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b

therefore x+2a+- is the mixed quantity required.

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Here the operation is performed according to the rule. (Art. 93), and the quotient x-2y+y' is the whole quantity required.

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to a mixed quantity.

Here, a is the integral, and

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is the mixed quantity required.

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Here the remainder b is placed over the denominator x+a, and annexed to the quotient as in (Art. 89).

Ex. 5. Reduce

3a2ba6ab-2x+2c
3ab

to a mixed quantity.

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CASE III.

To reduce a fraction to its lowest terms, or most simple
expression.

RULE.

149. Observe what quantity will divide all the terms both of the numerator and denominator without a remainder: Divide them by this quantity, and the fraction is reduced to its lowest terms. Or, find their greatest common divisor, according to the method laid down in (Art. 141); by which divide both the numerator and denominator, and it will give the frac tion required.

EXAMPLE 1.

Reduce

14x3+7αx2+28x
21x2

to its lowest terms.

The coefficient of every term of the numerator and denominator of the fraction is divisible by 7, and the letter x also enters into every term; therefore 7x will divide both the numerator and denominator without a remainder.

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Here the quantity which divides both the numerator and denominator without a remainder is evidently 6abc

30a2b2c-6abc2-12a2c2b

; then

36abcx

-=5ab. -C- 2ac; and

= 6x :

6abc 5ab-c

6abc

2ac

Hence

is the fraction in its lowest terms.

6x

Ex. 3. Reduce

to its lowest terms.

a4 -b4

Here, a1-b1=(a2+b2) × (a3 —b2), (Art. 107.); and, consequently, a2-b2 will divide both the numerator and de

a2-b2

nominator without a remainder; that is,

== new

62

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Here, by proceeding according to the method of (Art. 141), we find the greatest common measure of the numerator and denominator to be x2+2ax-2a2; thus,

XC4 · 3αx3-8α2x2+18a3x-8a4

x4

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-2αx3+12α3 x 8a4
−2ax3+ 2a2x2+16a3x—12a1

X3 -ax2-8a2x+6a3

Partial quot. x- 2a

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x2+2ax—2a2)x3 — ax2—8a3x+6a3 (x−3a

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And, dividing both terms by the greatest common measure, thus found, we have the fraction in its lowest terms; but the numerator, divided by the greatest common measure, gives x -3a, as above, equal to the new numerator; and the denominator, divided by the same, gives x2 5ax+4a3; thus,

x4-3ax3- 8a2x2+18a3x-8a4
x2+2ax3

2α2 x2

x2+2ax-2a2

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X- -3a

x2-5ax+4a2

150. In addition to the methods pointed out in (Art. 144 for finding the greatest common divisor of two algebraic quantities, it may not be improper to take notice here of another method, given by SIMPSON, in his Algebra, which may be used to great advantage, and is very expeditious in reducing fractions, which become laborious by ordinary methods, to the lowest expression possible. Thus, fractions that have in them more than two different letters, and one of the letters rises only to a single dimension, either in the numerator or in the denominator, it will be best to divide the numerator or denominator (whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other: this being done, let the greatest common divisor of these two parts be found, which will evidently be a divisor to the whole, and by which the division of the other quantity is to be tried; as in the following example.

Ex. 5. Reduce est terms.

x2+ax2+bx2-2a2x+bax-2ba3
x2 bx+2ax-2ab

to its low

Here the denominator being the least compounded, and rising therein to a single dimension only; I divide the same into the parts x2+2ax, and -bx-2ab; which, by inspection, appear to be equal to (x+2a)x, and (x+2a) X-b. Therefore x+2a is a divisor to both the parts, and likewise to the whole, expressed by (x+2a) ×(x −b); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator: but the former is found to succeed, the quotient coming out x2-ax+bx—ab, x2-ax+bx-ab exactly whence the fraction is reduced to

x-b

which is not reducible farther by x-b, since the division does not terminate without a remainder, as upon trial will be found.

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Here, the greatest simple divisor of the numerator and de.

5a5b+10a4b2+5a3b3

nominator is evidently, a2b; Now,

5a3+10a b+5ab2; and

a5b+2a+b2+2a3b3+a2b1
a2b

a2b

-=a3+2ab+

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