13. 3x3-3x2+2a2x-2a3, 3x3+12ax2+2a2x + 8a3. 14. 2x3-9ax2+9a2x-7a3, 4x3-20ax2+20a2x – 16a3. 15. 10x3+25ax2-5a3, 4x3+9ax2 - 2a2x - a3. 16. 6a3+13a2r - 9a.x2 - 10x3, 9a3+12a2x-11ax2 - 10.3. 17. 24x+y+72x3y2 — 6x2y3 — 90.xy1, 6x+y2+13x3y3 — 4x2y* — 15xy5. 18. 4x5a2+10x1a3 - 60x3a1+54x2α5, 24x3α3 + 30x3α3 – 126x2α®. 19. 4x+14x2+20x3+70.x2, 8x7+28.x2 – 8.x2 – 12μa+56x3. 20. 72x3-12ax2 + 72a2x - 420a3, 18x33+42ax2 - 282a2x+270a3. 21. 9.x1+2x2y2+y1, 3xa— 8x3y+5x2y2 — 2xy3. 102. The statements of Art. 97 may be proved as follows. I. If F divides A it will also divide mA. For suppose A=aF, then mA=maF. II. If F divides A and B, then it will divide mA ±nB. then mA±nB=maF±nbF Thus F divides mA±nB. 103. We may now enunciate and prove the rule for finding the highest common factor of any two compound algebraical expressions. We suppose that any simple factors are first removed. [See Example, Art. 98.] Let A and B be the two expressions after the simple factors have been removed. Let them be arranged in descending or ascending powers of some common letter; also let the highest power of that letter in B be not less than the highest power in A. Divide B by A; let p be the quotient, and C the remainder. Suppose C to have a simple factor m. Remove this factor, and so obtain a new divisor D. Further, suppose that in order to make A divisible by D it is necessary to multiply A by a simple factor n. Let q be the next quotient and E the remainder. Finally, divide D by E; let r be the quotient, and suppose that there is no remainder. Then E will be the HC.F. required. The work will stand thus: A) B(p D)nA (9 E)D(r First, to show that E is a common factor of A and B. By examining the steps of the work, it is clear that E divides D, therefore also qD; therefore qD+E, therefore nA; therefore 4, since n is a simple factor. Again, E divides D, therefore mD, that is, C. And since E divides A and C, it also divides pA+C, that is, B. divides both A and B. Hence E Secondly, to show that E is the highest common factor. If not, let there be a factor X of higher dimensions than E. Then X divides A and B, therefore B-pA, that is, C; therefore D (since m is a simple factor); therefore nA-qD, that is, E. Thus X divides E; which is impossible since, by hypothesis, X is of higher dimensions than E. Therefore E is the highest common factor. 104. The highest common factor of three expressions A, B, C may be obtained as follows. First determine F the highest common factor of A and B; next find G the highest common factor of F and C; then G will be the required highest common factor of A, B, C. For F contains every factor which is common to A and B, and G is the highest common factor of F and C. Therefore & is the highest common factor of A, B, C. CHAPTER XIII. LOWEST COMMON MULTIPLE. SIMPLE EXPRESSIONS. 105. DEFINITION. The lowest common multiple of two or more algebraical expressions is the expression of lowest dimensions which is divisible by each of them without remainder. The abbreviation L. C. M. is sometimes used instead of the words lowest common multiple. 106. In the case of simple expressions the lowest common multiple can be written down by inspection. Example 1. The lowest common multiple of aa, a3, a2, ao is ao. Example 2. The lowest common multiple of a3ba, ab5, a2b7 is a3b7; for a3 is the lowest power of a that is divisible by each of the quantities a3, a, a2; and b7 is the lowest power of b that is divisible by each of the quantities ba, b5, b7. 107. If the expressions have numerical coefficients, find by Arithmetic their least common multiple, and prefix it as a coefficient to the algebraical lowest common multiple. Example. The lowest common multiple of 21a4x3y, 35a2x4y, 28a3xy1 is 420a4x4y1; for it consists of the product of the numerical least common multiple of the coefficients; (2) the lowest power of each letter which is divisible by every power of that letter occurring in the given expressions. 108. We have shown how in the case of simple expressions the lowest common multiple can be written by inspection. The lowest common multiple of compound expressions which are given as the product of factors, or which can be easily resolved into factors, can be readily found by a similar method. Example 1. The lowest common multiple of 6x2 (a − x)2, 8a3 (a — x)3 and 12ax (a-x)5 is 24a3x2 (a – x)5. For it consists of the product of (1) the numerical L. C. M. of the coefficients; (2) the lowest power of each factor which is divisible by every power of that factor occurring in the given expressions. Example 2. Find the lowest common multiple of 3a2+9ab, 2a3 - 18ab2, a3+6a2b+9ab2. 9. x2+2x, x2+3x+2. 11. x2+4x+4, x2+5x+6. 10. 12. x2−3x+2, x2 − 1. - 13. x2-x-6, x2+x−2, x2-4x+3. 21. 12x2+3x-42, 12x3+30x2+12x, 32x2 - 40x – 28. 109. When the given expressions are such that their factors cannot be determined by inspection, they must be resolved by finding the highest common factor. Example. Find the lowest common multiple of 2x4x3- 20x2 - 7x + 24 and 2x4 + 3x3 - 13x2 - 7x+15. The highest common factor is x2 + 2x − 3. By division, we obtain 2x4x3- 20x2 − 7x+ 24 = (x2+ 2x − 3) (2x2 – 3x – 8). Therefore the L. C. M. is (x2+ 2x − 3) (2x2 - 3x - 8) (2x2 — x − 5). 110. We may now give the proof of the rule for finding the lowest common multiple of two compound algebraical expressions. Let A and B be the two expressions, and F their highest common factor. Also suppose that a and b are the respective quotients when A and B are divided by F; then A=aF, B=bF. Therefore, since a and b have no common factor, the lowest common multiple of A and B is abF, by inspection. |