Notice the second example. We first find how often x2, the first term of the divisor, is contained in a1 by the first rule, and we place the result a2 as the first term of the quotit. Next we multiply every term of the divisor by the x, and then subtract the product from the dividend, changing the signs of the quantities to be subtracted. The first term of the remainder is 223, and we now find how often the first term of the divisor is contained in it, and place the result +2x as the second term of the quotient. The same process is repeated with this term as the last, and a second remainder is found, viz., +4x2-8x+16. The 2 of the divisor is contained 4 times in the first term of the remainder, from which we see that 4 is the third term in the quotient. When the divisor has been multiplied by this term there is no remainder and the work is complete. The quotient is therefore x2+2x+4. The number 16 was not required in the first remainder, and need not have been brought down until a stage later. The following rule for the division of compound expressions may now be given, and should be compared 'with the corresponding rule in arithmetic. RULE. Arrange the expressions side by side in some common order of terms, divide the first term of the dividend by the first term of the divisor, and set down the result for the first term of the quotient. Multiply every term of the divisor by the first term of the quotient thus obtained, and subtract the product from the dividend. Bring down more terms of the dividend if necessary to join with the remainder and repeat the whole operation. Proceed in this way until all the terms of the dividend have been brought down. Another example is subjoined. Divide 2ax2 - a3—x3 by a2+ax — x2. Arranging according to powers of x we get — x2 +ax+a2) —x3+2ax2— a3(+x−a -x3+ ax2+a2x The answer is x-a. Divide, ax2-a2x-a3 ax2-a2x-a3 EXERCISE XVII. (1) x2-2x-35 by x-7. (2) x2+2x-99 by x+11. (4) 4x2-36 by 2x+6. (7) 12a2-7ab-10b2 by 3a+2b. (9) 2x1 — 2x3 — 9x2+41x-35 by x2-3x+5. (10) a3-3a2x+3ax2 - x3 by a-x. (11) x3-2α3ñ3+ao by x2−2ax+a2· (12) 16a-8161 by 2a-3b. (13) -32 by X- - 2. (14) 16x-32x3y+24x3y3-8xy3+y by 2x-y. (15) a3—b3+c3+3abc by a−b+c. (16) x-18x3+71x2+112x−220 by x2-10x+11. (17) -8y+2723+18xyz by x-2y+3z. (18) a+y+2x2y2-2z2-1 by x2+y2-22-1. (19) x2+3x3y2 - 2x1y*+-3x2y®+y3 by x1 — x2y2+ya. (20) 4x1-9x2y2+12xy3 — 4y1 by 2x2+3xy-2y2. (21) 6-4x-4x+6x3-15x2+7x-6 by x3-5x2+ 2x-3. (22) 6x7-x-7x+31-813+98x2-90x+20 by 2x4-3x2+5x2-8x+2. (23) Divide 1 by 1+x to 5 terms in the quotient. IMPORTANT RESULTS IN MULTIPLICATION. The following results may be obtained by multiplication. (See page 22.) (x+a)2= x2+2ax+a2. (x-a)2= x2-2ax+a3. (x+a)(x− a) = x2 — a2. i. ii. iii. iv. (x+a)(x+b)=x2+(a+b)x+ab. Hence from (i.) The square of the sum of two quantities is equal to the square of the first twice the product of two quantities the square of the second. And from (ii.) The square of the difference of two quantities is equal to the square of the first twice the product of the quantities + the square of the second. Twice Sq. of first. the product. Sq. of second. Thus (2x+y)2 = The product of the sum and difference of two quantities is equal to the difference of their squares. Sq. of first. Sq. of second. Thus (2a+b) (2a-b) = (2a)2 (b)2 = 4a2 (x+a)(x+b) = x2+(a+b)x+ab. Suppose a 3 and b=4. Then = (x+3)(x+4)= x2+7a+12. Notice that the first term in each of these products is a2 formed by multiplying the first term of the two factors together. The second term of the product is x, and its coefficient is found by adding together the second terms of the factors. The third term is the product of these second terms. These steps must be carefully observed. (x+5)(x+6)=x2+(5+6)x+5×6=x2+11x+30. The first term in the product is the square of x. The second term in the product is x, and its coefficient is 5 and 6 added together, viz.: 11. The third term is the product of 5 and 6, i.e., 30. The sum of +12-5=+7 which is the coefficient of x. The product of +12 and −5= −60, which is the 3rd term. (x-9)(x-8)= x2+(−9−8)+(−—9)× (−8) = x2-17x+72. Work the remaining examples in Ex. xviii. EXERCISE XVIII. Write down the result of (1) (x+8)2. (4) (5x+y)3. (7) (a-4) (13) (x+5)(x-5). (15) (x2+y)(x2-—y2). (5) (3x+10y). (6) (4a+6b)3. (8) (2x-5) (9) (a-b). (10) (10x-2y). (11) (12m—n)2. (12) (‡x—12y)2. (17) (2a+363)(2a2-363). (18) (şa-2b2)(a+2b2). (14) (2a+9)(2a-9). (16) (3x2+y)(3x2 — y). (19) (x+5)(x+10). (20) (x+15)(x-10). (21) (a-3)(a+30). (23) (x-17)(x-12). Add together, (22) (x-5)(x-9). (25) (x+2y)', (x-2y) and (x+2y)(x-2y). (x-5)(x+7), (x-5) and (a+10). (26) (3x+2y), (3x—y)(3x+y), (3x-2y)2. (28) (x+1)(x+2)(x-2) and (x-1). (31) (a+b+c) and (a+b-c). (32) (x+1)(x−1)(x2+1) and (x2+1) a2+b2 is divisible by neither a+b nor (a−b). From these examples we see that I. The sum of the odd powers of two quantities is divisible by the sum of the quantities, and II. The difference of the odd powers of two quantities is divisible by the difference of the quantities; also III. The difference of the even powers of two quantities is divisible both by the sum and difference of the quantities. IV. The sum of the even powers of two quantities is divisible by neither the sum or difference of the quantities. These results have been shown to be true when the indices are 1 and 2, and by reference to the examples above given they are most easily remembered. But they are also true for any even and odd powers whatever. Thus by a3 + b3 I. a+b =a2-ab+b3. II. as - bs a a2+ab+b2. |