147. The chief difficulty in finding the H. C. F. and the L. C. M. of two or more algebraic expressions consists in resolving the expressions into factors.

When two numbers in Arithmetic cannot readily be resolved into their prime factors, we divide the greater number by the smaller; then the divisor by the remainder; and so on until there is no remainder. The last divisor is the greatest common factor.

Likewise in Algebra when two given expressions cannot readily be resolved into their factors, we arrange the two given expressions in descending powers of a common letter, and divide the expression which is of higher degree in the common letter by the other expression. After the first division, we take the remainder for a new divisor and the divisor for a new dividend, and so proceed until there is no remainder. The last divisor is the highest common

factor.

NOTE. If the two expressions are of the same degree in the common letter, either expression may be taken for the divisor.

NOTE.

Each division is continued until the first term of the remainder is of lower degree than the first term of the divisor.

148. This method is of use only to obtain the compound factor of the H. C. F. Monomial factors of the given expressions must first be separated from them, and the H. C. F. of these monomial factors must be reserved to be multiplied into the compound factor obtained. Also, at any stage of the operation a monomial factor of either expression may be removed without affecting the compound factor sought. 1. Find the H. C. F. of

12x+30 x3- 72 x2 and 32x2 + 84 x2 - 176x.

12x2 + 30 x3 — 72 x2 = 6 x2 (2 x2 + 5 x − 12).
32 x3 + 84 x2 — 176 x = 4 x (8 x2 + 21 x − 44).
6x2 and 4x have 2x common.

2x2+5x12) 8 x2 + 21 x

8x2+20x

-44 (4

x+4) 2x2+5x - 12 (2 x − 3

.. the H. C.F. = 2 x (x + 4).

2 x2+8x

-3x-12

- 3 x 12

2. Find the H. C. F. of 4x2 - 8x-5 and 12x2 - 4x −65.

The first division ends here, for 20 x is of lower degree than 4x2. We take out the simple factor 10 from 20 x 50, for 10 is not a factor of the given expressions, and its rejection can in no way affect the compound factor sought, and proceed with 2x - 5 for a divisor.

3. Find the H.C.F. of

21x34x215x-2 and 21 x3- 32x2-54x-7.

21 x84x2 - 15 x − 2) 21 x3 — 32 x2 - 54 x -7(1

The difficulty here cannot be obviated by removing a simple factor from the remainder, for - 28x2 — 39 x - 5 has no simple factor. In this case, to avoid the inconvenience of fractions we multiply the expression 21 x3 — 4 x2 — 15 x - 2 by the simple factor 4 to make its first term exactly divisible by - 28 x2.

The introduction of such a factor can in no way affect the H.C.F. sought, for 4 is not a factor of either of the given expressions; and if we multiply only one of the expressions by 4 we do not introduce a common factor.

The signs of all the terms of the remainder may be changed; for if an expression A is divisible by - F, it is divisible by + F.

The process then is continued by changing the signs of all the terms of the remainder and multiplying the divisor by 4.

4. Find the H.C.F. of

8x2+2x-3 and 6 x3 + 5 x2 — 2.

The following arrangement of the work will be found.

NOTE. From the nature of division, the successive remainders are expressions of lower and lower degree. Hence, unless at some step the division leaves no remainder, we shall at last have a remainder that does not contain the common letter. In this case the given expressions have no H.C.F. that contains the common letter.

149. In the examples worked out we have assumed that the divisor which is contained in the corresponding dividend without a remainder is the H. C. F. required.

The proof may be given as follows:

Let A and B stand for two expressions which have no monomial factors, and which are arranged according to the descending powers of a common letter, the degree of B being not higher than that of A in the common letter.

Let A be divided by B, and let stand for the quotient, and R for the remainder. Then, since the dividend is equal to the product of the divisor and quotient plus the remainder, we have

A=BQ + R.

(1)

Since the remainder is equal to the dividend minus the product of the divisor and quotient, we have

Now, a factor of each of the terms of an expression is a factor of the expression. Hence, any common factor of B and R is a factor of BQ + R, and by (1) a factor of A. That is, a common factor of B and R is also a common factor of A and B.

Also, any common factor and by (2) a factor of R.

of A and B is a factor of A — BQ, That is, a common factor of A and B is also a common factor of B and R.

Therefore, the common factors of A and B are the same as the common factors of B and R; and consequently the H.C.F. of A and B is the same as the H. C. F. of B and R.

The proof for each succeeding step in the process is precisely the same; so that the H. C. F, of any divisor and the corresponding dividend is the H. C. F. required.

If at any step there is no remainder, the divisor is a factor of the corresponding dividend, and is therefore the H.C.F. of itself and the corresponding dividend. Hence, this divisor is the H. C. F. required.