111. There is an important relation between the highest common factor and the lowest common multiple of two expressions which it is desirable to notice. Let Fbe the highest common factor, and X the lowest common multiple of A and B. Then, as in the preceding article, Hence the product of two expressions is equal to the product of their highest common factor and lowest common multiple. hence the lowest common multiple of two expressions may be found by dividing their product by their highest common factor; or by dividing either of them by their highest common factor, and multiplying the quotient by the other. 112. The lowest common multiple of three expressions A, B, C may be obtained as follows. First, find X the L. C. M. of A and B. Next find Y the L.C.M. of X and C; then Y will be the required L. C. M. of A, B, C. For Y is the expression of lowest dimensions which is divisible by X and C, and X is the expression of lowest dimensions divisible by A and B. Therefore Y is the expression of lowest dimensions divisible by all three. EXAMPLES XIII. c. 1. Find the highest common factor and the lowest common multiple of x2-5x+6, x2-4, x3-3x-2. 2. Find the lowest common multiple of ab (x2+1)+x (a2+b2) and ab (x2 − 1) + x (a2 − b2). 3. Find the lowest common multiple of xy – bx, xy—ay, y2-3by+2b2, xy-2bx-ay+2ab, xy-bx-ay+ab. 4. Find the highest common factor and the lowest common multiple of 3+2x2 - 3x, 2x3+5x2 – 3x. 5. Find the lowest common multiple of 1−x, (1−x2)2, (1+x)3. 6. Find the lowest common multiple of x2-10x+24, x2 −8x+12, x2 – 6x+8. 7. Find the highest common factor and the lowest common multiple of 6x3+x2 – 5x − 2, 6x3+5x2−3x−2. 8. Find the lowest common multiple of (bc2 - abc)2, b2 (ac2 — a3), a2c2+2ac3+c1. 9. Find the lowest common multiple of x3-y3, x3y-y, y2 (x-y)2, x2+xy + y2. Also find the highest common factor of the first three expressions. 10. Find the highest common factor of 6x2 - 13x+6, 2x2+5x-12, 6x2 – x – 12. Also show that the lowest common multiple is the product of the three quantities divided by the square of the highest common factor. 11. Find the lowest common multiple of x1+ax3+a3x+a1, ¿c1+a2x2 + aa. 12. Find the highest common factor and the lowest common multiple of 3x3-7x2y+5xy2 −y3, x2y +3xy2 — 3x3 — y3, 3x2+5x3y+xy2-y3. 13. Find the highest common factor of 4x3- 10x2+4x+2, 3.x1 − 2x3 − 3x+2. 14. Find the lowest common multiple of 15. Find the highest common factor and the lowest common multiple of (222 — 3a2) y + (2a2 — 3y2)x, (2a2+3y2)x+(2x2+3a2) y. 16. Find the highest common factor and the lowest common multiple of 3 – 9x2 + 26x − 24, x3 – 12x2+47x – 60. 17. Find the highest common factor of x3-15ax2+48a2x+64a3, x2-10ax+16a2. 18. Find the lowest common multiple of 21x (xy-y2)2, 35 (x4y2 — x2y1), 15y (x2+xy)2. CHAPTER XIV. HARDER FACTORS. 113. In Chapter XI. we have explained several rules for resolving algebraical expressions into factors; in the present chapter we shall continue the subject by discussing cases of greater difficulty. 114. By a slight modification some expressions admit of being written in the form of the difference of two squares, and may then be resolved into factors by the method of Art. 86. Example 1. Resolve into factors x4+x2y2+y^. x1+x2y2+y1= (x2 + 2x2y2+y1) − x2y2 = (x2+ y2)2 - (xy)2 Example 2. Resolve into factors x4 - 15x2y2+9ya. x1 − 15x2y2+9y1= (x1 — 6x2y2+9y1) — 9x2y2 =(x2-3y)-(3xy) 115. Expressions which can be put into the form 3± ys may be separated into factors by the rules for resolving the sum or the difference of two cubes. [Art. 89.] 92 Example 3. Resolve ao - 64a3 - a6 +64 into six factors. The expression= a3 (a6 – 64) – (a® — 64) Example 4. =(a6-64) (a3-1) =(a+8) (a3-8) (a3 — 1) =(a+2) (a2 − 2a+4) (a− 2) (a2 + 2a+4) (a−1)(a2+a+1). a (a-1)x2-(a-b-1) xy-b (b+1) y2= {ax - (b+1) y} {(a-1) x + by}. NOTE. In examples of this kind the coefficients of x and y in the binomial factors can usually be guessed at once, and it only remains to verify the coefficient of the middle term. 116. From Example 2, Art. 53, we see that the quotient of a3+b3+c3 −3abc by a+b+c is a2+b2+c2 − bc − ca - ab. Thus a3+b3+c3-3abc=(a+b+c) (a2+b2+c2 − bc - ca - ab)...(1). This result is important and should be carefully remembered. We may note that the expression on the left consists of the sum of the cubes of three quantities a, b, c, diminished by 3 times the product abc. Whenever an expression admits of a similar arrangement, the above formula will enable us to resolve it into factors. Example 1. Resolve into factors a3-b3+c3+3abc. a3 - b3+c3+3abc=a3+ ( − b)3 + c3 − 3a ( − b) c, = (a − b + c) (a2+b2+c2 + bc - ca+ab), - b taking the place of b in formula (1). 1. Example 2. x3- 8y3-27-18xy=x3+ ( − 2y)3 + ( − 3)3 – 3x ( − 2y) (-3) =(x-2y-3) (x2+4y2+9 −6y+3x+2xy). EXAMPLES XIV. a. Resolve into factors: x4+16x2+256. 3. x+y4-7x2y2. 5. x2-6x2y2+ya. 7. 4m2+9n1 - 24m2n2. 9. 2419x2y+25y4. 2. 81a4+9a2b2+b4. 4. m2+n1-18m2n2. 6. 4x+91-93x22. 8. 9x+4y+11x22. 10. 16a+ba — 28a2b2. |