Page images
PDF
EPUB

CASE IT:

I'hen the divisor is a simple quantity, and the divi

dend a compound one.

RULE.

.

-6a2,

=2;

За.

[ocr errors]

3a

92. Divide each term of the dividend separately by the simple divisor, as in the preceding case; and the sum of the resulting quantities will be the quotient required. EXAMPLE 1. Divide 18a? +3a"b+6ab2 by 3a. 18a% 3a2b

bab? -Tere, =

=ab, and

За 1843 +3a6+6ab2 therefore,

=6a2 + ab +-26%.

3a Ex. 2. Divide '20ao x3_12a2x2 + 3ax? - 2a* x2 by 2ax.

20a223 Here, =10ax, -12a2 x---2ax?=-64,

-6a 2axa 8030:-2ax? = +4a, and -2a422-2ax?

20a2x3 — 12ao x2 +8a332 - 2a4x2 hence

2ax? a-t-4a"-a. Ex. 3. Divide 20ao x– 15axo +30axy? -5ax by

[ocr errors]

-a3 ;

[ocr errors]

-10ar

5ax.

-3x,

Here 20aRx---5ax=40, -15ax? ;-5ax=30avy --5ax=6y, and -5ax---5ax=-1; therefore, 20ao x 15axa +30axya--5ax

4a_3

5ax +-6y2-1.

Ex. 4. Divide 5a®-25a5a +50a-x3-50aox +25a --5a.co by bar,

-5a"X,

2

-ar

5a6 x
-25as ca

+500
Here as,

=*
5ax
5ax

5ax
50a3x4

+25a205 +10a3r”,

10a 13,

+ 5ax

50x -5ar6 5ax", and

75; therefore, as - 5q* x +

5ax 10aor?-10° 73+502" +25 is the quotient re?.

: quired. Ex. 5. Divide 3a* x2 -3a? x4 by 3a-x”.

Ans. xsEx. 6. Divide 21a33-7a- 14ax by 7ar.

Ans. 3ao xa -2. Ex. 7. Divide 12abc-48ax?y? +64a2b2c* - 16

Зc : 3r?у? ab2 by --16ab. Ans. ab

4+ --4abc?.

4 b Ex. 3. Divide 72ra ya za — 12axy-+24bcxyz by · 12xyz.

Ans. 6xyz--a+2bc. Ex. 9. Divide 4rays -- ~*y+3axoy by xoy.

4 Ans.

xy Ex, 10. Divide 50–76+60--3ac2 +9c3 by 34,

76 Ans. +2-act-3c.

3c 3c · Ex. 11. Divide -60x'y + 50.c"y-- 400y3 + 30x4y4–20.075 +10xyo--5xy? by -5xy. Ans. yo— 2.075 +4x+y* -6x3y3 +-8.c* y* ----10x5y

+12x6. CASE III,

[ocr errors]

2

ay + 3a.

When the dividend and divisor are both compound

quantities.

RULE.

93. Arrange both the dividend and divisor accord

ing to the exponents of the same letter, beginning with the highest, and place the divisor at the right hand of the dividend ; then divide the first term of the dividend by the first term of the divisor, as in *Case I., and place the result under the divisor.

Multiply the whole divisor 'by this partial quotient, and subtract the product from the dividend, and the remainder will be a new dividend.

Again, divide that term of the new dividend, which has the highest exponent, by the first term of the divisor, and the result will be the second term of the quotient. , Proceed in the same manner as before, repeating the operation till the dividend is exbausted, and nothing remains, as in common arithmetic. This rule is evident from (Art. 88).

d

EXAMPLE 1. Divide 1205 ha - 6a4b378a2b5---4a? 34 --22aRb-+-5a7 by 4a2 ha -20b+ 504, * It can be readily perceived that the letter a is • the one to be chosen, in order to arrange the terms * of the dividend and divisor according to its powers, beginning with the dividend, 5a? is the term which contains the highest power of a; placing 5a7 for the first term, -22ab, for the second, and so on; the terms of the dividend, arranged according to the powers of a, are written thus ;

507 - 22aobrf12a5b2-6a+63-4a364 +-8a2b5. And the terms of the divisor, arranged according to the powers of a, are written thus;

5a 501-29b-4a6%.

Dividend.

Divisor: 5a7-2240b+12a: b2---6a9b3--1a3b+ +-8a2bi 5a——2a36+4a26

, 5a? + 2ab+3a,

Quotient. ai - 4aPb+263

[merged small][ocr errors][merged small][merged small][merged small][merged small]

7

UC

[ocr errors]

The sign of the first term 5a7 of the dividend being the same as that of 5a", the first term of the divisor, the sign of the first term of the quotient is +, which is omitted (Art. 14)." Dividing 5a? by 5a", the quotient is a", which is written under the divisor. Multiplying successively the three terms

the divisor by the first term a3 of the quotient, and writing the product under the corresponding terms of the dividend; subtracting 5a? - 2a*b+ 4a'ba from the dividend, the remainder is

- 20a%b+89562 -6a4b3_-4a3b4+8ab5. Dividing -20aRb the first term of this new dividend by 5a“, the result will be -4a2b, this quotient. having the sign -, because the dividend and divisor have different signs : Multiplying all the terms of the divisor by -4aRb; we have - 20aRb+8a5b2

16a4b3 ; subtracting this result from the partial dividend, the remainder will be 10ab3_-40364 + 3a2b5, dividing the first term of this new partial dividend, 16a-63, by the first term 5a* of the divisor, multiplying all the divisor by the result +20%,

and subtracting the product from the last partial, dividend, nothing remains; therefore the last term of the quotient sought is +267, and the entire quotient is a? -4a2b +263.

94: It is very proper to observe that in division, the multiplications of different terms of the quotient by the divisor, produce frequently terms which are not found in the dividend, and which it is necessary to divide afterward by the first term of the divisor. These terms are such as are destroyed when the dividend is formed by the multiplication of the quotient and divisor.

See a remarkable example of these reductions :

[ocr errors]

3

« PreviousContinue »