a+b) 2 ×3×ay3 (a+b)2 Hence the least common multiple required is 2a ×3×a2y2 (a+b)2=12a3y2(a+b)2. Ex 3. Required the least common multiple of 8a, 4a2, and 12ab. Ans. 24a2 b. Ex. 4. Required the least common multiple of a2-b2, a+ b, and a2+b2. Ans. a4-b4. Ex. 5. Required the least common multiple of 27a, 15b, 9ab, and 3a2. Ans. 135a2b. Ex. 6. Required the least common multiple of a3+3a2b+ 3ab2+b3, a2+2ab+b2, a2-b2: Ans. a+2a3b — 2ab3—ba ̧ Ex. 7. Required the least common multiple of a+b, a—b, a2+ab+b2, and a2 —ab+b2. Ans. a-bo. § IV. REDUCTION OF ALGEBRAIC FRACTIONS. CASE I. To reduce a mixed quantity to an improper fraction. RULE. 147. Multiply the integral part by the denominator of the fraction, and to the product annex the numerator with its proper sign under this sum place the former denominator, and the result is the improper fraction required. 26 5a Ex. 1. Reduce 3x + to an improper fraction. The integral part 3x, multiplied by the denominator 5a of the fraction plus the numerator (2b), is equal to 3x × 5a+2b =15ax+26: Here 5a Xy5dy; to this add the numerator with its proper sign, viz. -3x; and we shall have 5ay-3x. 5ay-3x Hence, is the fraction required. Here, x2 Xx=x3; adding the numerator a2-y3 with its proper sign: It is to be recollected that the sign affixed to the fraction subtracted, and consequently that the sign of each term of the numerator must be changed, when it is combined with x3, x3 —a2 + y2 means that the whole of that fraction is to be hence the improper fraction required is x Or, as ; (Art. 67), the proposed , may be put under the form x2+ which is reduced as Ex. 1. Thus, x2xx+y—a2— Here, 5a2×2ax=10a3x; adding the numerator 3x2-a+7 to this, and we have 10a3x+3x2-u+7. 10a3x + 3x2-u+7 Hence, is the fraction required. 2αx 3ab+c Ex. 5. Reduce 4x2 to an improper fraction. 2ac Here, 4x2X2ac=8acx2, in adding the numerator with its 3ab+c proper sign; the sign - prefixed to the fraction -signi fies that it is to be taken negatively, or that the whole of that fraction is to be subtracted; and consequently that the sign of each term of the numerator must be changed when it is 8acx2-3ab-c 108); hence the reason of changing the signs of the numera tor is evident. To reduce an improper fraction to a whole or mixed quantity. RULE. 148. Observe which terms of the numerator are divisible by the denominator without a remainder, the quotient will give the integral part; and put the remaining terms of the numerator, if any, over the denominator for the fractional part; then the two joined together with the proper sign between them, will give the mixed quantity required. |