Dividing the remainder by 12, 3x - 2) 6x2 - 25 х + 14(2x - 7 Hence, 3x - 2 is the H.C.F. required. Note 4. If the first term of a remainder is negative, the sign of each term may be changed. 3. Find the H.C.F. of 2x2 - 3x-2 and 2x2 5x-3. 2x2 - 3x - 2) 2x2 - 5x - 3 (1 -2x-1 Changing the sign of each term of this remainder, 2x+1)2x2 - 3x - 2 (x - 2 2x2+ x -4x-2 -4x-2 Hence, 2x+1 is the H.C.F. required. Note 5. If the first term of the dividend or of any remainder is not divisible by the first term of the divisor, it may be made so by multiplying the dividend or remainder by any quantity which is not a factor of the divisor. 4. Find the H.C.F. of 2x - 7x2 + 5x-6 and 3x - 7 x2 - 7x+3. Since 3x is not divisible by 23, we multiply the second quantity by 2. 2x2 - 7x2 + 5x – 6) 6 x3 - 142 - 14x + 6(3 6x2-21 x2 + 15x - 18 7 x2 - 29 x + 24 Since 2 is not divisible by 7x2, we multiply each term of the new dividend by 7. 7 x2 - 29 x + 24) 14 x – 49x2 + 35x - 42(2x 14x2 - 58x2+48x 9x2-13x - 42 Multiplying this by 7 to make its first term divisible by Hence, x - 3 is the H.C.F. required. Note 6. If the given quantities have a common factor which can be seen by inspection, remove it, and find the H.C.F. of the resulting expressions. This result, multiplied by the common factor, will give the H.C.F. of the given quantities. 5. Find the H.C.F. of 6x - ax2 - 5 a2x and 21 x2 - 26 ax2 + 5 a2x. Removing the common factor x, we find the H.C.F. of 6 x2 ах 5a2 and 21 x2 - 26 ах + 5a2. Multiplying the latter by 2, 6x2-6 ах 5 ах 5 a2 5 ax 5 a2 Multiplying x-a by x, the common factor, we have x(x - a) or x2 ax as the H.C.F. of the given expressions. EXAMPLES. 131. Find the highest common factors of the following: 1. x2+x-6 and 2x2-11x+14. 2. 6x2 - 7x - 24 and 12x2+8x -15. 3.2a2-5a + 3 and 4a3 — 2 a2 — 9 a + 7. 4. 24x2+11 ах 28 a2 and 40x2 — 51 ах +14a2. 5. 8α – 22a2+5a and 6 a2b – 23 ab + 206. 6. x3-5mx2 + 4m2x and x2 - mx3 + 3m2x2 - 3m3x. 7.5m2n2+58 mn2 + 33 n2 and 10m3 +31 m2 — 20 m 21. 8. 2α+ +3 αx - 9 a2x2 and 6a3-17a2x +14 ax2 - 3x3. 9. x - 8 and x2 - 6x2+11x-6. 10. 2x - 3x2 - 2+1 and 6x - x2+3x-2. 11.8m2-22mn+5n2 and 6 m2-29m3n +43 m2n2 - 20 mn3. 12. ax2 + 2 ax2 + ax + 2a and 35 - 12x2 – 3x2 - 6x. 13. ax* -ax2-2ax2+2ax and ax3-3ax2+2ax+ax2 — ах. 14. 2x2-2x+4x2+2x+6 and 3x2 + 6 x − 3 x — 6. 15. α2 + a3-6a2+a+3 and a+ + 2a3 - 6a2 - a + 2. 16. x - x − 5x + 2x2 + 6x and x + x - x2-2x2 2x. 17. 15 a2x – 20a2x2 – 65 а2x - 30а2 and 12 bx +20bx2 -16 bx – 166. 18. a2 + a3x + a2x2 + ax3 - 4x4 and a2 + 2 ax + 3a2x2 + 4 ax3 — 10 x2. 19. x2+x+x2-1 and x + 3x2+2x. 20. x - xy - 3x2y2 + 5 xy - 6 y and 3x4 - 5 xy - x2y2 - 7 xy23 + 10y. 21. 2x-5x+5x2-5x+3 and 2x2-7x+4x2+5x-3. 22. 3a2-2ab + 2a2b2 - 5 ab3 - 26 and 6a - ab + 2a2b2 – 2 ab3 - b. 132. To find the H.C.F. of three or more quantities, find the H.C.F. of two of them; then of this result and the third quantity, and so on. The last divisor will be the H.C.F. of the given quantities. EXAMPLES. Find the highest common factors of the following: 1.2x2 - 5x - 42, 4x2 + 8x -21, and 6x2 + 23x +7. 2. 12x2 - 28x – 5, 14x2 - 39x+10, and 10-11x-35. 3.6m2+7mn + 2n2, 3m3 — 7 m2n — 12 mn2 - 4 n3, and 15m2 + 4mn - 4n2. 4. 6a2+13α-5, 6a2+19a2+8a-5, and 3a3+2a2 + 2α-1. 5. x + 3x2-6x - 8, x2 +5x2+2x-8, and x-3x2 - 16x + 48. 6. x-7x+6, x2 + 3x2-16х+12, and x- 5x2 + 7x-3. 7. 2α3-3a2 - 5a + 6, 2a3+3a2 - 8α - 12, and 2 a3- a2 - 12a - 9. X. LOWEST COMMON MULTIPLE. 133. A Common Multiple of two or more quantities is a quantity which can be divided by each of them without a remainder. Hence, a common multiple of two or more quantities must contain all the prime factors of each of the quantities. 134. The Lowest Common Multiple of two or more quantities is the product of their different prime factors, each being taken the greatest number of times which it occurs in any one of the quantities. It is evident from this definition that the lowest common multiple of two or more quantities is the expression of lowest degree which can be divided by each of them without a remainder. Thus, the lowest common multiple of xy2, yz, and xzt is When quantities are prime to each other, their product is their lowest common multiple. 135. In determining the lowest common multiple of algebraic quantities, we may distinguish three cases. CASE I. 136. When the quantities are monomials. 1. Find the L.C.M. of 36ax, 60a2y2, and 84cx. Hence, the L.C.M. = 2.2.3・3・5・7・acxy (Art. 134) = 1260 a3cxy2, Ans. |