Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Ex. 2. Find the G.C.M. of 10 23 + 31 cm - 63 2 and 14 cl + 51 mm

54 x.

[ocr errors]

We may tell by inspection that w is a common factor, which we therefore strike out of both, only taking care to reserve it. The quantities then become

10 x + 31 aC 63, and 14 x2 + 51 54. We may now proceed according to rule, taking the former as divisor. We see, however, that the coefficient of the first term of the dividend is not exactly divisible by the coefficient of the first term of the divisor. Multiply therefore (to avoid fractions) the dividend by such a number as will make it so divisible, viz., by 5. This will not affect the G.C.M., as 5 is not a factor of the first expression, viz., 10 202 + 31

63. It may as well be here mentioned that the G.C.M. of two quantities cannot be affected by the multiplication or division of one of the quantities by any quantity which is not a measure of the other. We shall

, for a similar reason, reject certain factors or introduce them into any of the remainders or dividends during the operation. (See Art. 47).

14 x2 + 51 x 54

5 10x + 31 x - 63)70 2c + 255 x 27007

70 cm + 217 x 441

38 x + 171

Rejecting the factor 19 of this remainder, we have

22c + 9)10 x + 31 x - 63(52 - 7
102 + 45 x

14 2 – 63
14 a 63

Hence, 2 + 9 is the last divisor, and multiplying this by 3, the common measure struck out at the commencement, we find the G.C.M to be x (2 x + 9) or 2 ac + 9 2.

21) 206

[ocr errors]

21 x

[ocr errors]

847 a

Ex. 3. Find the G.C.M. ofc-7.00 -3.0-500+ 42cc-34.-21, and -11c* + 2500+ 19 -49 -21.
205 - 11 * + 25 202 + 19 cm - 49 x

7 26 3 24 5 008 + 42 m - 34 x - 21 (+ 4
11 2010 + 25 c* + 19 2? 49 cm
4 225 28 24 24 263 + 91 cca 13X – 21
4 ar

44 x4 + 100 203 + 76 cm 1962 - 84

16 x* - 124 23 + 15 cm + 183 x + 63
Multiplying the preceding divisor by 16, and taking the result for a dividend, we have
16 c* – 124 30% + 15 + 183 x + 63)16 206 176 x4 + 400 20+ 304 cm 784 x 336(oc
16 25 124 x4 + 15 203 + 183 cm +

63 x
52 2c* + 385 203 + 121 oca

336
(Multiplying this remainder by 4)

4
208 209 + 1540 23 + 484 oca

1344( - 13
208 2* + 1612 203 195 x2

2379

819
72 203 + 679 wca

1009x

525-
Multiplying the preceding divisor by 9, and taking the result for a dividend, we have-
72 203 + 679

1009 x
525)144 **

1116 203 + 135 20* + 1647 x + 567( – 2 x 144 2* 1358 208 + 2018 202 + 1050 x

242 23 1883 x + 597 x + 567 (Multiplying this remainder by 36)

36 87122 -67788 x + 21492 x + 20412(121 8712 203 82159 x2 + 122089 x + 63525

14371 2 - 100597 x 43113

33883

[ocr errors]
[ocr errors]

Dividing this remainder by 14371, and taking the quotient for a new divisor, we have— ** 7x – 3) – 72 x + 679 202 – 1009 x 525( - 72 x + 175

- 72 3 + 504 + + 216 x

175 - 1225 x - 525
175 m2 1225 x

525 3 is the G.C.M. It will be seen that we have introduced and rejected factors during the operation in order to avoid fractional coefficients. This, as will be seen from the general theory, will not affect the result, provided that no factor thus introduced or rejected is a measure of the corresponding divisor or dividend, as the case may be.

[ocr errors]
[ocr errors]
[ocr errors]

Theory of the Greatest Common Measure. 46. Let A and B be the two algebraical quantities, and the operation as indicated by the rule (Art. 45) be performed. Thus, let A be divided by B, with B)A(P quotient p and remainderc. Then

pB let B be divided by C, with quotientq, C)B(2 and remainder D. Lastly, let C be divided by D, with quotient r, and

DCT remainder zero.

TD Then we are required to show that

0 D is the G.C.M. of A and B.

(1.) D is a common measure of A and B.

Now, we have C rD, B = qC + D, A pB + C. Hence, D is a measure of C, and therefore of qc. It is therefore a measure of qC + D or B. Hence, also, D is a measure of pB, and since it is also a measure of C, it must be a measure of pB + C or A. But we have shown it to be a measure of B. Hence, D is a common measure of A and B.

(2.) D is the G.C.M. of A and B.

For every measure of A and B will divide A - pB or C; and hence every measure of A and B will divide B - QC or D. Now, D cannot be divided by any quantity higher than D, and, therefore, there cannot exist a measure of A and B higher than D. Hence, D is the G.C.M. of A and B.

mB' suppose,

nC suppose,

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)A(P

C')B(2

qC Ć

DC"(T

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(2, where m has no factor that B has ;

pB
C)nBlq, where n has no factor that C has ;

qC
DCir
r]

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 x is common to the quantities 4 203 — 20 202 + 24 x, and 4 200 + 16 cm

Rejecting it, we get aca 5x + 6, and aca + 4 whose G.C.M. is easily found to be a 2.

Multiplying by 42, we find the required G.C.M. to be 4 cm - 8x.

84 2.

21,

.

7 aC

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 23 + 4 cm 10 3 + 3, and 15 23 + 47 cm + 13 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 208 + 4 – 10 + 3 = 3 203 3 x2 + 7 x2

3 x + 3 = 3 * (2 – 1) + 73 (3 - 1)-3(x - 1)

(3 32 + 7 ac 3) (2 – 1). Now, 3 2 + 7 00 3 is not further resolvable, and X 1 is evidently (Art. 30) not a factor of 15 23 + 34 2a + 13 x 12. It is, therefore, very probable that 3 x + 7 – 3 is the G.C.M. required.

We may test it thus— 152c8 + 47 x2 + 13x – 12 = 15 203 + 35.2 – 15 x + 12 + + 28x – 12

= 5 (3.c + 7 x - 3) + 4(3 + 7. - 3)

=(5x + 4) (3x + 7 x – 3). Hence, 3 m2 + 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 following-
1. aca 5x + 6 and a + 3 x 18.
2. 23 + 6 oct + 11 x + 6 and 23 + 5 x + 7 x + 3.
3. 2 2x3 + 10 cm

90 and 3 23 + 16 cm
4. x + (a + b c + ab and x + (a + c) x + ac.
5. a3 73 and a3 + a b + ab?.

18

[ocr errors]

141.

[ocr errors]
« ΠροηγούμενηΣυνέχεια »