Hence we see that the whole division might be performed by any one term of the divisor, if all the terms of the dividend which depend on that term and the quotient could be ascertained. This cannot often be done by inspection; for in many products, though at first there are as many terms as there are units in the product of the number of terms in the multiplicand by the number of terms in the multiplier, some of the terms are united together by addition or subtraction, and some disappear entirely. Even if all the terms did remain entire, they could not be easily distinguished.

However, one term may always be distinguished, and from it one term of the quotient may be obtained.

First, it is evident that the highest power of either letter in the dividend, must have been produced by multiplying the highest power of that letter in the divisor by the highest power of the same letter in the quotient; for in order to produce the dividend, each term of the divisor must be multiplied by every term of the quotient. Therefore, if 4 a1 be divided by 2 a2 it must give a term of the quotient. Or, if b1 be divided by b2 it must give

a term of the quotient. Let the quantities be arranged according to the powers of the letter a.

Dividend.

Divisor.

4 a1 — 9 a2 b2 + 6 a b3 — b1 (2 a2 — 3 a b + b2

--

2a2+3 abb2 quotient.

I divide 4 a1 by 2 a2, which gives 2 a2 for the first term of the quotient. Now in forming the dividend, every term of the divisor was multiplied by this term of the quotient, therefore I

multiply the divisor by this term, by which means I find all the terms of the dividend, which depend on this term. They are

4 a 6 a3 b + 2 a2 b2.

Here is a term 6 a3 b which is not in the dividend, this must have disappeared in the product. The term 2 a2 b2 is not found alone, but it is like 9 a2 b2 and must have disappeared by uniting with some other term to form that. I subtract these three terms > from the dividend, and there remains

6 a3 b-11 a2 b2 + 6 a b3 — bt.

which does not depend at all on the term 2 a2 of the quotient, but which was formed by multiplying each remaining term of the quotient by all the terms of the divisor. This then is a new dividend, and to find the next term of the quotient we must proceed exactly as before; that is, divide the term of the dividend containing the highest power of a, which is 6 a b, by 2 a2 of the divisor, because this must have been formed by multiplying 2 a2 by the hghest remaining power of a in the quotient. This gives for the quotient +3 a b. I multiply each term of the divisor by this, and subtract the product as before, and for the same reason. The remainder is

which depends only on the remaining part of the quotient. The highest power of a, viz. 2 a2 b2, must have been produced by multiplying some term of the quotient by 2 a2 of the divisor; therefore I divide by this again, and obtain - b2 for the quotient. I multiply by this and subtract as before, and there is no remainder, which shows that the division is completed.

By the above process I have been enabled to discover all the terms of the dividend produced by multiplying the first term of the divisor by each term of the quotient. If both be arranged according to the powers of the letter b, and the same course pursued, the same quotient will be obtained, but in a reversed older. In the division the term 2 a2 b2 has the sign Here we must observe that the divisor and quotient multiplied together must reproduce the dividend.

If a b be divided by +a, the quotient must be + b, because + a + b gives + a b.

a b be divided by +a, the quotient must be

If

-b, be

- a b.

The rule for signs therefore is the same as in multiplication. When the signs are alike, that is, both + or both

the sign of the quotient must be ; but when the signs are unlike, that is, and the other the sign of the quotient must be By the reasoning above we derive the following rule for division of compound numbers.

one

-.

Arrange the dividend and divisor according to the powers of some letter. Divide the first term of the dividend by the first term of the divisor, and write the result in the quotient. Multiply all the terms of the divisor by the term of the quotient thus found, and subtract the product from the dividend. The remainder will be a new dividend, and in order to find the next term of the quotient, proceed exactly as before; and so on until there is no remainder.

Sometimes, however, there will be a remainder, such that the first term of the divisor, will not divide either term of it; in which case the division can be continued no farther, and the remainder must be written over the divisor in the form of a fraction, and annexed to the quotient as in arithmetic.

In this example, the division may be continued until the remainder is 4 b3, which cannot be divided by a, therefore it must be written over the divisor 2 a b as a fraction and added to

The above rules are sufficient to solve all equations of the first

First, clear it of fractions by multiplying by the denomina

tors.

* Let the learner prove his results by multiplication.

Expressing the multiplication, we have

=

(a bx-2c) (3 a—b) (3) — (2 a c) (5 a) (3)

(a bx) (5 a) (3 a—b) (3) — (b2 x) (5 a) (3 a—b). Performing the multiplication it becomes

9a2b2 x
45 a3 b x

18 ac-3 a b3 x + 6 b c 30 a2c

15 a b2x 15 a2 b2 x + 5 a b3 x.

Transposing all the terms which contain x into the first member, and those which do not contain it into the second member, it becomes

9 a2 b2 x

3 a b3 x

5 a b3 x

45 a3 b x + 15 a b2 x + 15 a2 b2 x18 a c 6 b c +30 a2 c.

Uniting the terms which are alike

39 a2 b2 x 8 a b3 x 45 a3 b x= 18 a c Separating the first member into factors

6 b c +30 a3 c.

2. Find the value of x in the following equation;

3. What is the value of x in the following equation?

4. What is the value of x in the following equation?

5. What is the value of x in the following equation?