65. When the two terms of a binomial are cubes, and have contrary signs, the binomial may be factored by Formula 4. Thus,

66. When the terms of a binomial are cubes, and have like signs, the binomial may be factored by Formula 5. Thus,

67. When the terms of a binomial are 4th powers, and have contrary signs, the binomial may be factored by Formula 6. Thus,

What are the factors of

(1) a-b1?

(2) 81a-166*?

(3) 16ab81c'd'?

Ans. (a+b)(a - b)(a2 + b2).

Ans. (3a+26)(3a — 2b)(9 a2+462).

Ans. (2ab+3cd)(2 ab - 3 cd)(4a2b2+ 9c2ď2).

(4) 8 a*x* - 625 c*y*?

Ans. (2ax+5cy)(2 ax — 5 cy)(4 a2x2 + 25 c2y2).

67a. Verify this formula by multiplication:

(x+a)(x+b) = x2 + (a + b)x+ab.

By it the following expressions may be factored:

(1) x2+17x+70.

(2) 22-18x+80.

(3) x +5 – 36.

Ans. (x+7)(x+10).

Ans. (x-8)(x - 10).

Ans. (x+9)(x − 4).

68. A common divisor of two quantities is a quantity that will divide each of them without a remainder. Thus, 3a2b is a common divisor of 9a2b2c and 3a2b2 — 6 a3b3.

69. A simple or prime factor is one that cannot be resolved into any other factors.

Every prime factor common to two quantities is a common divisor of those quantities. The continued product of any number of prime factors common to two quantities is also a common divisor of those quantities.

70. The greatest common divisor of two quantities is the continued product of all the prime factors which are common to both.

71. When both quantities can be resolved into prime factors by the method of factoring already given, the greatest common divisor may be found by the following rule:

Resolve both quantities into their prime factors.

Find the continued product of all the factors which are common to both, and it will be the greatest common divisor required.

Exercises.

NOTE. For convenience, G. C. D. will be used to denote greatest common divisor.

and, since the quotients have no common factor, they cannot be further divided.

5. 3a2b-9ac-18 a'xy and b'c-3bc6bcxy.

Ans. b-3c-6xy.

6. 4ac-4acx and 3 a'g - 3agx. Ans. a(a — x), or a2 — ax.

7. 4c2 - 12 cx+9x2 and 4c2 – 9 x2.

8. x3-y3 and x2 - y2.

9. 4c4bcb2 and 4c2-b2.

10. 25 ac2 -9x'y' and 5acd2+3d2x2y2.

Ans. 2c-3x.

Ans. xy.

Ans. 2c+b.

Ans. 5 ac+3x2y2.

71 a. When the quantities cannot be readily factored, another method of finding the G. C. D. is used, one of successive division, depending on the following principles:

(1) Any COMMON factor of two quantities is a factor of their G. C. D.: hence, in finding the G. C. D., any common factor that is apparent may be suppressed, and set aside as a factor of the G. C. D.

(2) No factor that is NOT common can be a factor of the G. C. D.: hence any factor in either polynomial that is not common may be suppressed and disregarded. So, also, either polynomial may be multiplied by any factor that is not contained in the other.

(3) Any quantity that will exactly divide two other quantities will divide the difference between any two multiples of those quantities.

Thus, a b will exactly divide the two quantities

a2-2ab+b2 and a2 - b2.

-

Multiply each of these quantities by some number (that is, take multiples of them), - the first by 5, and the second by 3, giving 5a2 - 10 ab +56 and 3a2-36. The difference between these multiples is 2a2 — 10 ab+8b2, and this is exactly divisible by a-b.

(4) Any quantity that will exactly divide the difference between any multiples of two quantities and one of the quantities will exactly divide the other quantity.

Thus, take the two quantities.

c3+ c2d + cd2 + d3 and c2+2cd+d2.

Take a multiple of each, multiplying the first by 6, and the second by 2c, giving

6c+6cd6cd2+6d3 and 2c+4cd+2cd2.

The difference of these multiples, 4 c3 +2cd+4cd3 +6 ď3, is divisible by c+d, and so is the quantity c2+2cd+d2; c+d will also divide the other quantity, c+c'd+cd2+d3. To show how these principles may be applied, take the two polynomials

18x18x'y+6xy-6y3 and 122-15 xy + 3y2.

These are arranged according to the descending powers of the same letter, x.

The factor 3 may be suppressed in each, and, as it is common, is to be set aside as a factor of the G. C. D. (Principle 1), leaving

6 x3 — 6 x3y + 2xy2 - 2y3 and 4x2 — 5 xy + y2.

Multiply the first of these by 2 (Principle 2), to make its first term exactly divisible by the first term of the second polynomial, and avoid fractional coefficients, and divide the result by the second polynomial. In performing this division, multiply the first remainder by 4 to avoid fractional coefficients.

19 xy2 - 19 y3 is a difference between multiples of

12x3 – 12x2y + 4 xy2 – 4 y3 and 4x2 − 5 xy + y2.

Any quantity that will divide these latter quantities must divide 19 xy2 – 19 y3 (Principle 3), and any quantity that will divide

19xy2 - 19 y3 and 4x2 - 5xy + y2

must also divide 12x3 - 12x2y + 4xy2 - 4 y3. What, then, is the G. C. D. of 19xy2 – 19 y3 and 4x2 – 5 xy + y2 ?

19y may be suppressed and disregarded (Principle 2).

4x-y

x -- y is the G. C. D. of 19xy2 – 19 y3 and 4x2 - 5xy + y2; and hence it is the G. C. D. of 12x3- 12x2y + 4xy2 - 4 y3 and 4x2 -5xy + y2. This multiplied by 3, the common factor set aside in the beginning, giving 3x-3y, will be the G. C. D. of the polynomials given.