Divide a successively by a again, rigidly adhering to the principle that to divide any power of a by a, the exponent becomes one less, and we have

a3, a3, a1, ao, a ̄1, a ̄2, a3, &c.

Now these quotients must be equal, that is, a3 in one series equals a3 in another, and

Another illustration. We divide exponential quantities by subtracting the exponent of the divisor from the exponent of the dividend. Thus, a divided by a2 gives a quotient of a5-2=a3. a5 divided by a2= a5-7-a-2. We can also divide by taking the dividend for a numerator and the divisor for a απ 1 denominator, thus, therefore, —=a2, (Axiom 7). a2 a29

a2

From this we learn, that exponential terms may be changed from a numerator to a denominator, and the reverse by changing the signs of the exponents.

Observe, that to divide is to subtract the exponents.

(ART. 23.) When the dividend is a compound quantity, and the divisor a simple (or single) quantity, we have the following rule, the reason of which will be obvious if the preceding part of division has been comprehended.

RULE.--Divide each term of the dividend by the divisor, and the several results connected together by their proper signs will be the quotient sought.

9. Divide 3abc+12abx-3a2b by 3ab. Ans. c+4x-a. 10. Divide 25a2bx-15a2cx2+5abe by -5ax.

Ans. -5ab+3acx-bcx-1.

11. Divide 20ab3+15ab2+10ab+5a by 5a.

Ans. 463+362+26+1.

(ART. 24.) We now come to the last and most important operation in division, the division of one compound quantity by another compound quantity.

The dividend may be considered a product of the divisor into the yet unknown factor, the quotient; and the highest power of any letter in the product, or the now called dividend, must be conceived to have been formed by the highest power of the same letter in the divisor into the highest power of that letter in the quotient. Therefore, both the divisor and the dividend must be arranged according to the regular powers of some letter.

After this, the truth of the following rule will become obvious by its great similarity to division in numbers.

RULE.-Divide the first term of the dividend by the first term of the divisor, and set the result in the quotient.*

* Divide the first term of the dividend and of the remainders by the first term of the divisor; be not troubled about other terms.

Multiply the whole divisor by the quotient thus found, and subtract the product from the dividend.

The remainder will form a new dividend, with which proceed as before, till the first term of the divisor is no longer contained in the first term of the remainder.

The divisor and remainder, if there be a remainder, are then to be written in the form of a fraction, as in division of numbers.

EXAMPLES.

Divide a2+2ab+b2 by a+b.

Here, a is the leading letter, standing first in both dividend' and divisor: hence no change of place is necessary.

OPERATION:

a+b)a2+2ab+b2 (a+b

a2+ ab

ab+b2

ab+b2

That the pupil may perceive the close connection between

multiplication and division, we

Now take this product for a dividend, and one of the factors, (a2+2ab+462), for a divisor, and of course the other factor, (2a2—2ab+b2), will be the quotient, and the operation will stand thus:

a2+2ab+462)2a1+2ab+5a2b2—6ab3+4b1(2a2—2ab+b2

The several partial products which make up the dividend, and marked (1), (2), (3), are again found in the operation of division, and there marked (1), (2), (3), the same as in Arithmetic.

Some operators put the divisor on the right* of the dividend, as in the following example:

Divide a3-63 by a-b.

a3-b3 a-b

}

a2—a2b} a2+ab+b2, Quo.

a2b-b3
a2b-ab2

ab2-b3
ab2-b3

GENERAL EXAMPLES.

1. Divide a2+2ax+x2 by a+x.
2. Divide a3-3a2y+3ay2—y3 by a—y.

Ans. a+x.

Ans. a2-2ay+y2.

3. Divide 24a2b-12a3ch2—6ab by -6ab.

Ans. -4a+2a2cb+1.

4. Divide a+5a2b+5ab2+63 by a+b.

Ans. a2+4ab+b2.

*NOTE. This is in imitation of the French, and being a mere matter of taste, involving no principle, we have no right to find fault with those who adopt it; and others must not complain of us because we prefer the English custom.

5. Divide a3+2ab+2ab2+b3 by a2+ab+b2. Ans. a+b.

6. Divide x3-9x2+27x-27 by x-3.

7. Divide 6x1-96 by 6x-12.

Ans. x2-6x+9. Ans. x+2x2+4x+8.

(ART. 25.) When a factor appears in every term of both dividend and divisor, it may be cast out of every term without affecting the quotient; thus, in the last example, the factor 6 may be cast out by division; and 1-16 divided by x-2 will give the same quotient as before.

8. Divide 6a+9a2-15a by 3a2-3a.

(Observe Art. 25).

Ans. 2a2+2a+5.

9. Divide 25—x3—2x2-8x by 5x2-4x.

Ans. 5x+4x2+3x+2.

10. Divide 18a2-862 by 6a+46.
11. Divide 2x3-19x2+26x-16 by x-8.

Ans. 3a-2b.

Ans. 2x2-3x+2.

Ans. y1-y3+y2-y+1. Ans. y3+y1+y3+y2+y+1.

12. Divide +1 by y+1.
13. Divide yo-1 by y—1.
14. Divide x2-ɑ2 by x—ɑ.
15. Divide 6a3-3a2b-2a+b by 3a2-1.

Ans. x+a.

Ans. 2a-b.

16. Divide yo-3y1x2+3y2x1—xo by y3-3y2x+3yx2—x3.

Ans. y3+3y2x+3yx2+x3.

17. Divide 64a1b-25a2b8 by 8a2b3+5ab1.

18. Divide 2a1-2x1 by a-x.

Ans. 8a2b3-5ab4.

Ans. 2a+2a2x+2ax2+2x3.
Ans. (a-x).

19. Divide (a-x)5 by (a-x)2.
20. Divide a3-3a2x+3ax2-3 by a-x.

21. Divide a+1 by a+1. 22. Divide 6o-1 by b—1.

Ans. a2-2ax+x2. Ans. a1-a+a2a+1. Ans. b5+b2+b3+62+6+1.