three, if the quotient figure had been taken 5. In this manner a division is avoided each time that the nearest quotient figure is in excess, and the general rule is to take that figure for quotient which will give the remainder less than the half of the divisor.

50. If each of the terms of either of the quantities contains a factor which is not contained in each term of the other quantity, such factor forms no part of the common measure, and ought to be suppressed. This principle should be recollected at every step of the operation; but if the given quantities have a factor common to both, and that factor be suppressed in the operation, it must not be neglected altogether, because it forms a part of the common measure, and it must necessarily be introduced finally as a factor of the greatest common measure.

51. Also if the first term of any dividend is not exactly divisible by the first term of the divisor, it may be made so by multiplying the dividend by the least factor which will avoid fractional quotients. This multiplication will not affect the greatest common measure of the two quantities, because the factor thus introduced into the dividend is not found as a factor in the divisor, which has no simple factor, or if it had a simple factor, it has been divested of it.

EXAMPLES.

1. Find the greatest common measure of x3 2x2-7x+14.

Since the highest power of x is the same in both quantities we may take either as a divisor, and the work will be nearly the same in both cases.

Here in the first remainder, the simple factor common to all the terms is 2 which is suppressed as no part of the common measure of 2x2 12x+16 and -2x2-7x+14. In like manner, the simple factor 9 is rejected as no part of the common measure of x2 6x+8 and 9 x 18. The last divisor, x 2, is therefore the greatest

common measure.

2. Find the greatest common measure of 3x2 - 2xy -y and

2 x3

2 x2y +7x y2 — 7 y3.

3x2-2xy - y2) 2 x3- 2x2y + 7 x y2 — 7 y3

3

Multiplying by 3,

3x2-2xy-y) - 6 x y +69 x y° - 63 y3 (-2y

65 x y 65 y3 = 65 y2 (x − y).

Rejecting the factor 65 y2 as no part of the common measure, and dividing by x

--

y,

y is the greatest common measure of the two proposed

3. Required the greatest common measure of the two polynomials 6 x3 — 6 x2y + 2xy - 2y3 and 12x2

15 x y + 3y2.

Suppressing the factor 3 in the latter of these, and introducing the factor 2 in the former, to avoid fractions, we have

Rejecting the factor 19 y3 as no part of the common measure, we have

Hence x

y is the greatest common measure.

4. Find the greatest common measure of the polynomials

a3 − 2 a2 x + a x3 + a x2 — x1 and a3 + a x3 − a x2 + xa. Arranging the terms according to the ascending powers of x, we have a3 — 2 a2 x + a x2 + a x3 — x* ) a3 +0 a x2 + a x3 +x* (1

Hence a2

ax + x3 is the greatest common measure required.

5. Find the greatest common measure of ao· ax and a + α3 x

Here a2 is a simple factor of the former quantity, and a3 is a simple factor of the other, and the greatest common measure of a2 and a3 is

obviously a2, which must be reserved as a simple factor of the greatest common measure. We have then the following operation:

Hence a2x (a2 — x2) or a1

a2x2 is the greatest common measure.

52. Let a, b, c denote three algebraic quantities, and let d be the greatest common measure of a and b, and m the greatest common measure of c and d; then because d is a common measure of a and b, every measure of d is also a common measure of a and b; therefore every common measure of c and d is also a common measure of a, b, c; and therefore the greatest common measure of c and d, that is, m, is also the greatest common measure of a, b, c. In a similar manner the greatest common measure of four or more quantities may be found.

EXAMPLES FOR PRACTICE.

1. Find the greatest common measure of 3x2y and 12 x y2; of 6 a2x2 and 9 a x3; of 12 x2 y3 z* and 8 x* y3 zo. Ans. 3xy, 3 a x2, and 4x2 y3 z2. 2. Find the greatest common measure of 2 a2x2, 4 x2 y2, and 8x3 y; and also of 3 a"x"−1y"+1, 6 a2 x"+1y"—', and 21 a"

Find the greatest common measure

-1

2

Ans. 2x2, and 3 a"-1"-1y"-1.

+ 2 a2x2 + 2 a x3 + x1 and 5 a3 + 10 a*x + 5 a3 x2.

6. Of a3 7. Of a3 x

a3 b2 and a1

- b1.

9. Of 2x2-xy-6y2 and 3x2

10. Of 7 x 23 x y +6xy and 5- 18 x y +

11. Of a2 + b2 + c2 + 2 a b + 2 a c + 2 b c and a2 —

12. Of x2+2x+1 and x3 + h x2 + h x + 1.

13. Of y3 +5 y3 +6 and 3y+ 120 y + 117. 14. Of x3 3x2+5x - 2 and x

2x+1. 2y.

Ans. x

11xy-6y". Ans. x-3y.

2x4 Ans. x3

4x+1.

3x2 + 3x - 1.

15. Of 3+5x2 + 7x+3, x2 +3x-x-3, and 2+2-5x+3.

Ans. x + 3.

THE LEAST COMMON MULTIPLE OF TWO OR MORE QUANTITIES. 53. A multiple of a quantity is one which can be divided by it without remainder; thus 8ax is a multiple of 2a, and 16 x y3 is a multiple of 4 x y2, or of 2 x y3.

54. A common multiple of two or more quantities is one which can be divided by each of them without remainder. Thus 36 ax is a common multiple of 2 a and 9x.

55. The least common multiple of two or more quantities is the least quantity which can be divided by each of them without remainder. Thus the least common multiple of 8 a3 and 12 a* x3 is 24 a* x3; and the least common multiple of 6a2x2y3, 15 a3 x y3, and 21 ax3y, is 210 a3 x3 y3.

56. Let a and b be two quantities, m their greatest common measure, and their least common multiple; then we have,

a=hm and b = km,

where h and k have no common measure, since m is the greatest common measure of a and b; hence hk is the least common multiple of h and k; therefore the least common multiple of hm and km is ħ km; consequently the least common multiple of a and b is

Hence the least common multiple of two quantities a and b is found by dividing their product ab by their greatest common measure; or, which amounts to the same thing, divide either of the quantities by their greatest common measure, and multiply the quotient by the other quantity.

This is evident, since

ахь a

m

b

= xb== ха.

57. Let a, b, c be three quantities, and the least common multiple of a and b; then the least common multiple of 7 and c is the least common multiple of a, b, and c. For any common multiple of a and b contains l, their least common multiple, and therefore every multiple of I is a common multiple of a and b, and every common multiple of aud e is a common multiple of a, b, and e; consequently the least common multiple of and c is the least common multiple of a, b, and c.

To find the least common multiple of three quantities: find the least common multiple of two of them, and then the least common multiple of this last multiple and the third quantity will be the least common multiple of all three, and so on, if there are four or more quantities.

EXAMPLES.

1. Find the least common multiple of 15 x2 y2, 6 x3y, and 12 x y3. The greatest common measure of 15 x2 y2 and 6 x3y is (47) 3x2 y ; 15xy × 6x3 y = 5y × 6 x3 y = 30 x3 y2 the least common mul3xy tiple of 15 x2 y2 and 6x3y. Again, the greatest common measure of 30xy and 12 x y3 is 6xy; hence the least common multiple of all 30 x3 y3

three is

6 x y3

× 12xy = 5 x x 12 x y = 60 x3 y3.

.. 3xy × 2 × × × y × 5 × × × 2y = 60 x3 y3, as before.

This last method is similar to that employed in arithmetic for finding the least common multiple of any number of quantities.

2. Find the least common multiple of 8x (xy), 15 x (x − y)2 and 12 x3 (x2- y2).

.. x2 (x − y) × 3

= 120 x (x2 - y)

× 4 × 2 × 5 x2 (x − y) × (x + y)

(xy), the least common multiple.

3. Find the least common multiple of 3 x3-2x2 -x and 6x2-x-1. Suppressing the simple factor x in the former quantity, we have

Hence 3x+1 is the greatest common measure, and therefore

x x (6 x2-x-1) = x (x − 1) (6 x2 — x − 1)

= 6 x1 — 7 x3 + x = least common multiple.

=

Find the least common multiple

4. Of 8x, 10 x3y, and 12 x2 y3.

5. Of 10 (x2 + xy), 8 (xy — y2), and 5 (x2 — y2).

Ans. 120 x*y*.

Ans. 40 x y (x − y3).

6. Of 2 a2 (a + x), 4 a x (a − x), and 6 x2 (a2 − x2).

Ans. 12 ax (a2 — x2).

and a3 + a x3

x y* + y3.

a x2 + x*.

a2x2 + (a2 + a) x3 − x3.