since P and Q have no common factor, it must therefore contain all the factors of MPQ together, and MPQ is the least quantity that contains all these. Hence MPQ is the L. C. M. of X and Y. Note. In the fractional expression for the L. C. M. the denominator should always be cancelled against some of the factors in the numerator, before the multiplication is effected. Also, great care should be taken to work out all the examples in a form similar to the two just given. (2.) Find the L. C. M. of a3-3 and 4a4+4a2x2+4x1. Removing the factor 4 from the latter, we proceed to find the G. C. M. (a-x) being obtained by cancelling (a2+ax+x2) against a3-x3. By multiplication the result is known to be L. C. M. = · 4a5-4a1x+4a3x2-4a2x2+4ax1-4x5. All the examples should be worked out in this form. For examples of L. C. M. the reader is referred to those in the G. C. M., each of which is well adapted for an exercise in finding the least common multiple. CHAPTER VIII. ABRIDGED MULTIPLICATION AND DIVISION. 50. By actual multiplication we find that (a+b)x(a−b) = a2 — b2. Here (a+b) represents the sum of the two quantities a and b. And (a2-b2) represents the difference of the squares of a and b. Hence the above formula, put into words, gives the following Rule— The product of the sum and the difference of any two quantities equals the difference of their squares. (1.) Write down the products of the following, without the work of multiplication:— 4a+1 and 4a-1; 6a2-1 and 6a2+1 7ab+4xy and 7ab-4xy; x3+y2 and 23-y2 D a3+b3 and a3-b3; -a4+b1 and a1+b4 - a2-b2 and -a2+b2; 3a2x2-4b2y2 and 3a2x2+4b2y2 5x2y+2xy2 and 2xy2 — 5x2y. (2.) Find the product of a+b+c and a+b-c. Working. Regarding a+b in the first instance as a single quantity, and placing a vinculum over it, we have (a+b+c) (a+b-c) = (a+b+c) (a+b − c). Again, the product a+b+c+d and a+b-c-d may be thus found: (a+b+c+d) (a+b−c−d) = (a+b+c+d) (a+b-c+d) =(a+b)2 - (c+d)2= a2+2ab+b2 — c2 — 2cd — d2. Note. Any two factors which contain the same letters and the same coefficients may be arranged as a sum and difference, and their product found by the above Rule. Thus the two factors 3a-4b-4c+5d and 3a+4b-4c-5d are equal, to (3a-4c-4b−5d) and (3a-4c+4b-5d) respectively. Also of a+b+c+d and a-b-c-d b+c-d-a and a-b+c-d 1−x+x2-x3+x4 and 1+x+x2+x3+x+ Find the continued product of x+y+z, x+y-z, and x2+y2-z2+2xy 51. By actual multiplication of the factors (x+a) and (x+b) as below, we find their product to be— Whence it appears that the coefficient of x in the product is the sum of the last terms of each factor, and the last term of the product is the product of the two last terms of each factor. Thus (x+7)(x+9) = x2+(7+9)x+7x9 = x2+16x+63 Again, (x+2) (x−5) = x2+ (2−5) x+2× (−5) = x2-3x-10 (x-4) (x−7) = x2+ (−4−7)x+(-4) × (-7) Similarly, may the products of the following be found : 22-4ab and y2-7ab, x3-10y3 and x3+2y3. Also, the continued products of the following factors:x+4y, x-4y, and x2+y2 For, And, (x+4y) (x−4y) = x2-16y2 by Rule (p. 49) (x2-16y2) (x2+y2) = x1+ (1−16) x2y2 – 16y1 = x1 — 15x2y2 — 16y1. = a+3b, a- -3b, and a2-3b2 7a-b, 7a2+b2, and 7a+b x2-2y2, x1—4y1 and 2y2+x2 (x+y) and (x− y)2, (a+b)3 and (a - b)3 52. It has been shown (p. 29) that am-m is always divisible by a-b, and the form of the quotient was there given. Hence the quotient of any quantity of the form am—bm÷a−b may at once be written down without working out the division. Thus, aa—b4÷a−b=a3+a2b+ab2+b3 Also, since a1 — b1 = (a−b) (a3+a2b+ab2+b3), it is always possible to write down the factors of such a quantity. Examples XIX. Resolve into their factors the following quantities: a3-1, x3-1, 4x2-9y2, 32x3-243, x7—y11, 8a3x3 — 27b®, It was also shown in the same Article that am-bm is divisible by a+b, when m is even, am+bm is divisible by a+b, when m is odd. .*. am—bm is divisible by both (a−b) and (a+b) in the former Resolve into all their factors the following quantities:- 25-a3, 64a6 — xo, a3x3 —b3y3, 256x1-16y4, 729a*b*c®+x3y3z3, 53. To resolve into factors a trinomial, as x+5 +4 In Article 51 it was shown that x2+(a+b)x+ab is the product of two factors (x+a), (x+b). Now, since x2+5x+4 is of the some form as x2+(a+b) x+ab, let us assume the two factors of it to be (x+a) and (x+b). Then we must have— a+b=5 And we must now find by inspection such values of a and b as shall fulfil these two conditions; we thus find |