(5) Multiply + a by x + b, also x + 2 by x+3, x-4 by x+6, and x-5 by x-7. NOTE. Some useful theorems are furnished by the foregoing results: we shall here express them in common language, recommending the learner to bear them in remembrance, and to avail himself of them whenever occasions occur. 1. From Ex. 1, it follows, that the sum of two numbers, or quantities multiplied by their difference, gives the difference of the squares of those numbers or quantities. Thus, (5+2) (5—2)—5°—22; for a and b, in the Ex. referred to, are open to receive any numerical interpretation whatever. It is plain that the numerical relation just expressed is true, for 5+2 is 7, and 5-2 is three: the product of these is 21, and 52—22 or 25-4, is also 21. Again (9+3)(9-3)=9°—3°; (7+2) (7—2)—7—22; and so on for every pair of numbers. 2. From Ex. 2, it follows that the square of a quantity of two terms, separated by +, is equal to the sum of the squares of the individual quantities PLUS twice their product. And from Ex. 3, that the square of a quantity of two terms, separated by is equal to the sum of the squares of the individual quan tities MINUS twice their product. Thus, (3+5)=32+52+ 2(3x5), and (5-3)=52+32-2(3x5): also, (6+2)2= 62+22+2(2×6), and (6—2)2=62 + 22—2(2 × 6). 3. From Ex. 5, we see that the product of two such factors as x+a, x+b; or x+a, x—b, &c. may always be written down at once: the first term in the product is x3, the coefficient of x in the next term is the sum of a and b, with their proper signs, and the third term is the product of a and b with their proper signs. Thus, (x-2)(x+5)=x2+3x—10; (x-3)(x-6)=x2-9x+18, (y+1)(y—7)=y3—6y—7, (≈—4)(z+3)=z2-z-12, and so on. (6) Multiply a2-3x2 by a2+x2. a2-3x2 a2x2-3x1 a-2a2x2-3x1 (7) Multiply 3a2x-5by2 by 2a+b2y2. 3a2x-5by also, (8) Multiply 2x3-3x2+4x-5 by x+2x-3. (9) Multiply ax2+(b-c)x+1 by x-a+2. ax2+(b−c)x+1 x-a+2 ax3+(b-c) x2+x —a2x2—a(b—c) x—α 2ax+2(b-c)x+2 ax3+{(b−c)—a2+2a} x2 + { (2—a)(b−c)+1}x+2—a EXAMPLES FOR EXERCISE. (1) Multiply 4x2-3x+5 by 2x2—x. 2 (5) Multiply 3-ax2+bx+c by x2-ex+f. 9) Multiply x-2x3y+3x2y2—2xy3+y1 by (x+y)2, first removing the brackets. (10) Multiply (a—b)2, (a+b)(a—b), and a2+b2 together. [See NOTE, page 43.] (11) Required the product of (x−7)(x+5), and (x+2)(x+6). [See NOTE 3, page 44.] (12) Required the product of (x+2)2, (x+2)(x+2), and (x-3)(x-4). (13) Multiply x2—y2, x2+y2, and x1+y1 together. (14) Multiply 2x2-4x+2 by x2—2x+1. (15) Multiply ax + by-cz, bx-ay, and cx-by together. 4 * 2 (16) Multiply 2-(a—b)x+c by x2-(c+d)x-ab. (17) Multiply ax2+(a+b)x+c by bx2—(b—c)x—d. (18) Multiply 2a r·3—(a—b)x2+4 by 3bx3—ex2+2a. (19) Multiply -7ax1 ÷ 3bx3—5cx2+6ex−2 by 4ax27bx-5. + (20) Multiply x2—(ax+b), x2+ax-b, and x2+ax+b together. (21) Prove that {(a2+x2)2. — a2x2} (a2 — x2)= (a3 +x3) (a3—x3). (22) Prove that {x2—(a+b)x+a(a−b)+b2}(x+x+b)= a+b3+x3-3abx. NOTE.-In performing multiplication operations, such as those proposed in the foregoing examples, algebraists are always on the watch for occasions to apply the principles noticed at page 43, to abbreviate the labour, and are, therefore, in the habit of looking carefully to the constitution of the factors with which they have to work before actually commencing the process. And, as respects the process itself, they do not always follow the plan of arranging the products furnished by the several terms of the multiplier in rows one under another, but sometimes make one row suffice for all, abridging this row, when like quantities occur in it, by collecting the like terms into one, and then writing down the abridged form. Thus, in such a case as (x+3)(x-4)(x−5)(x+6), for example, it would be noticed, from a glance at the factors, that the principle (3) page 44, applies to both the first and second pair of those factors; so that the four factors, by aid of the principle referred to, would be at once replaced by the two factors (x2-x—12) (x2+x—30); and, remembering the principle (1), the multiplication of these would be conducted thus: (x2-x)-12 x4—x2—12(x2+x)—30(x2—x)+360 x4-43x2+18x+360, which is the product of the proposed factors, found with the least amount of trouble. DIVISION. 28. CASE I.-When dividend and divisor are both simple quantities:— RULE.-First determine the sign of the quotient from the principle that, in both multiplication and division, like signs give plus, and unlike signs minus. Then, to the quotient of the coefficients, annex the quotient of the letters, and the complete quotient will be expressed. To find the quotient of the coefficients is a mere matter of common arithmetic; and, to find the quotient of the letters, we have only to see what letters, joined to those already in the divisor, are wanted to make up the letters in the dividend: these wanting letters are those of the quotient; for they are such as that, when joined to the letters of the divisor, that is, when multiplied by those letters, they produce the letters in the dividend; and the quantity which, multiplied by the divisor, produces the dividend, must of course be the true quotient. Thus, let it be required to divide 16ax3y by -2axy3; then the signs, being unlike, the sign of the quotient must be minus; also the quotient, as far as the coefficients are concerned, is 8; and lastly, looking at the letters in the divisor, we see that, to make up those in the dividend, there are wanting one a, two x's (or x2), and one y: that is, there is wanting the quantity axy, which is, therefore, the quotient of the letters: hence, 16a2x3y* =-8axy. Again, if we have to divide 35 ab'xy3 -2axy by-5abxy, we see that the sign of the quotient must be plus, the coefficient of it 7, and the letters by; since these joined to, that is, multiplied by, the letters abxy, make up the letters -35ab3xy3 ab2xy3 in the dividend; consequently, -5 abxy -=7by2. Should it happen that letters appear in the divisor which are not to be found in the dividend, then, as the actual division by these cannot be performed, they must still be retained, in their character of divisors, in the final result. Thus, if the example last -35ab'xy3 given had been where the letters c, z, in the divi-5abcxyz' sor, do not occur in the dividend, the quotient would have been 7 by; the divisors, c, z, being retained. Similarly, if the Cz only one a and one x is found in the dividend, the remaining a and 2 must be retained as still unappropriated divisors in the result, which will be 7 by From what has now been stated, with reference to the four fundamental rules of algebra, the learner will recognise the truth of the statement at page 15, namely: that the only trouble worth mentioning, in applying them to examples, arises solely |