To avoid fractional coefficients, we multiply the partial remainder of the first division by 2, divide the remainder of the first division by 5. In beginning the second stage of the work, the dividend is the first divisor multiplied by 5. To avoid fractional coefficients, we multiply the partial remainder of the second division by 5, and divide the remainder of the second division by 72. The required H. C. F. is xy (x-1).

1st divisor, x3-7 x2+11 x −5)x1 —– 10 x3 +35 x2 — 50 x +24 (x−3

(1) (2)

The remainder x2-4x+3, = (x-1)(x-3), is readily factored.

Dividing 237 23 +11x-5 by x 1, we have

2-3-7 x2+11 x − 5 = (x − 1)(x2 - 6 x + 5) = (x − 1)2(x — 5).

The H. C. F. of the first remainder and (2), and therefore the required H. C. F., is x

1.

10. The examples worked in the preceding articles illustrate the following method of finding the H. C. F. of two expressions:

(i.) Remove from the given expressions any monomial factors, and set aside their H. C. F. as a factor of the required H. C. F.

(ii.) Divide the expression of higher degree in a common letter of arrangement by the one of lower degree; if the expressions be of the same degree, either may be taken as the first divisor.

(iii.) Divide the first divisor by the first remainder, the first remainder (second divisor) by the second remainder, and so on, until a remainder 0 is obtained. The last divisor will be the required H. C. F.

If a remainder which does not contain the letter of arrangement, and which is not 0, is obtained, the given expressions do not have a H. C. F. in this letter of arrangement.

(iv.) At any stage of the work the dividend may be multiplied by any number which is not a factor of the corresponding divisor; or the divisor may be divided by any number which is not a factor of the corresponding dividend.

(v.) If the divisor and dividend at any stage of the work can be factored readily, it is better to find their H. C. F. by factoring than by continuing the method of division.

11. To find the H. C. F. of three or more integral algebraic expressions find the H. C. F. of any two of them, next the H. C. F. of that H. C. F. and the third expression, and so on.

For any common factor of three or more expressions, and hence their H. C. F., must be a factor of the H. C. F. of any two of them.

EXERCISES XIV.

Find the H. C. F. of the following expressions:

1. x3 + 4x 5, x3- 2x2 + 6 x − 5.

10.

9. 3 x38x2 - 36 x + 5, 9 x3- 50 x2 + 27 x
10. 4 x3y3 3x2y2 - 4 xy + 3, 5 x3y3 + 8 x2y2 + xy
11. x3-3x2 - 2 y3, 2 x3- 5 x2y — xy2 + 6 y3.

12. a3
13. 14x3 41 x2y + 17 xy2 — 5 y3,

15. 30x825 ax2 + 8 a2x — a3, 18 x3 — 24 ax2 + 15 a2x — 3 a3.

16. 36 a6 +9 a3 - 27 a4 — 18 a5, 27 a5b2 — 9 a3b2 — 18 a1b2.

17. 3x5 10 x3 + 15 x +

-

8, x5 — 2 x1 — 6x3 + 4x2 + 13x + 6.

25. 2x4x3 + 3 x2 + x + 4, 2 x1 — 3 x3 — 2 x2 + 9 x − 12,

4x4 - 16 x3 + 25 x2 – 23 x + 4.

12. The words Highest Common Factor in Algebra refer to the degree of the common factor. Thus, the H. C. F. of

That factor is of higher degree in x than any other common factor, as x 1, x + 1.

The words Greatest Common Measure refer to the greatest numerical common measure when particular numerical values are substituted for the letters.

and

If we substitute 6 for x in the above expressions, we have

x8 - 2 x2 x + 2 = (x2 − 1) (x − 2) = 35 × 4 =

140,

x34x2.

x + 4 = (x2 − 1)(x − 4) = 35 × 2

= 70.

The arithmetical G. C. M. of 70 and 140 is evidently 70.

Now notice that when x = 6, the G. C. M. of the expressions is not the same in numerical value as the H. C. F.; for when x = = 6,

x2 - 135, not 70.

The reason for this is that while x 2 and x 4 do not have an algebraic common factor, their numerical values for particular values of x may have a common numerical factor.

Thus, when x = 6, x − 2 and x − 4 have the values 4 and 2, respectively, and therefore have the common factor 2.

The words Greatest Common Measure should not therefore be used in the same sense as the words Highest Common Factor.

13. The following principles will be of use in subsequent work :

(i.) In the process for finding the arithmetical G. C. M. of two integers, M and N, the remainder at any stage of the work can be expressed in the form

+(mM-nN),

wherein m and n are positive integers, and the upper sign goes with the first, third, etc., remainders, and the lower sign with the second, fourth, etc. Let M be greater than N. Then, in the process for finding the G. C. M., let Q1 be the quotient and R1 the remainder of the first division, Q2 and R2 the quotient and the remainder, respectively, of the second division, and so on. It is to be kept in mind that the Q's and the R's are positive integers.

Then, by Ch. III., § 4, Art. 13, we have

=(Q2Q3+1) M− (Q1 Q2 Q3 + Q1 + Q3) N. (6)

In like manner, the value of each succeeding remainder, in terms of M and N, can be derived.

In (4), m = = 1, n = Q1 ; in (5) m = Q2, n = Q1 Q2+1, and so on.

(ii.) If M and N be two positive integers, prime to each other, then two positive integers, m and n, can be found, such that

mM-nN±1.

Since M and N are prime to each other, 1 is their G. C. M. Therefore, the next to the last remainder will be 1 (the last being 0). Consequently, by (i.) two positive integers, m and n, can be found such that

±(mM― nN)= 1; or mMnN=± 1.

(iii.) If M and N be two positive integers, prime to each other, then any common factor of M and NR must be a factor of R.

Since, by Art. 7 (i.), any common factor of M and NR is a factor of mR.M-n. NR, the last equation shows that this factor is a factor of R.

The following principles follow directly from (iii.).

(iv.) If M be a factor of NR and be prime to N, it is a factor of R. (v.) If M be prime to R, S, etc., it is prime to RS

(vi.) If each of the integers M, N, P be prime to each of the integers R, S, T, then MNP is prime to RST.

(vii.) If M be prime to N, then Mr is prime to No, wherein p is a positive integer.

§ 3. LOWEST COMMON MULTIPLES.

1. The Lowest Common Multiple (L. C. M.) of two or more integral algebraic expressions is the integral expression of lowest degree which is exactly divisible by each of them. E.g., the L. C. M. of ax2, bx3, and ca1 is evidently abcx*.

L. C. M. by Factoring.

2. Ex. 1. Find the L. C. M. of a3b, a2bc2, and ab2c*.

The expression of lowest degree which is exactly divisible by each of the given expressions cannot contain a lower power of a than a3, a lower power of b than b2, and a lower power of c than c1. Therefore, the required L. C. M. is a3b2c*.

Observe that the power of each letter in the L. C. M. is the highest power to which it occurs in any of the given expressions. If the expressions contain numerical factors, the L. C. M. of these factors should be found as in Arithmetic.

Ex. 2. Find the L. C. M. of

3 ab2, 6 b (x + y)2, and 4 a2b (x − y)(x + y).

The L. C. M. of the numerical coefficients is 12.

The highest power of a in any of the expressions is a2; of b is b2; of x+y is (x + y)2; and of x-y is xy.

Consequently the required L. C. M. is 12 a2b2(x + y)2(x − y). In general, the L. C. M. of two or more expressions is obtained by multiplying the L. C. M. of their numerical coefficients by the product of all the different prime factors of the expressions, each to the highest power to which it occurs in any of them.