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When the quantities to be multiplied together have literal coefficients, proceed as before, putting the sum or difference of the coefficients of the resulting terms into a parenthesis, or under a vinculum, as in addition.

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prod. x2-(a—b)x3+(p—ab+3)x2 +(bp—3a)x+3p.

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prod. ax-(b+ac) x3+(c+be+a)x3—(c2+b)x+c

Ex. 15. Required the continual product of a+2x, a-2x', and a2+4x2.

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It may be necessary to observe, that it is usual, in some cases, to write down the quantities that are to be multiplied together, in a parenthesis, or under a vinculum, without performing the whole operation; thus, (a+2x)×(a—2x) ×(a2+ 4x). This method of representing the multiplication of compound quantities by barely indicating the operation that is to be performed on them, is preferable to that of executing the entire process; particularly when the product of two or more factors is to be divided by some other quantity; because, in this case, any term that is common to both the divisor and dividend may be more readily suppressed; as will be evident, from various instances, in the following part of the work. Ex. 16. Required the product of a+b+c by a-b+c.

Ans. a2+2ae-b2 +c2.

Ex. 17. Required the product of x+y+z by x-y-z.
Ans, x2 y2-2yz-z2.

Ex. 18. Required the product of 1-x+x2-x3 by 1+x.

Ans. 1-x4.

Ex. 19. Multiply a3+3a2b+3ab2+b3 by a2+2ab+b2. Ans. a+bab+10a3b2+10a2b3+5ab1+b".

Ex. 20. Multiply 4x2y+3xy-1 by 2x2-x.

Ans. 8x y+2x3y-2x-3x2y+x. Ex. 21. Multiply x3+x2y+xy2+y3 by x-y. Ans. x2-y. Ex. 22. Multiply 3x3-2a2x2+3a3 by 2x3-3a2x2+5a3. Ans. 6x6-13a2x2+6a2x2+21α3x3 — 19a3x2+15ao. Ex. 23. Multiply 2a2 – 3αx+4x2 by 5a2-6ɑx--- 2x2.

Ans. 10a-27a3x+34a2x2-18ax3-8x1. Ex. 24. Required the continual product of a+x, a-x, a2 +2ɑx+x2, and a2 – 2ax+x2. Ans. ao-3αax2 +3a2xa—x®. Ex. 25. Required the product of x-ax+bx-c, and a -2x+8.

Ans. x3-(a+2)x2+(b+2a+3)x3 —−(c+2b+3a)x2+(2c+36)

x-3c.

-rx

Ex. 26. Required the product of mx2-nx-r and nx―r. Ans. mnx3-(n2+mr)x2+r2. Ex. 27. Required the product of pr2-rx+q and x2-q Ans. px-(r+pr)x3 +(q+r2 —PI)x2 — q2. Ex. 28. Multiply 3x2—2xy+5 by x2+2xy-3. Ans. 3x+4x3y—4x2 × (1+y2)+16xy—15. Ex. 29. Multiply as+3a2b+3ab2+b3 by a3—3a2b+ 3ab2-b3. Ans. a-3a4b2+3a2 ba—bo. Ex. 30. Multiply 5a3-4a2b+5ab2-3b3 by 4a2-5ab+ 262.

Ans. 20a41a4b+50a3b2-45a2b3+25ab4-6b5.

§ IV. Division of Algebraic Quantities.

80. In the Division of algebraic quantities, the same circumstances are to be taken into consideration as in their multiplication, and consequently the following propositions must be observed.

81. If the sign of the divisor and dividend be like, the sign of the quotient will be +; if unlike, the sign of the quotient will be -.

The reason of this proposition follows immediately from multiplication :

+ab

Thus, if ax+b+ab; therefore

=+b:

+a

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82. If the given quantities have coefficients, the coefficient of the quotient will be equal to the coefficient of the dividend divided by that of the divisor.

Thus, 4ab2b, or

Aab

26

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2a.

For, by the nature of division, the product of the quotient, multiplied by the divisor, is equal to the dividend; but the coefficient of a product is equal to the product of the coefficients of the factors (Art. 70). Therefore, 4ab÷÷2b

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83. That the letters of the quotient are those of the dividend not cominon to the divisor, when all the letters of the divisor are common to the dividend: for example, the product abc, divided by ab, gives c for the quotient, because the product of ab by c is abc.

84. But when the divisor comprehends other letters, not common to the dividend, then the division can only be indicated, and the quotient written in the form of a fraction, of which the numerator is the product of all the letters of the dividend, not common to the divisor, and the denominator all those of the divisor not common to the dividend; thus, abc divided by amb, gives for the quotient in observing that we suppress the common factor ab, in the divisor and dividend without altering the

C

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quotient, and the division is reduced to that of, which

m

admits of no farther reduction without assigning numeral values to c and m.

85. If all the terms of a compound quantity be divided by a simple one, the sum of the quotients will be equal to the quotient of the whole compound quantity..

ab ac ad ab+ac+ad

Thus, + + =

α

α

α

a

=b+c+d.

For, (b+c+d) xa=ab+ac+ad.

86. If any power of a quantity be divided by any other power of the same quantity, the exponent of the quotient will be that of the dividend, diminished by the exponent of the divisor.

am

Let us occupy ourselves, in the first place, with the division of two exponentials of the same letter; for instance, m and n being any positive whole numbers, so that we can have,

an,

m>n, m=n, m<n.

It may be necessary to observe that, according to what has been demonstrated (1), with regard to exponentials of the same letter, the letter of the quotient must also be a, and if the unknown exponent of a be designated by x, then a* will be the quotient, and from the nature of division,

am an Xaa=ant¤;

from which there necessarily results the following equality between the exponents,

m=n+x;

And as, subtracting n from each of these equal quantities, the two remainders are equal (Art. 49), we shall have,

m-n=x....

(1).

Therefore, in the first case, where m is >n, the exponent

of the quotient is m—n; thus.

a3÷a3=a5—5—a3, and a3÷÷a=a31=a3.

Also, it may be demonstrated in like manner,

that (a+x)3÷

(a+x)2 = (a+x)5—2=(a+x)3 ; and

(2x+y)7
(2x+y)5

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(2x+y)2.

In the second case, where m=n, we shall have,

аan Xα=am Xax=am+ x;

From which there results between the exponents the equality, m=m+x,

and subtracting m from each of these equals (Art. 49), м—‚—‚¤, or x=0 . . . . (2) ;

therefore, the exponent of the quotient will be equal to 0, or a=ao, a result which it is necessary to explain. For this purpose, let us resume the division of a by a which gives

unity for the quotient, or

am

am

= 1 ; and as two quotients, arising from the same division, are necessarily equal; therefore,

a=1.

Hence, as a may be any quantity whatever, we may conclude that; any quantity raised to the power zero, must be equal to unity, or 1 and that reciprocally unity can be translated into This conclusion takes place whatever may be the value of a; which may also be demonstrated in the following

ao.

manner,

Thus, let aoy; then, by squaring each member, a° Xao== yXy, or ao=y2 ;

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y y

or y = 1;

but ay; consequently a=1.

In the third case, where m is less than n; let n=m+d, d being the excess of n above m; we shall always have,

am = am+dXa* = a +d+a,

and equalising the exponents, because the preceding equality cannot have place, but under this consideration,

m=m+d+x.

subtracting m+d from both sides, the final result will be

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