1. Find the sum of amx-4a/mx+7a ↓ mx—3a/mx α +4a√mx- √mx. 3 Ans. 4a/mx. 2. Find the sum of 7mpx2y3-4mpx2y3-8mpx2y3 + 4mpx2y5-6mpx2y3+2mpx2y3. Ans. -5mpx y3. 3. Find the sum of 3(x2—y?)*+7(x2—y?)*—4(x2— y2)*—8(x2—y?)*+5(x2—y2)\_2(x2—y2)$. Ans. (x2-y2) Ans. 2(b—c)x3· 4. Find the sum of 5(b—c)x3—7(b—c)x3+4(b—c)x3 —4(b—c)x3 +7(b—c)x3-_ —c)x3—3(b—c)x3. 23. CASE III. When both the signs and the quantities are unlike, or some like and others unlike. RULE. Find the sum of each parcel of like quantities by the last rule, and write the several results after each other, with their proper signs. NOTE. The above rules will be obvious from the following considerations:-The first rule is simply this, that any number of quantities, as 4, and 5, and 7, of the same kind, will make 4+5+7, or 16 quantities of the same kind. In the second, it must be remembered, that minus quantities are subtractive, while plus ones are additive, and that to add and then subtract the same quantities, is the same as to perform no operation at all; therefore to add a greater quantity, and then subtract a less, is the same as to add their difference; and to subtract a greater, and then add a less, is the same as to subtract their difference, which is the rule. Again, it is evident that the third rule just enables us to combine several accounts in the second into one sum. EXAMPLE. 3xy+4az -3cx -4ax2+4xy-3az 7cx-3αx2-2xy -5az +4cx-ax2 3xy +5cx +3αx2 8xy-4az+13cx-5αx2 Here we begin with the first term, which contains xy. We therefore collect all the (xy)'s into one sum; then find the sum of the (az)'s, and so on till we have collected all the terms. 1. Find the sum of 4x2+8xy+4y2, 3x2+4xy+y2, 2x2-4xy+2y2, and x2—y2. Ans. 10x2+8xy+6y2. 2. Find the sum of 3ab+4ac-cd, -6bc+8cd+4ab, +6bc-4ab+5cd, and 4bc+2ab. Ans. 5ab4ac+12cd+4bc. 3. Find the sum of 4a2+10ab+7b2, 3a2+4ab+b2, 2a2+6ab+462, and a2—6ab-762. Ans. 10a2+14ab+5b2. 4. Find the sum of a-3a+2b, 3a-4a3b—b, —7a+14a36, and a—4a3¿1⁄2. -4ab2+4b. Ans. —2a+3a22+26. SUBTRACTION. 24. RULE. Change the signs of the quantities to be subtracted, or suppose them changed, and then collect the quantities, as in addition. NOTE. This rule will appear evident from the following considerations :-If we are required to subtract 3a-b from 5a, this means that we are to take away (3a-b) from 5a; if then we take away 3a, the remainder will be 5a-3a; but it is evident we have here taken away too much by b; we must then add b, and we will have for the true remainder 5a-3a+b=2a+b, where it is evident that the signs of the subtrahend have been changed, and then the quantities collected by addition. The same process of reasoning will apply to any other case. EXAMPLE. From 7a+5ac2-3bc-4bc2, Rem. 5a+8ac2—7bc+5bc2. 1. From a2+2ab+b2, take a2—2ab+b2. 2. Fromx+3y+3xy2+y3, take x3-3x2y+3xy2-y3. Ans. 4ab. Ans. 6x2y+2y3. 3. From x3-a3, take a2-2ax+x2. Ans. 23-2+2ax-a2—a3. 4. From 3bx2-4cx+5x3, take 5x3-4bx2+3cx. Ans. 7bx2-7cx. MULTIPLICATION. 25. In algebraic multiplication three things are to be attended to; first, the sign; second, the coefficient; and third, the literal part of the product. RULE. When the signs of the factors are like, the sign of the product is plus, and when the signs are unlike, the sign of the product is minus. The product of the coefficients of the factors is the coefficient of the product. And the letters of both factors, written after each other as the letters of a word, form the literal part of the product, the letters being commonly arranged in the order of the alphabet. NOTE 1. That like signs give plus, and unlike give minus, can be shown in the following manner:-When +b is to be multiplied into +a, the meaning is, that +b is to be added to itself as often as there are units in a, and therefore the product is +ab; if now (b—b), which is evidently =0, be multiplied by a, the product must be =0; but +bx+a has been shown to be +ab, and that the whole product may be =0, the other part must be ab; therefore -bx+a=—ab. Again, since the product of two factors is the same, whichever be considered as the multiplier, +ax--b——ab, and if (a-a), which equals 0, be multiplied into b, the product must be 0; but it has been shown, that +ax-b――ab, therefore that the product may be 0, -ax-b must be equal to +ab. Hence like signs give plus, and unlike signs give minus. NOTE 2. When the same letter appears in the multiplicand and multiplier, it will appear in the product with a power equal to the sum of its powers in each factor; for a3xa2 denotes (aaa)×(aa)= aaaaaaa 3—ɑ(3+2), that is, its power in the product is the sum of its powers in the multiplier and multiplicand. Multiplication naturally divides itself into three cases. 26. CASE I. When the multiplicand and multiplier are both simple quantities. RULE. Multiply the coefficients together for the coefficient of the product, and the letters together for the literal part, and prefix the proper sign. EXAMPLE. 5ac3ab2-15a2b2c. 1. Multiply 3a2bc, by 7ab2c3. Ans. 21a3b3c4. Ans. 15abc2d2. Ans. 28a2bc4d. Ans. -21a2c2xy. 27. CASE II. When the multiplicand is a compound and the multiplier a simple quantity. RULE. Multiply each term of the multiplicand by the multiplier, and write the several products after each other, with their proper signs. EXAMPLE. Multiply 3ac+2bd+5c", by 4ab. Multiplicand, Multiplier, Product, 3ac+2bd+5c2. 4ab. 12a bc+8ab2d+20abc. 1. Multiply 7a2+4ab2+b2, by 3ac. Ans. 21a3c+12a2b2c+3ab2c. 2. Multiply 3a3+4a2b-7ac, by 5a2b. Ans. 15a5b+20ab2—35a3bc. ac+3a2b-face, by 4ab2. Ans. 2a2b2c+23a3b3—3a2b2c2. 3. Multiply Ans. 2ab+4a2b2+}ab3. 5. Multiply 7a2c2-5bc2-4ab, by abc. Ans. 5a3bc3-3ab2c3—3a2b2c. 28. CASE III. When both multiplicand and multiplier are compound quantities. RULE. Multiply all the terms of the multiplicand by each term of the multiplier, and collect the several products into one sum by addition. EXAMPLE. Multiply 3a2-6ab+362, by a-b. Multiplicand, 3a2-6ab+362. Product by b, Total product, 1. Multiply a+b, by a+b. 2. Multiply a-b, by a-b. 3. Multiply a+b, by a-b. 4. Multiply a2+ab+b2, by a—b. 5. Multiply a2-ax+x2, by a+x. 6. Multiply 7ab+3ac+4d, by 3a-2c. Ans. a2-b2 Ans. a3-b3. Ans. a3+x3. Ans. 21a b+9a2c+12ad-14abc-bac-8cd. 7. Multiply 5a+3x+y, by 3a+2x. Ans. 15a2+19ax+3ay+6x2+2xy. 8. Multiply x+4y-2, by 1x+3y. Ans. 4x2+3xy—4x+14y2—бy. 9. Multiply x3+x2y+xy2+y3, by x-y. 10. Multiply —x3+x2-x+1, by x2+x-1. Ans. x6x4x3—x2+2x−1. 11. Multiply ax + bx2+cx3, by 1+x+x2+x3. Ans. ax+(a+b) x2+(a+b+c) x3 +(a+b+c) x2+(b+c) x5+cx6. 29. THEOREM I. The square of the sum of two quantities is equal to the sum of the squares of each of the quan tities, together with twice their product. (See Art. 28, Example 1.) 30. THEOREM II. The square of the difference of two quantities is less than the sum of their squares by twice their product. (See Art. 28, Example 2.) 31. THEOREM III. The product of the sum and difference of two quantities is equal to the difference of their squares. (See Art. 28, Example 3.) NOTE. The above theorems are very important, and should be committed to memory. 12. Write by Theorem I. the squares of the following quantities, and verify the results by multiplication, (x+y)", (2a+x)2, (a+2b)2, (a2+x2)2, (ac+x)2, (3x+4y)2. Ans. (x2+2xy+y2), (4a2+4ax+x2), (a2+4ab+4b2), (a1+2a2x2+x1), (a2c2+2acx+x2), (9x2+24xy+16y2). 13. Write by Theorem II. the squares of the following quantities, and verify the results by multiplication, (a-x)3, (2a-b)2, (3x-2y)2, (a2—b2)2, (ac—y)2, (3a2—4c)2. Ans. (a22ax+x2), (4a2-4ab+b2), (9x-12xy +4y2), (a+-2a2b2+b), (a2c2-2acy+y2), (9a1-24a2c+16c). 14. Write by Theorem III. the product of the following quantities, and verify the results by multiplication, (a+x) (a—x), (2a+x)(2a—x), (3x+2y)(3x—27), (a2— b2)(a2+b2), (4a2-9c2) × (4a2+9c2). Ans. (a2x2), (4a2 — x2), (9x2 — 4y2), (a1 —b1), (16a1-81c1). 15. Find the continued product of (a−x) (a+x) (a2+x2) (a1+x1). Ans. a8-x8. 16. Find the continued product of (2a+x) (2a—x) (4a2 +x2) (16α1+x1). Ans. 256a8x8. 17. Find the continued product of (x2+xy+y2) (x—y) (23+y3). Ans. x6-y 18. Find the continued product of (x-x2y+xy2—y3) (x+y) (x+y). Ans. 10x6y+x+yε—y1o ̧ DIVISION. 32. DIVISION being the converse of multiplication, naturally gives the following Rules. 1st, If the signs of the dividend and divisor are like, the sign of the quotient is +; and if unlike, the sign of the quotient will be 2d, Divide the coefficient of the dividend by that of the divisor, for the coefficient of the quotient. 3d, Since any quantity divided by itself gives 1 for quo |