braic quantities, "first divide each of them by their greatest simple common measure (if they have one); arrange their terms according to the dimensions of the same letter, and divide either, or both of them, by the greatest simple factor which it may contain; then perform on them the same operation as that for finding the greatest common measure of two numbers, observing only, that the remainders which arise are to be divided by their greatest simple factors, and that the dividends may, if requisite, be multiplied by any simple quantity which will make the first term of the dividend a multiple of the first term of the divisor. Lastly, multiply the compound common measure thus obtained by the simple one originally taken out, and the product will be the greatest common measure required." CASE IV. To find the greatest common measure, or divisor, of the terms of a fraction, consisting of compound quantities, or to reduce a fraction to its lowest, or most simple terms. RULE I. Range the quantities according to the powers, or dimensions of some letter, as in Division, and by inspection expunge the common factor or factors, if any, either in the divisor or dividend; then divide that quantity which has the highest power by the other, whether it be the numerator or denominator; and divide the last divisor by the last remainder, and so on till nothing remains; the last divisor will be the greatest common measure: but if such a divisor cannot be found, the fraction has no common measure. Having found the greatest common measure, divide the terms of the fraction by it, and the resulting fraction will be in its lowest terms. NOTE. If any of the divisors become negative, they may have their signs changed without altering the truth of the result. Also, if the first term of a divisor be not contained an exact number of times in the first term of the dividend, the latter may be multiplied by any quantity that will make the division complete. And lastly, any quantity which is common to all the terms of the dividend or divisor, may be expunged before the division is commenced. 1. Find the greatest common measure of then reduce the fraction to its lowest terms. 6a2+7ax-3x2)6a2+11ax+3x2(1 6a2+7ax-3x2 6a2+7ax-3x2 6a2+11ax+3x2 and Now dividing this 4ax+6x remainder by 2x, the quotient is 2a+3x; and dividing the divisor 6a+7ax-32 by this quotient, we have 2a+3x)6a2+7ax-3x2(3a-x 6a2+9ax -2αx-3x2 -2ax-3x2, and since there is no remainder, therefore 2a+3x, the last divisor, is the greatest common measure 6a2+7ax-3x2 3a-x required. 2a+3x) 6a2+11ax+3x2 3a+x Ans. In this example the dimension of the numerator is the same as that of the denominator; therefore, it makes no difference whether we divide the numerator by the denominator, or the denominator by the numerator. This will be evident from the following ope ration: 6a+11ax+3x2)6a2+7ax-3x2(1 6a2+11ax+3x2 Now, dividing this -4ax-6x remainder by -2x, the quotient is 2a+3x; or, as the remainder above is now, negative. And, by changing all the signs, we obtained 4ax+6x2, or 2a+3x, for the new divisor. 2a+3x)6a2+11ax+3x2(3a+x 6a2+9ax 2ax+3x2 2ax+3x2 Hence 2a+3x is found to be the greatest common measure, as before. Required to find the greatest common measure of 7a2-23ab+662 2. -or or x2-xy-2y x2-3xy+2y ax+a 5a3—18ab11ab2—6b3 and then reduce each fraction to their lowest terms. xy-2y)x-3xy+2y2(1 and Divide this -2xy+4y2 remainder, each term of which can be divided by 2y, and the signs of the terms being changed, we obtain x-2y for a new divisor. x-2y)x2—xy—2y'(x+y is no remainder, therefore x-2y, the common measure required. x-2y) x2-2xy +xy-2y +xy-2y and since there last divisor, is the greatest x2-xy-2y2 x+y x2-2xy+3y2 x-y Answer. In this example, the dimensions of the numerator is the same as that of the denominator; therefore it makes no difference whether we divide the numerator by the denominator, or the denominator by the numerator. The same remark will apply to the next example. This will be evident from the following operation. x2-3xy+2y2)x2-xy-2y (1 x2-3xy+2y 2xy-4y the remainder, which is now positive, and divide by +2y, and we obtain x-2y, for our new divisor, the same as before. Here the greatest common measure is 2-8x-3; therefore, 2x3-16x-6 2 x3-8x-3) Ans. as required. x+1 Or thus, 3x3-24x-9 axta, or ( x2-1(x-1, the least common measure. 2(x3-8x-3) 2 3(x3-8x-3) 3 - as before. } x2 + x -X -1 ax+a a 3d expression solved. Multiply the denominator by 7, we shall have 7a-23ab+662)35a-126ab77ab2-426 (5a-11b 35a-115a b+30ab2 -11ab47ab2 -42b3 Multiply by 7 -77ab329ab2—29463 -77a2b-253ab2-66b3 Divide this remainder by 7662 and 76ab2-22863 we have for the divisor a 36)7a-23ab-662(7a2b 5a3-18a2b+11ab2—6b3 5a2-3ab+262 Now, for the last fraction in example second, the greatest common measure of the fraction is x-a) x2+ax+a2 -Ans. lowest terms. x+a3ñ3—a3x—a1)x2+a2x2+a1(1 This remainder, —ax3+a2x2+a3x+2a1 which, being divided by a, and the signs of the terms being changed, becomes xaxa2x2a1) x1ax3-a3x-a1(x+2a x-ax3-a2x2-2a3x 2ax3 + a2x2+a3x-a1 Divide this remainder, 3ax+3ax+3a by 3a, we obtain x2+ax+a for our new divisor. x2+ax+a2) x2-ax2-a2x-2a3 (x-2a ́x3—ax2+a2x -2ax2-2a2x-2a3 Divide this remaind. a*x+ax2-ax3-x by x a3+a2x-ax2-x3 (a3—a2x—ax2+x3(1 a3—a2x-ax2x3 Divide this remainder -2a2x+2x3 which is now negative, by —2x. -2a2x+2x3 |a3+a2x—ax2—x3 | (a -2x a3ax2 Divide this remainder a'x-ax3 by x, and we obtain a2-x2 for our new divisor. a2x2)a2+x2(—1 -a2+x2` a3—a2x-ax2+x03 a-x 4. Reduce the three following fractions, which are from Simpson's Algebra,' to their lowest terms, viz: 5a+10ab5a32 x3+ax2+bx2-2a2x+bax-2ba3 or a3b+2a2b2+2ab3+b13 x2-3ax-8a2x2+18a3x-8a* x3-ax2-8a2x+6a and x2-bx+2ax-2ab Here, dividing first by the greatest simple divisors, 5a and b, we have a2+2ab+b2, and a3+2ab+2ab2+b3: and if the latter of these be divided by the former, the work will stand thus : a2+2ab+b2)a3+2a2b+2ab2+b3(a where the remainder is +ab+63; which being divided by 62, its greatest simple divisor, gives a+b; by this divide a+2ab+b2, and the quotient will come out a+b, exactly; therefore, the last divisor, a+b, will exactly measure both quantities. .. the fraction given will be reduced to 5a*+5a3b a2b+ab2+b3 •. Again, a3—a3x—ax2+a3)a2 + 0 + 0 + 0 —x2(a+x —a2x+0+x3. From whence it appears that a2+0 or a2-r2 will measure both a*-2*, and a3—a2x-ax2+x; and, by a2 + x2. dividing thereby, the fraction proposed is reduced to a-x x3-ax2-8a2x+6a3)x1—3ax3-8a2x2+18a3x-8a1(x-2a -2ax2+0+12a3x- 8a1 —2ax3+2a2x2+16a3x—12a1 remainder —2a2x2-4a3x-+-4a1; which, divided by -2a2, gives x2+2ax-2a2 for the next divisor. x2+2ax2-2a2x 3ax-6a2x+6a -3ax2-6a2x+6a3 x2+2ax-2a2)x-3ax3-Sa2x2+18a3x—Sa1(x2—5ax+4a2 x2+2ax3-2a2x2 -5ax-6a2x2+18ax -5ax3-10a2x2+10a3x +4a2x2+8a3x-8a* Now if, by proceeding in this manner, no compound divisor can be found, that is, if the last remainder be only a simple quantity, we may conclude the case proposed does not admit of any, but is already in its lowest terms. Thus, for instance, if the fraction a3+2a2x+3ax2+4x3 proposed were to be it is plain by inspection, a2+ax+x2 that it is not reducible by any simple divisor; but to know whether it may not, by a compound one, I proceed as above, and find the last remainder to be the simple quantity 72; whence I conclude that the fraction is already in its lowest terms. Another observation may be here made, in relation to fractions that have in them more than two different letters. When one of the letters rises only to a single dimension, either in the numerator or in the denominator, it will be best to divide the said numerator or denominator (whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other; this being done, let the greatest common divisor of these two parts be found; which will evidently be a divisor to the whole, and by which the division of the other quantity is to be tried; as in the following example, where the x3+ax2+bx2-2a2x+bax-2ba2 Here the deno fraction given is xbx+2ax-2ab minator being the least compounded, and b rising therein to a sin |