OPERATIONS.] Ex. 3. Divide a 4 + 6 4 + 6(a + Z>) 3 by a 1 + 1- - ah, where a is large compared "with b. We must arrange according to powers of a. a 4 - 3 6 + a -i 2 "We have spoken as if magnitude alone was the circum stance which should determine the precedence of the letters in a division. In the more advanced processes of algebra there are other circumstances which give precedence to certain letters, such, for example, as the fact that x may and often does stand for the phrase " quantity," whilst a stands for some determinate numerical quantity. This leads us to exhibit a proposition in division of the greatest value and most extensive application. It is as follows : jmainder 20. PilOPOSlTiON. If any function of x, consisting of ^ r . powers of that letter with numerical multipliers, is divided 1011 by x - , the remainder, when all the x & are divided out, is the same function of a that the dividend is of x; in other words, the remainder is the dividend altered by writ ing a in place of x. To prove this proposition we shall employ the following AXIOM : If two expressions in x are identical in form and value, but one multiplied out farther than the other, we may write any numerical quantity we please in place of x in both, and the results will be equal. For example, (x 1) 2 + (x- 1) - 3 is identical with x 1 - 2(x + 1) + x 1 ; and it is evident that if we write any number (say 1) for x, the results are the same in both. We now proceed to prove the proposition. Let the dividend be x n +px n ~ l + qx"* 2 , &c., where n is a whole number, and p, q, &c., positive or negative numerical quantities. Let the quotient, when this is divided by x a, be Q, the remainder, which does not contain x, II ; then x* +j)x n ~ 1 + qyr-* + , &c. = Q(z - a) + R by the definition of Division. Now this equality is in reality an identity in terms of the axiom. If then we write a in place of x, the results will be equal ; this gives a" +pa n ~ 1 + qa n ~* + &c. =Q.O + R which is the proposition to be proved. Examples. Ex. 1. If n be any whole number, x* - a" is divisible by x a without remainder. For the remainder, by the proposition, is a" - a" = . Ex. 2. If n be an even number, x* - a" is divisible by x + a without remainder. For the remainder is ( - a)" - a" = 0, since n is even. Observe that the divisor here has to be changed to x-(- a), so that - a stands in place of the a of the pro position. Ex. 3. If n be an odd number, x n + a" is divisible by x + a without remainder. For the remainder is ( - a)" + a" = 0, because n is odd. Ex. 4. To prove that 46V - (b* + c 2 - a 2 ) 2 is divisible by -a + b + c; and heuco to resolve it into simple factors. Here the x - a of the proposition is replaced by a-(b + c) (the negative sign of the whole divisor being of no conse quence). To determine the remainder, therefore, we write 6 + c in 525 piece of a in the dividend, or thing to be divided; the result is, 4&V 5 - (//> + c- - 6 + c) 2 = , hence 46V - (6- + c 2 - a 2 ) 2 is divisible by -a + b + c. Now, since the dividend contains only squares of a, and b } and c, any change in the sign of a, or b, or c, produces no change in the dividend. What we have just proved then becomes (putting a for a) the following : 46 2 c 2 (6 2 + c 2 a 2 ) 2 is divisible by a + b + c . This last becomes (putting - b for b, and then - c for c) : 46 2 c 2 (b 2 + c 2 a 2 ) is divisible by a b + c, and by a + b c. Hence finally, 46 2 c 2 (6 2 + c 2 a 2 ) 2 = (a + b + c) The above example is a good exercise for the student. The result may be more simply arrived at by employing a proposition of very great value and frequent use that the difference of the squares of two quantities is the product of the sum and difference of the quantities. Ex. 5. To prove that (1 - a 2 ) (1 - b-) (1 - c 2 )- (c + ab) (b + ac) (a + be) is divisible by 1 + abc. It is simpler here to write a single letter x for abc, whereby the given quantity becomes (1 - a 2 ) (1 - 6 2 ) (1 - c 2 ) - )(x + b*-)(x + c 2 ), which is obviously under the form pp, when 1 is written for x, and . . is divisible by 1 + x. Ex. G. Prove that (x~-x + 1) (x i -x^ + I)(;r 8 -a: 4 + l) (.t 16 x s + 1 ) . . . (x 2n x n + 1 ) is the quotient of a; 4 " + # 2n + 1 by x 2 + x + 1 ; n being any power of 2. The divisor (x^ + x + l) being multiplied by a; 2 r+1 gives x 4 + x z + 1 ; which, being again multiplied by a; 4 - x" + 1, gives x s + x* + 1 ; and so on to the end. Additional Examples in Division. Ex. 1. Divide 1 - Wx 3 + I5x i - Qx 5 by (1 - x}*. We must first multiply out (1 x) 3 , and then divide the given expression by the product, 1 3x + 3x" x 3 . The quotient is 1 + 3x + 6x-. Ex. 2. Divide 65x"~f - (^ + G4^) by x 1 - Ixy - Sf. We must arrange dividend and divisor in terms of powers of one of the letters, say x the division will then assume the fonn giving # 2 - Ixy + 8y 2 . Ex. 3. Divide x 3 + y 3 + z 3 3xyz by x + y + 2. We must give exclusive attention to some one letter, say x, in dividing out ; thus - x(y + z) 2 the quotient being x" 1 + y"- + z- xy xz - yz. Ex. 4. Divide the product of x" + 3x + 2, x" -5^ + 4, x* + 5x- - 14, by the product of a- 2 1, x- 2. Here we observe that x 2 1 is the product of x + 1 , x - 1. Now (Art. 20), x" + 3x + 2 is divisible by x+1, and 2 - 5x + 4 by x - 1. Hence, if the product is divisible by x" - 1, x- - 2, without remainder, the third factor, * 4 + 5# 2 - 14 musi be divisible by x 2 - 2, which is found to be the f-ase. The quotient required is therefore the product of
(x + 2) (x - 4) (x- + 7) = * 4 - 2x 3 - x 1 - Ux - 56.
