2. 4ab)8ab√12a382 +4ab(24 x —3a2b+1 quo

When the divisor and dividend are both compound quantities.

RULE

1. Range the terms according to the powers of some letter in both of them, placing the highest power of it first, and the rest in order.

2. Divide the first term of the dividend by the first term of the divisor, and place the result in the quotient.

3. Multiply the whole divisor by the quotient term, and subtract the product from the dividend.

4. To the remainder bring down as many terms of the dividend as are requisite for the next operation; call the sum a dividual, and divide as before; and so on, as in Arithmetic.

EXAMPLES.

EXAMPLES.

2

1. Let it be required to divide a3-3a1x—3ax*x*

by afx.

a+x)a 3 —3a * x—3ax2 +x3 (a*—4ax+x2

a3 +a2x

-4a2x-3ax2 first dividual,

-4a2x—4ax*

axx second dividual. + ax2+x3

2. Divide

The process may be explained thus:

First, a divided by a gives a2 for the first term of the quo tient, by which we multiply the whole divisor, viz. a+x, and the product is a' tax, which, being taken from the two first terms of the dividend, leaves -4a2x; to this remainder we bring down -3ax, the next term of the dividend, and the sum is —4a2x—3ax2, the first dividual; now dividing -4a2x, the first term of this dividual, by a, the first term of the divisor, there comes out -4ax, a négative quantity, which we also put in the quotient; and the whole divisor being multiplied by it, the prod uct is —4a1×—4ax2, which being taken from the first dividual, the remainder is +ax'; to which we bring down x', the last term of the dividend, and the sum is +ax2+x3, the second dividual; and +ax', the first term of the second dividual, divided by a, the first term of the divisor, gives x2 for the last term of the quotient; by which we multiply the whole divisor, and the product is 4-ax2+x3, which being taken from the second dividual leaves nothing; and the quotient required is a1----qux + x3.

*Here a, the first term of the dividend, being divided by a, the first term of the divisor, gives for the first term of the quotient. For a=a1, and a = a, and the difference

of the exponents is I-, or; therefore a divided by a gives a1a =√, as above. Or it may be considered thus: ask what quantity being multiplied by ✔a will give a, and the answer is a; then the divisor being multiplied by √, the product is a-ab; but there being no term in the dividend,

that

4. Divide à3—3a2c+4ac2 —c3 by a2—2ac-fc3·

3

ac2-c3

«3—2ac+c2)a 3 —3a2c+4ac3——2c2 (a—c+a2—2äc+c2

Here it is obvious, that the division cannot terminate without a remainder; therefore we write the divisor under the remainder with a line between them, and add the fraction to ac, the other two terms, to complete the quotient.

But when the dividend does not precisely contain the divisor, then we generally express the whole quotient as a fraction, having reduced it to its lowest terms, or rejected the letters and factors, that are found in every term of the dividend and divisor.

5. Thus,

that corresponds to ab, the second term of this product, we subtract aab from a-b, the dividend, and the sign of the quantity ab being changed, the remainder is +✔ab -b. Now +ab, the first term of this remainder, divided by ã,

the first term of the divisor, gives — for the second term of the quotient, by which we multiply the divisof, and the prod

uct, viz. abb, being subtracted from the aforesaid re mainder, nothing remains; and the quotient is

5. Thus, a'bx+acx+ax3 divided by adx+an

a*bx+acx2+ax3 ab+cx+x

gives

, or

adx+anx

d+n

Here the quotient

a'bx+acx2+ax3
adx+anx

is reduced to

ab+cx+x2

d-f-n

-, by dividing every term of its numerator and

denominator by ax.

6. And a+ab+d2 divided by aac+a2c2 gives a+ab+d2

a2 — ac+a2c2

Here the quotient cannot be reduced to lower terms, be cause the factor a is not to be found in the term d2.

But it is to be observed, that though a fraction cannot be reduced to lower terms by a simple divisor, yet it may sometimes be so reduced by a compound one; as will appear in the reduction of fractions.

8. Divide a3----3a2+3ay”—y3 by a—y.

Ans. a2-ax+x2.

Ans. a*—2ay+y'.

9. Divide 6x4-96 by 3-6.

Ans. 2x3-4x+8x+16.

2

10. Divide a3-5a4x+10a3x2-10a1x3+5ax-x5 by

FRACTIONS.

ALGEBRAIC FRACTIONS have the same names and rules

of operation, as fractions in Arithmetic.

PROBLEM