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tient, and a quantity multiplied by 1 gives that same quantity for product, it follows, that the letters which are common to the divisor and dividend, with the same exponent in each, will not appear in the quotient.

4th, The letters which are in the dividend, and not in the divisor, will appear as multipliers in the quotient; while letters which are in the divisor, and not in the dividend, will appear as denominators of a fraction in the quotient.

5th, When the same letter is in both dividend and divisor, with different exponents, it will appear in the quotient with an exponent equal to the difference of its exponents, and in the denominator of a fraction, when the exponent of the divisor is the greater; thus, a@cat=a2, a --a’=a3, a'ra=a-3, or

a" -- a" =a 33. CASE I. When the dividend and divisor are both simple quantities.

RULE. Place the dividend as the numerator of a fraction, and the divisor as its denominator, then divide by the above general rules. EXAMPLE. Divide 14a2b3c, by 2aobc.



2a2bc Proof. 2a2bc X 762=14a2b3c. 1. Divide 8a^x, by —2ax.

Ans. -4ax. 2. Divide —27p?mx*, by 9pmx. Ans. —3px. 3. Divide -84x+y52, by — 12x3yoz. Ans. 7xyz.

4ay 4. Divide 120a-xy4, by 30axy3.

Ans. 5. Divide —40x2972

, by 8x2,2,1. Ans. -5x22. 6. Divide 36x?y3z, by 20.x

34. CASE II. When the dividend is a compound, and the divisor a simple quantity.

RULE. Divide each term of the dividend by the divisor, as in the last case, and the sum of the separate quotients will be the answer. EXAMPLE. Divide 3a+b+ 4a3bc, by 4ab.

= ataʼc. Ans.

Divide 12a’léc—24a123_10a+12%, by 3abc.
-—= fabł_


Ans. 3abc


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1. Divide 3a4_6a3+4a2_8a, by 2a”.

Ans. 3 a2–3a+2 2. Divide 1203–36a2b +36ab2_1213, by 4ab.

Ans. -9a +96–

b 3. Divide 48a-b2r436a2622+1401Px4, by 2abx.

Ans. 24a3b-18abx + 762x3. 4. Divide 40p2x2y2–30p*r3y2–60p2x3y4, by-10p2ry.

Ans. -4cy +3p%,'y + 6xy. 5. Divide 723622–12a3vc-14a12., by 4a yet.


4 6. Divide 4a2c2_8a2cx +4a2x”, by 4a2.

Ans. (2_2cx + x2. 35. Case III. When both dividend and divisor are compound quantities.

Rule. Place the quantities as in division in arithmetic, arranging both dividend and divisor according to the powers of the same letter. Divide the first term of the dividend by the first term of the divisor, and put the result with its proper sign for the first term of the quotient. Multiply the terms of the divisor by this quotient, and subtract the product from the dividend; to the remainder bring down as many terms of the dividend as may be necessary, then divide as before, and so on till the work is finished.


Divisor. Dividend.

a+x)a+ 3a2x+3aco +x5(a2 + 2ax+xo
a taox

2aRx+ 2axo

ax2 + x3

ax2 + x3 Divisor. Dividend.

Quotient. a'—2ax+xo)a4_4a5x+60°x?_4axs +x(a2--2ax + x2

a4--2axta xo
-2ax +5ax"-4ax:
-2a3x + 4aoxo_2ax:

axo2ax+ x4
apx" - 2ax3 + x4

Divisor. Dividend. Quotient.
x-y)x+-yo(x3 +xy + xy +y3
24—25 y


x+yo —y4


xy3 — y4 36. Sometimes the quotient never terminates, but may be carried out in an infinite series. Divisor. Dividend.

1-21 (1+3+x2 + x3 + x4 + &c.



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х5 Here it is evident the quotient would never terminate, but go on to infinity, the power of x increasing by unity at

This quotient gives several interesting series, by substituting fractional values for x; for example, if we ma successively equal to , , , , , we will have the following results :(1.)

=2=l+}+*+&tio, &c. to infinity. (2.)

=1+3+1+2+3+ &c. to infin.

=1+1+16+01+zło + &c. to infin. (4.)

-3=1+3+3 ++++ + &c. to infin. (5.)

-4=l+*+ i +67+01't + &c. to infin. (6.) {4th—2nd gives {=}+1+11+}i,+&c. toinfin. (7.) 5th—3rd gives =i+&+81+ **+ &c. to inf.




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1. Divide a5 +5a4x+10a3x2 +10a x5 + 5ax4+x5 by 2° +2ax +%.

Ans. as +3aRx+3axo+x3. 2. Divide a5_5a4 + 10a5-10a% +5al by ay—3a2 +31–1.

Ans. a-2a +1. 3. Divide x4-a4 by x3 +xa txao tay.

Ans. I-a. 4. Divide x5 +y5 by x4-2y + xoyo-wys +yo.

Ans. x+y.

5. Divide 926–46x5 +95x2 + 150x by x2—4x–5.

Ans. 94-10x5 +5.0-30x. 6. Divide 25x6—24-203-8x2 by 5x3—4x2.

Ans. 503 +4c? +3x+2. 13

4 7. Divide x4. -23 + x2 + 3x—2 by zx


. 6

Ans. xs-42° +1. 37. Verify by division the truth of the following expressions :

(1.) =x3 + xa txa? + ab.



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(2.) cco_coacao<ao.

sta 25

-a5 (3.) =x4+xa+xa® +xa3+a. -a5

2a5 (4.) =x4—230 +a?_xa3 +a4xta

xta 24+a4

2a4 (5.) =x3 + x2 +xa? tas +, xhtat

2a4 . (6.)

=x_c2a +xalmas + xta

xta 25 +a5 (7.)

=x4 + x3a + xoa? +xa3 +24+ zoo tas (8.) =x4-25atxoa?_xas + a4.

ata The eight expressions given above are particular illustrations of the four following

Theorems : 38. THEOREM I. The difference of the same powers of two quantities is always divisible by the difference of the quantities themselves, whether the exponents be even or odd. See 1st and 3d.

39. THEOREM II. The difference of the same powers of two quantities is divisible by the sum of the quantities without a remainder, if the exponent be even, but not if the exponent be odd. See 2d and 4th.

40. THEOREM III. The sum of the same powers of two quantities is never divisible by the difference of the quan


tities without remainder, whether the exponents be odd or

See 5th and 7th. 41. THEOREM IV. The sum of the same powers of two quantities is divisible without remainder, by the sum of the quantities, if the exponent be odd, but not if it be even. See 8th and 6th.


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42. The Rules for the management of Algebraic Fractions are the same with those in common arithmetic.

Case I. To reduce a mixed quantity to the form of a fraction.

RULE. Multiply the integral part by the denominator of the fraction, and to the product annex the numerator with the same sign as the fraction; under this sum place the former denominator, and the result is the fraction required. EXAMPLE. Reduce 2a +ő to the form of a fraction.

2axb+x 2ab+x

b Here 2a, the integral part, is multiplied by b, and x, the numerator, is added to the product, the sum of which forms the numerator; under which we write the former denomi

2ab+x nator b, which gives the fraction required.

b 43. If the numerator be a compound quantity, and the fraction be preceded by the sign minus, the signs of all its terms must be changed by the rule for subtraction, (Art.

-3x ac+cx—2a+3.x 24); thus, a+x

x2 1. Reduce 3a + to the form of a fraction.

3a" +

Ans. 2. Reduce a_b +

to the form of a fraction. d

adbd+ac Ans.

d 3. Reduce 150° + x- to the form of a fraction.



ab 2ax?_23 4. Reduce 3ax + xo.

to the form of a fraction. bc

3abcx+bcx®--2ar? +23 Ans.







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