is to be divided by some other quantity; because. in this case, any term that is common to both the divisor and dividend may be more readily suppressed; as will be evident, from various instances, in the following part of the work.

Ex. 16. Required the product of a+b+c by ab+c. Ans. a2+2ac-b3 +c2. Ex. 17. Required the product of x+y+z by x-y-z. Ans. xy-Qyz-22. Ex. 18. Required the product of 1-x+x2-x2 by 1+x. Ex. 19. Multiply a3+3a2b+3ab2+b3 by a2 + 2ab+b2.

Ans. 1

Ans. a5+5a+b+10a3b2+10a2b3+5ab+b5. Ex. 20. Multiply 4x2y+3xy-1 by 2x2. -X.

4

Ans. 8x+y+2x3y-2x3-3x2y + x. Ex. 21. Multiply x3+x2y+xy+y3 by x-y. Ans. x-y. Ex. 22. Multiply 3x3-2a2x2+3a3 by 2x3-3a a2+5a Ans.6x-13a2 x5 +6a2x2+21a3x3--19a3 x2+15α.

2

Ex. 23. Multiply 2a-3ax+4x2 by 5a2-6x -2x2. Ans. 10a4-27a3x+34a2x2-18ax3-8x1. Ex. 24. Required the continual product of a +x. a-x, a2+2ax+x2, and a2-2ax-x2.

4

Ans, a-3a1 x2+3a2 x1-x®. Ex. 25. Required the product of x-ax+bx -c, and x2-2x+3.

4

3

Ans. x5-(a+2) xa1 + (b +2a+3) x 3 —(c+2b+3a) x2+(2c+36)x-3c. Ex. 26. Required the product of mx2-nx-r Ans. mnx3-(n2 +mr)x2 +r2. Ex. 27. Required the product of px3-rx+q and x2-rx-q•

and nx-r.

Ans. px-(r+pr)x3+(q+r2—pg)x2 —q2,

Ex. 28. Multiply 3x2-2xy+5 by x2+2xy-3. Ans. 3x+4x3y-4x3 × (1+y2)+16xy-15. Ex. 29. Multiply a3+3a2b+3ab2+b3 by a3 3a2b+3ab2-b3. Ans. a3a b2 +зa2b1—b®. Ex. 30. Multiply 5a3-4a2b+5ab2-363 by 4a2 -5ab262.

Ans. 20a41a4b+50a3b2-45a2b3+25ab46b5..

IV. Division of Algebraic Quantities.

80. In the Division of algebraic quantities, the same circumstances are to be taken into consideration as in their multiplication, and consequently the following propositions must be observed.

31. If the sign of the divisor and dividend be like, the sign of the quotient will be +; if unlike, the sigu of the quotient will be -.

The reason of this proposition follows immediately from multiplication:

+ab

Thus, if +a+b+ab; therefore

=+b:

ta

-ab

+ax-b=-ab ;

-b

+a

- ab

-ax+b=ab;

---

82. If the given quantities have coefficients, the coefficient of the quotient will be equal to the coefficient of the dividend divided by that of the divisor.

Thus, 4ab2b, or

4ab

2a.

For, by the nature of division, the product of the quotient, multiplied by the divisor, is equal to the dividend; but the coefficient of a product is equal to the product of the coefficients of the fac4. ba tors (Art. 70). Therefore, 4ab2b=× =2a. 2 b

83.. That the letters of the quotient are those of the dividend not common to the divisor, when all the letters of the divisor are common to the dividend: for example, the product abc, divided by ab, gives c for the quotient, because the product of ab by c is abc.

34. But when the divisor comprehends other letters, not common to the dividend, then the division can only be indicated, and the quotient written in the form of a fraction, of which the numerator is the product of all the letters of the dividend, not common to the divisor, and the denominator all those of the divisor not common to the dividend: thus, abc divided by amb, gives for the quotient, in observing that we sup

m

press the common factor ab, in the divisor and dividend without altering the quotient, and the division is reduced to that of, which admits of no farther

m

reduction without assigning numeral values to c and m.

35. If all the terms of a compound quantity be divided by a simple one, the sum of the quotients will be equal to the quotient of the whole compound quantity.

Thus,

86. If any power of a quantity be divided by any other power of the same quantity, the exponent of the quotient will be that of the dividend, diminished by the exponent of the divisor.

Let us occupy ourselves, in the first place, with the division of two exponentials of the same letter;

am

for instance, m and n being any positive whole

ans

numbers, so that we can have,

m>n, m=n, m<n.

It may be necessary to observe that, according to what has been demonstrated (71), with regard to exponentials of the same letter, the letter of the quotient must also be a, and if the unknown exponent of a be designated by x, then a will be the quotient, and from the nature of division,

am=an Xa2=an+x;

from which there necessarily results the following equality between the exponents,

m = n + x;

And as, subtracting n from each of these equal quantities, the two remainders are equal (Art. 49), we shall have,

Therefore, in the first case, where m is >n, the exponent of the quotient is mn; thus,

5-3

3

3-1

3=a2, and a3÷÷a—a3—1—a2. Also, it may be demonstrated in like manner, that (a + x)3÷÷(a + x)a = (a + x)5−2 = (a + x)3 ; and (2x+y)?

7

(2x+y) 5

= (2x+y) ̃—5—(2x+y)2.

In the second case, where m=n, we shall have, ama" xa*am Xa2=am+x;

From which there results between the exponents the equality,

m=m+x,

and subtracting m from each of these equals (Art:49), m―m=x, or x=0.. . . (2) ;

therefore, the exponent of the quotient will be. equal to 0, or a=a, a result which it is necessary to explain. For this purpose, let us resume the division of am by a", which gives unity for the quo

ат

tient, or =1; and as two quotients, arising from

the same division, are necessarily equal; therefore,

a°-1.

Hence, as a may be any quantity whatever, we may conclude that; any quantity raised to the power zero, must be equal to unity, or 1, and that reciprocally unity can be translated into ao. This conclusion takes place whatever may be the value of a; which may also be demonstrated in the following manner. Thus, let ay; then, by squaring each member, a° Xa0=y Xy, or a=y";

therefore, (47), y2 =y,

or y = 1;

but ay; consequently a°=1.

In the third case, where m is less than n; let n=m+d, d being the excess of n above m; we shall always have,

am=am+dXa2=am + d + x

and equalising the exponents, because the preceding equality cannot have place, but under this consideration,

m=m+d+x.