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47. A factor which does not contain any factor common to both A and B may be rejected at any stage of the process. Let the operation stand thus:

B = mB' suppose,
B)A(P

C')B(2

907
С nC"
suppose, DC"(

rD

0 where neither m nor n contains any factor common to A and B.

It will be an exercise for the student to show that D is the G.C.M. of A and B.

48. A factor, which has no factor that the divisor has, may be introduced into the dividend at any stage of the process. The operation may stand thus B)mA(p, where m has no factor that B has;


C)nB(q; where n has no factor that C has ;

9C
DC

r]

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As in Arts: 46, 47, it may be easily shown that D is the G.C.M.

Both the above principles are made use of in working out Ex. 3 Art. 45.

49. When a common factor can be found by inspection, it is advisable to strike it out of the given expressions. Then, having found by the ordinary process the G.C.M. of the resulting quantities, we must multiply the G.C.M. so found by the rejected factor.

Thus, 4 x is common to the quantities 4 x3 20 oca + 24 cm and 4 203 + 16 cm

Rejecting it, we get 32 - 5x + 6, and 22 + 4x - 21, whose G.C.M. is easily found to be a · 2.

Multiplying by 4x; we find the required G.C.M. to be

*

84 20.

7.C

50. By a little ingenuity on the part of the student in breaking up the given expressions into factors, the ordinary and often tedious process of finding the G.C.M. may be avoided. The limits of our space will allow us only one example.

Ex. Find the G.C.M. of 3 c3 + 4 22 10x + 3, and 15 x3 + 47 + 13 x 12.

The first expression contains a - 1 as a factor (Art. 30), for the sum of its coefficients is zero. The other factor may be obtained thus3 203 + 4x2 – 10 x + 3 = 3 x3 3 x2 + 7 x2

3 x + 3 = 3 x* (C – 1) + 7 x (C-1) - 3(x - 1)

(3 * + 72 – 3) (x - 1). Now, 3 ac + 7 oC 3 is not further resolvable, and x 1 is evidently (Art. 30) not a factor of 15 23 + 34 2+ 13 x 12. It is, therefore, very probable that 3 x + 7 x – 3 is the G.C.M. required.

We may test it thus15x + 47 x + 13x – 12 = 1523 + 35x - 15x + 12x + 28x - 12

= 5 x (3 xco + 7 x - 3) + 4 (3x2 + 7 x - 3)

= (5x + 4) (3.2c2 + 7 x - 3). Hence, 3.202 + 7 x – 3 is the G.C.M. required.

+

G.C.M. of Three or More Quantities. 51. The G.C.M. of three or more quantities may be found thus—

RULE.—Find the G.C.M. of any two of the quantities, then the G.C.M. of the G.C.M. so found and a third quantity, and

The last found G.C.M. will be the G.C.M. required.

So on.

Ex. XII. Find the G.C.M. of the following1. ca 52 + 6 and " + 3 x

18.
2. 23 + 6 2 + 11x + 6 and 3 + 5 x + 7 x + 3.
3. 2 202 + 10 x 18 x - 90 and 3 23 + 16 22 - 26 x
4. 2c + (a + b x + ab and ac2 + (a + c) x + ac.
5. a3 63 and as t a b + ab?.

141.

6. 2013 4 x + 3 and 203 + 4 22 5.
7. 4 203 32 2 + 85 x 75 and 3 203 15 m2 + 15 x + 9.
8. 9 x2 – 3 xy + 2 y - 4 and 6 * - 4 x3 – 9 ay® + 6 .

9. 48 204 + 8 203 + 31 32 + 15 3 and 24 c4 + 22 x3 + 17 + 5 x.

10. 15 a3 + a b 3 ab2 + 2 73 and 54 a?l2 - 24 64.

11. 32 – (3 c + d + 1)2 – (2 a + b - 3c - d + 2) # + 2 a + b and 2 ? (a + b + 2) x + a + b. 12. 6 25

424 11 22 3 x2 3 x 1 and 4 24 + 2 23 18 x + 3 x 5. 13. ab + 2 ao 3 62

4 bc

ca and 9 ac + 2 a 5 ab + 4 c + 8 bc 12 62.

14. em oca t ex + 2a + 1 and e2x24 - 222 + 2C4 1.

15. ax* + (b + c) aca ax - b - cand ex (f - g) * + (f - e) * - g.

16. 4 x4 + 2 203 + 4 x + 39 x 9, 8 4 + 20 m2 + 51 x + 9, and 2 204 + acé + 3 ca + 18 X. 17. a.cz (c 1) x2 + (c + 1)

(6 + d) 2c + (c + d) ac2 (c + e) x + e, and (c + 1) 200 + d + 2) ** (d + 1) 203 - (c + 2) .?.

18. a3 73 + c3 + 3 abc and a 62 + ca + 2 ac.

ас

+

a, bx+

Least Common Multiple. 52. When two or more algebraical expressions are arranged according to the powers of some letter, the expression of lowest dimensions which is divisible by each of the given expressions is called the L.C.M.

53. The L.C.M. of monomials and of expressions whose factors are apparent may be found by inspection.

Ex. 1. Find the L.C.M. of ab, ac, ad, bc, bd, cd.

If we form an expression, whose elementary factors contain each of the elementary factors of the given quantities, we shall evidently have a common multiple ; and if no ele, mentary factor of this expression is of a higher power than the highest power of the same factor in the given quantities, we shall get the L.C.M.

Hence, the required L.C.M. = abcd.

a

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Ex. 2. Find the L.C.M. of— (a 6) (6 - c), (a - b) (c – a), (6 - c) (c – a). Ans. (a - b) (6 - c) (c - a).

Ex. 3. Find the L.C.M. of a (oc + 1), b (aca 1), c (c2 + 2 x – 3), d ( + 4x – 3). We may write the given expressions thus

a (x + 1), 6 ( + 1) (x - 1),
c (36 - 1) (2 + 3), d (2 + 1) (x + 3).
Hence, the required L.C.M. = abcd (:

abcd (- 1) (+ 1) (x + 3).

Ex. 4. Find the L.C.M. of as ax + 2*, a2 + ax + ac", ad + *, a- ab Now (Art. 29) a3 + 208 (a + 2) (a? ax + 2?), and as ಟಿ

a) (as + ax + aco). Hence the required L.C.M.(a + 2) (a 2) (a2 + ax + x^) (a? ax + 20°) 206.

54. The L.C.M. of two quantities is found by dividing their product by their G.C.M.

Let a and b be the two quantities, and d the G.C.M.;
And
suppose a

qd. It is evident that p and q contain no common factor. Hence

PI

is the L.C.M. of p and q; and, therefore, no expression of lower dimensions than pod can possibly be divisible by pd and qd.

Hence pgd is the L.C.M. of pd and qd, or of a and 6. Now pqd = pd x qd :d = a × b = d, and hence the rule: 55. To find the L.C.M. of three or more quantities. · RULE.—Find the L.C.M. of two of the quantities, then the L.C.M. of the expression thus obtained and a third quantity, and so on. The last expression

The last expression so found is the Î.C.M. required.

We shall prove this rule in the case of three quantities.

Let a, b, c be the quantities, and m be the L.C.M. of a and b.

Then the L.C.M. of m and c is the L.C.M. required.

For every common multiple of m and c is a common multiple of a, b, c. And every common multiple of a and b

pd and 6

must contain the m, their least common multiple. Hence, every common multiple of a, b, c must be a common multiple of m and c, and the converse is also true. Hence, the L.C.M. of m and c is the L.C.M. of a, b, c.

Ex. XIII. Find the L.C.M. of 1. axy, 3 aʻwoy, 4 aoy3, 6 x*y. 2. 5 a*b*, 6 a'c', 46°c. 3. (a - b) (6 c), (6 - a) (a – c), (c = a) (c - b). 4. ax (2c + a), a'2 - a), c- a'. 5. 2c + 3x + 2, 2c2 + 4x + 3, ac + 5 x + 6. 6. ac?

22 11ac + 30, aca 25. 7. 6 2 + 372 + 56, 8x2 + 38 x + 35, 12 c2 + 47 x

30,

+ 40.

+ 28.

8. 5 (22 - 2 + 1), 6 (202 + 1), 7 (208 + 1). 9. 20* + amaca + a', aʻaca + a'rc + , axa – axc + a'. 10. a + (a + b) x + ab, xa + (a + c) x + ac, ac? + (b + c) x + bc. 11. 1

X,

1 x, 1 + 2*, 1 + 2c", 1 12. 20% + 6 uca + 11x + 6, 23 6 m2 25 x + 150. 13. a3 – 3 ab (a - b) - 63, a3 - 63, a + a2b + ab. 14. 204 1, 6206 + 5 24 + 8 23 + 4x2 + 2x 1.

15. a 2 a?l2 + 64, at + 4 ao6 + 6 a+ 4 ab3 + 64, a4 4 ab + 6 ao82 4 ab3 + 64. 16. 3 203

4x + 1,

2 203 7 x t 5, 4 x4 + 6 m2

+

10 %.

48, 5**

20, 32

17. 3 c2 + 6 x – 24, 20% – 12 + 16, 5 a* - 22 x - 36. 18. a- ab?, 13 - apb, ab? 69, a b - a'. 19. 3 2c*

16 x + 20. 20. 2018 yo, c - 2 ° + y, z + 2y + xy + y.

21. Qco + ax* + awc + a’sca + atac + a' and 2015 AX* + aacs awca + atx a'i

22. aa + 62 c? da + 2 ab 2 cd and am - 62 - c + d + 2 ad 2 bc.

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