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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
B)AP

C')B (2

mB' suppose,

pB

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C =

nc"
suppose,

DC"

rD

0 where neither m nor r 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(2, where m has no factor that B has;

C)nBly, where n has no factor that C has;

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DCT

rD

0 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 w is common to the quantities 4 203 20 202 + 24 ac, and 4 203 + 16 oca

Rejecting it, we get 22 5 8C + 6, and a2 + 4 00 21, whose G.C.M. is easily found to be a 2.

Multiplying by 4ă; we find the required G.C.M. to be 40 - 8x

1

84 2.

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 x3 + 4 zca 10 x + 3, and 15 x + 47 cm + 13 x

12. The first expression contains x - 1 as a factor (Art. 30),

. for the sum of its coefficients is zero. The other factor may be obtained thus3 203 + 4x2 - 10x + 3 3 23 3 x2 + 7 C2

3 x + 3 = 30° (C – 1) + 7 (x - 1)-3(x - 1)

x 3 x (3 * + 7 x - 3)(x - 1).

1 Now, 3 x2 + 7 2 3 is not further resolvable, and x 1 is evidently (Art. 30) not a factor of 15 x3 + 34 xo + 13 x 12. It is, therefore, very probable that 3 x2 + 7 x – 3 is the G.C. M. required.

We may test it thus15x + 47 x2 + 13x – 12 = 15x + 35x? - 15x + 12x + 28x - 12

= 5 x (3 x + 7 x - 3) + 4(3.2? + 7 x - 3)

= (5x + 4) (32? + 7 x - 3). Hence, 3 a* + 7 2 – 3 is the G.C.M. required.

3 x

coC

a

=

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. x2 5x + 6 and a2 + 3 x 18. 2. 23 + 6 cm + 11x + 6 and acl + 5 x + 7 x + 3. 3. 2 2003 + 10 x*- 18 X – 90 and 3 23 + 16 x* - 26 x 4. dc + (a + b c + ab and x2 + (a + c) x + ac. 5. a3 63 and as + a2b + ab?.

141.

a

6. 203 - 4x + 3 and 203 + 4 22 5.
7. 4 203 32 2 + 85 x 75 and 3 203 15 + 15% + 9.
8. 9 .x2 – 3 xy + 2 y - 4 and 6 x* – 4 23 – 9 xy + 6 yo.
922

2 4 6 9. 48 x4 + 8 203 + 31 32 + 15 x and 24 c4 + 22 2x3 + 17 2c2 + 5 x.

10. 15 a3 + a b 3 ab2 + 2 53 and 54 a 62 - 24 64. 11. 303 - (3 c + d + 1)2 – (2 a + b - 3c - d + 2) 3 x2

* + 2 a + b and 2x2 - (a + b + 2) x + a + b. 2

+ . 12. 6 205 4 xt 11 23 3 22 3 @ 1 and 4x4 + 2 x3 18 2 + 3 - 5. 13. ab + 2 a? 3 62 4 bc

C and 9 ac + 2 az. 5 ab + 4c2 + 8 bc 12 62

14. etc + et + oct + 1 and eax_4 eat + oct 1. 15. ax+ (b + c) x2 6 cand ex: – (f - g) x2 +

e) a g. 16. 4 x4 + 2 23 + 4x + 39 x 9, 8 x4 + 20 x2 + 51 x + 9, and 2204 + 2x3 + 3 c2 + 18 x.

17. amci (c + 1) x2 + (c + 1) : a, back (b + d) ac? (c + d) ac (c + e) s + e, and (c + 1) 200 + d + 2) ** (d + 1) 30 - (c + 2) x2.

?. 18. as 73 + c3 + 3 abc and a? - 72 + c + 2 ac.

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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.

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Ans. (a

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

c 6) (6 - c) (c - a). Ex. 3. Find the L.C.M. of a (x + 1), 6 (aca 1), c (x2 + 2x - 3), (* + 4 x - 3). We may write the given

d ac expressions thus

a (x + 1), 6(x + 1) (0 - 1),
c (oc – 1) (0 + 3), d (2 + 1) (0 + 3).
Hence, the required L.C.M. = abcd (

OC

1) (0 + 1) (3C + 3).

Ex. 4. Find the L.C.M. of a ax + 2*, a2 + ax + x*, as + 20, a: aco. Now (Art. 29) a3 + gold (a + c) (a? ax + ac),

and as 2c3 (a c) (a + ax + 2x®). Hence the required L.C.M.-(a + x) (a – 2) (as + ax + xc9) (az ax + 2x2) am 26.

54. The L.C.M. of two quantities is found by dividing their product by their G.C.M. Let a and 6 be the two quantities, and d the G.C.M.;

qd. It is evident that p and q contain no common factor. Hence pq 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 pqd is the L.C.M. of pd and qd, or of a and b.

pd x qd id = 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, theri the L.C.M. of the expression thus obtained and a third quantity, and so on. 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 6

And suppose a

pd and i

a

Now pad

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 a-x*y, 4 ay, 6 x*yo. 2. 5 a2b3, 6 aʻc?, 46%c?. 3. (a - b) (6 - c), (6 - a) (a

6 ( - c), (c = a) (c ). 4. ax (c + a), a?( - a), .* - a.

+
5. @c2 + 3x + 2, 2c2 + 4x + 3, ac2 + 5x + 6.
6. ac?

oca

11x + 30, aca 25. 7. 6 x3 + 37 20 + 56, 8.202 + 38 c + 35, 12 x2 + 47 x

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30,

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of 40.

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8. 5 (2 - x + 1), 6 (2a + 1), 7(202 + 1).

C + 9. ** + aʻxca + a“, aʻoca + a’s + a*, axa a-x + a'. 10. 22 + (a + b) x + ab, + (a + c) * + ac, ac? + (b + c) * + bc. 11. 1 X,

1
+ 2C,

1 + 2*, 1 + 2*, 1 + 28. 12. Q3 + 6 x + 11x + 6, 23

6 x2 25 x + 150. 13. a3 - 3 ab (a - b) - 69, a3 - 6,2 + ab + ab?.

- b

+ 14. 24 1, 6206 + 5 x4 + 8 203 + 4x2 + 2 x 1.

15. at 2 a2b2 + 64, a* + 4 a3b + 6 aRb2 + 4 ab3 + 5, a4 4 ab + 6 a^l2 4 ab3 + 64.

16. 3 203 4x + 1, 2 23 7 x it 5, 4 cm 6 x 2

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+

+

10 .

48, 5 22

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17. 3 cm + 6 x - 24, 23 - 12x + 16,5 * - 22 x - 36. oc

. 18. a ab, 13 - apb, ab? - 6, a b - a.

? 63- ' 19. 3 24

20, 32c2

20. 20. 208 - °, 204 - 2xy + y, 23 + a*y + xy + y.

* *x21. Oct + ax + a x3 + ax + ato + ab and 205 ac + a 23

axca + atac a: 22. a2 + 62 c? da + 2 ab 2 cd and a 72 ca + d + 2 ad 2 bc.

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