3. 25 a* 30 a’b + 19 a’62 - 6 ab3 + 84. 4. 1 4x + 10 22 20 203 + 25 24 24 25 + 16 20%. 5. a2 + 2 abx + (2 ac + 6*) x2 + 2 (ad + bc) 208 + (2 bd + c^) ** + 2 cdac + d^xcl. 6. a*x2n 6 a 22n-1 + 17 aʻx2n - ? 24 axen – 3 + 16 22n-4. 7. * + 2 + 2 -, aʻx-:- 2 + a-??. a' 8. 92m 3 aʻxcm + 25 ao 30 ax" + + 5 a?. 4 Find the square roots of 9. 1296, 6241, 42849, 83521. 10. 10650-24, 000576, 1, 4. 2 11. /17, 71.5, 13 + 1 15 + 1' 73 - 1 Give the values correct to four places of decimals of — of •31416 25 + 12 110 – 2 + 3.1416 of 193 15 - 2 12. 110 + 2 4 0 32.16 Find the cube roots of 13. 8 aby, 125 12/?, a+ 6 a b + 12 ab? + 8 63. 14. 212 + 9 10 + 6 208 99 26 42 2c4 + 441 22 343. 15. 2 + 3 x+y + 3 xy + 903 3 cxa 3 cy. + 3 c^x + 3 cʻy c?. 16. x3 + x^3 + 3 (2 + 2-?), x*y–3 + 3 x*y-' + 3 xy-1 + 1. Find the cube roots of 17. 5849513501832, 1371.330631. 18. 20-346417; 037, 109. Give the value of the following correct to four places of decimals : 05.12 + 3:03375 1 19. 980 9.01 94 + 32 + 1. 5 + 2 of -05 + 7:04 7 21. ( 17 + 2) ( 17 - 1), (5 + 13) (4 + /12). 20. 9-625 + 3:04 22. Vil + 6 12, 36 + 15 13. 25 + 1 4 V.3, and c = - 9-027. b CHAPTER IV. GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE. Greatest Common Measure. 44. In Arithmetic (page 24) we defined the G.C.M. of two or more numbers as their highest common factor. In Algebra the same definition will suffice, provided we understand by the term highest com:non factor, the factor of highest dimensions (Art. 18). This, it need hardly be remarked, does not necessarily correspond to the factor of highest numerical value. 45. To find the G.C.M. of two quantities. RULE.—Let A and B be the quantities, of which A is not of lower dimensions than B. Divide A by B, until a remainder is obtained of lower dimensions than B. Take this remainder as a new divisor, and the preceding divisor A as a new dividend, and divide till a remainder is again obtained of lower dimensions than the divisor; and so on. The last divisor is the G.C.M. Before giving the general theory of the G.C.M. we shall work out a few examples. Ex. 1. Find the G.C.M. of 27 and 2.2 - 11x – 63, According to the above rule, the operation is as follows: ac - 62 27)222 - 11x 63(2 222 12 x – 54 - 6x – 27(x + 3 aca 3x - 27 3x - 27 :: The G.C.M. is x – 9, 9) 22 Ex. 2. Find the G.C.M. of 10 203 + 31 22 63 2 and 142c8 + 5122 54 x. We may tell by inspection that x is a common factor, which we therefore strike out of both, only taking care to reserve it. The quantities then become 10 x2 + 31 a 63, and 14 x2 + 51 2 – 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 x2 + 31 x 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). Rejecting the factor 19 of this remainder, we have 63 Hence, 2 x + 9 is the last divisor, and multiplying this by x, the common measure struck out at the commencement, wę find the G.C.M to be x (2 x + 9) or 2 x2 + 9 a. 49 20 21) 26 212 63 2 Ex. 3. Find the G.C.M. of x® - 7.30 -3 *-5.200 + 42x* - 342-21, and 2-11c* + 25 208 + 19 2? -49x-21. 203 - 11 2* + 25 203 + 19 22 7 2c5 3 204 5 203 + 42 ca 34 x – 21 (2 + 4 26 11 205 + 25 24 + 19 x? 4922 4 205 28 2c* 24 23 + 91 cm 133 - 21 16 * 124 23 + 15 cm + 183 + 63 16 25 124 24 + 15 23 + 183 x2 + 336 4 33883 1344( - 13 208 2* + 1612 203 195 x2 2379 x 819 72 203 + 679 1009x 525 Multiplying the preceding divisor by 9, and taking the result for a dividend, we have 72 203 + 679 22 1009 2 525)144 x4 1116 3 + 135 c + 1647 x + 567( - 2 c 144 *c* - 1358 3 + 2018 x2 + 1050 x 242 23 1883 m + 597 x + 567 36 14371 cm 100597 C 43113 Dividing this remainder by 14371, and taking the quotient for a new divisor, we havecca – 7 x – 3) - 72 x2 + 679.xco 1009 2 525( - 72 x + 175 72 23 + 504 + + 216 x 525 525 7 X – 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. 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)AP quotient p and remainder C. Then pВ let B be divided by C, with quotient q, C)B(2 and remainder D. Lastly, let C be qC divided by D, with quotient r, and DCT remainder zero. rD 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 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. |