CASE II.

(48.) To divide one polynomial by another, we have this

RULE.

I. Arrange the dividend and divisor with reference to a certain letter, then divide the first term on the left of the dividend by the first term on the left of the divisor, the result is the first term of the quotient; multiply the divisor by this term, and subtract the product from the dividend.

II. Then divide the first term of the remainder by the first term of the divisor, which gives the second term of the quotient ; multiply the divisor by this second term, and subtract the product from the result after the first operation. Continue this process until we obtain 0 for remainder, or when the divisor does not terminate, which is frequently the case, we can carry on the above process as far as we choose, and then place the last remainder over the divisor forming a fraction, which must be added to the quotient.

EXAMPLES.

1. What is the quotient of 2ab+b3+2ab2+a3 divided by a2+b2+ab?

Arranging the terms according to the powers of a, and operating agreeably to the above rule, we have

OPERATION.

Dividend-a3+2a2b+2ab2+b3a2+ab+b2=divisor

2. What is the quotient of a2b-3a2+2ab-6a-46+22

6

3

10 remainder.

3. Divide x-x+x3-x2+2x-1 by x2+x-1.

4. What is the quotient of x3-3x2+3a2x-a3 divided

5. Divide 14af-21bf+7cf+6ag-9bg+3cg by 7f+3g.

Ans. 2a-3b+c.

6. Divide 4x3+4x2-29x+21 by 2x-3.

7. Divide

Ans. 2x2+5x-7.

119c2-200cd+408ce-113ch-39d2+72de

+37dh-96eh+20h2 by 17c+3d-4h.

Ans. 7c-13d+24e-5h.

8. Divide 72x4-78x3y-10x2y2+17xy3+3y by 6x2

—4xy—y2.

9. Divide 36a2b-63ab2+2063 by 12ab-562.

Ans. 12x2-5xy—3y2.

Ans. 3a-4b.

Ans. a+b.

4

11. Divide a⭑-ba by a−b.

Ans. a3+ab+ab2+b3.

(49.) The following examples can not be accurately performed, there being still a remainder, however far the division be carried,

12. Dividing 1 by 1-6, we have in succession

b 1=(1-6)=1+

1-6

2r3

CHAPTER II.

ALGEBRAIC FRACTIONS.

(50.) In our operations upon algebraic fractions, we shall follow the corresponding operations upon numerical fractions, so far as the nature of the subject will admit.

CASE I.

To reduce a monomial fraction to its lowest term, we have this

RULE.

I. Find the greatest common measure of the coefficients of the numerator and denominator. (See Art. 11, Arithmetic.)

II. Then, to this greatest common measure, annex the letters which are common to both numerator and denominator, give to these letters the lowest exponent which they have, whether in the numerator or denominator : the result will be the greatest common measure of both numerator and denominator.

III. Divide both numerator and denominator by this greatest common measure, (by Rule under Art. and the resulting fraction will be in its lowest terms.