(7.) x3+10x2+25x-28 and x2+14x+49. Ans. x+7 (8.) x3-9x2+15x-2 and x2 - 4. Ans.x-2. 3 (9.) a2+4a-5, a2+7a+10 and a2-5. Ans. a +5. (10.) a3+1 and a3 +ba2+ab +1. Ans. a +1. (11.) a2-3x+2, a2-a-2 and a2-4a+4. Ans. a-2 (12.) b+by-y2-y3 and b2+by+b+b3. Ans. y+1 (13.) a2-a-2 and 1+a3. Ans. a+1. (14.) 3a2+4a-4 and 3a2+a-2. Ans. 3a-2. (15.) 8a2+14a-15 and 8a3 +30a2+13a-30a. Ans. 8a2+14a-15. (16.) 6a2+7a-3 and 12a2+16a-3. Ans. 2a+3. (17.) 2a3-10a2+12a aud 3a1—15a3 +24a2-24. Ans. a-2. (18.) a3 +ab+ab2+b3 and a1+a3b+ab3 —b1. (19.) a1- 2a3b + 2ab3-b1 and a1 2ab+b4. (20.) a3-8a+ 3 and a6 +3a5+a+3. (21.) at-qa3-qa+q3 and qa2-q3. Ans, a2+b2. 2a3b+2ub? — Ans. (a-b) Ans. a +3. Ans. a-q. (22,) 18a3-33a2+44a-35 and 6a3 -19a2+38a-28 23. 12a +29a2+14a and 12a3-a2-6a. 6 ga+and Ans. 6a-7. 11 7 4 a2 За (25.) a2 Баз 11a2 α +1 and at. 4 + 4 (26.) 2a5-a1+2a3-10a2-4a-5 and 10a2+5a+5. Ans. 2a2+a+1. (27.) 8a4b+2a3b-2a2-3a2b+a and 14a2-7a. Ans. 22-a. (28.) a1-x1 and a3—a2x-ax2+x3. Ans. a2-x2. LEAST COMMON MULTIPLE.-L.C.M. Find the L.C.M. of all the co-efficients. To this affix all the algebraical quantities occurring in all the expressions, with their highest powers. Thus find the L. C. M. of 16x2y2x, 8x3 y2x2 and 12ya. Here the L.C.M. of 16, 8, and 12, is 48; and the letters common to all the three expressions arew, y, z, a: so the L.C.M. of the three expressions is 48.xyz . Find the L.C.M. of the following expressions: (1.) a-b and a3 —b3. Ans. a+ab — ab3 —b1. (2.) x3-bx-ex+be and x2. ―ax-ex+ae. Ans. (-a) (x—b) (x-e.) (3.) 5x5xy +5.xy-5y3 and 6x+6x2y+6xy2 +6ys. (4.) 4-4wy; 12xy +12ys; 8x3-8x2y. Ans. 30-30y*. Ans. 24ty2-24x2y1. Ans. 36a3b2—36ab1. (5.) Gab+6ab2; 9a-9ab2; 4a3-4ab2. (6.) a*-* and a3-a3x-ax2+x3. Ans. a5-ax-ax1+x5. (7.) a3¬aax−ax2+a3 and x1—a1, and ax3 +a3x— Ans. ax5-ax1—a3x+a®. (8.) a, and a2-2. Ans. a+a3x-ax3-x. (9.) w2-4; w-2; -1; and x-1. (10.) 1-a; 1+2 and 1-æ. Ans. x4-5x2 +4. Ans. 1-x2. Ans. 16a4-1. To find the LC.M. of two compound expressions, first find their G.C.M., and divide either of the quantities by it. The product of this quotient and the quantity will be the L.C.M. required. If there are three compound expressions, find the L.C.M. of any two of them, then that of the result, and the third; and so on for any number of expressions. (12.) 3-11a+6, 2xa −7x+3; and 6x2 −7x+2. Ans, 6-25x2+23x-6. (13.) 6-13x+6; 12x2-5x-2 and 15x2+2x−8. Ans. 1204-134x3-129x2+74x +24. (14.) a-b: a3-b2; and a2+2ab+b2. Ans as+a2b-ab2-b3. (15.) a+b; a-b2; and a2+2ab+b2. Ans. a3+a2b—ab3—b3. (16.) x+2; x2-4; x2+4x+4. Ans. x3+2x2-4x-8. (17.) x2+4x+4; x−2; x2 −4. Ans. x3 +22 -4x-8. (18.) x2-9; x2+6x+9; x+3. Ans. x3+3x-9x-27. (19.) x-xу; x2 —x; ay—ay2. 2 Ans. axy-ax3y-axy+axy". (20.) x-2x; x2-x; 2x-4α. Ans. x2-5x+6y2. (22.) a-3b; a3-5ab+6b; a-2b; a-b. Ans. x+3y-xy3-y1. (25.) (y-2) (y+2); y−2; (y-1) (y+1); y−1. Ans. y-5y+4. FRACTIONS. For arithmetical principles involved in dealing with vulgar fractions, see arithmetic. CASE 1.-To express an improper fraction as a mixed quantity; 19x 3x (1.) Express as a mixed quantity. Ans. 2x+ 8 8 7 Ans. 2xy + 7 CASE. To reduce a mixed quantity to an improper fraction proceed as in arithmetic. 762 За 3a2+7b2 Reduce a+ -to an improper fraction. Ans. За Reduce the following mixed quantities to improper fractions: (x+y) (xy)=22-y2 and conversely x2-y2=(x+y) (x-y). (x+y) (x+2)= x2 + (y + z) x+yz and conversely x2+(y+ z) x+yz=(x+y) (x+2) Applying the last axioms to an illustration, 2+7x +16, we deduce that this=2+(y+z) x+yz=(x+y) (x + z) = x2 (4+4)=x+4×4=(x+4) (x+4). Resolve into factors (1.) a2+2ab+b2. Ans. (a+b) (a+b). GREATEST COMMON MEASURE.-G.C.M. RULE.-Proceed as in ordinary arithmetic, or where possible resolve the quantities of which the G.C.M. is required into their factors; thus find the G.C.M. of -5a3y-z and 10x+y3z2. =5a3yzx1 and of these quantities 10x+y3z2=5x3 y2z × 2xyz.) 5x3yz is common. or 5x sy2z)10x4y3z2(2xyz The last divisor in this 10x4y 3 z2 case, being the first, forming the G.C.M. a2+2a+1)a3+2a2+2a+1(a a+1)a2+2a+1(a+1 a2 + a a+1 a2+2a+1=(a+1)+(a+1) (1.) 3x2y2+6x2y3 and 12xy2+9x2y2. a+1 Ans. 3xy3. |