divisible by a-b; also a-be, and so on for any positive value of n.

The quotients obtained by dividing the difference of the same powers of two quantities by the difference of the quan tities follow a simple law. Thus,

+bn-1.

etc.

(an—bn) ÷ (a—b)=an-1+an-2b+an-3b2+....+a2bn¬3+abn¬3

The exponents of a decrease by unity, while those of b increase by unity.

84. It may also be proved that the difference of like even powers of any two quantities is always divisible by the sum of the quantities.

Thus, (a2-b2)÷(a+b)=a—b.

(a1—b1) ÷ (a+b)—a3—a3b+ab2—b3.

(a®—b®)÷(a+b)—a3 — a1b+a3b2 — a2b3+aba—b3,

Also, the sum of like odd powers of any two quantities is always divisible by the sum of the quantities.

Thus, (a3+b3)÷(a+b)=a2—ab+b2.

(a3+b3)÷(a+b)=a*—a3b+a2b2— ab3+ba.

(a2+b2)÷(a+b)—ao — a3b+a1b2 — a3b3+a2b1—ab3+bo,

The exponents of a and b follow the same law as in Art. 83, but the signs of the terms are alternately plus and minus.

85. When exact division is impossible.-One polynomial can not be divided by another polynomial containing a letter which is not found in the dividend; for it is impossible that one quantity multiplied by another which contains a certain letter should give a product not containing that letter.

C

A monomial is never divisible by a polynomial, because every polynomial multiplied by another quantity gives a product containing at least two terms not susceptible of reduction. Yet a binomial may be divided by a polynomial containing any number of terms.

Thus, a-b1 is divisible by a2+ab+ab2+b3, and gives for a quotient a-b.

To resolve a Polynomial into Factors.

86. When a polynomial is capable of being resolved into factors, the factors can generally be discovered by inspection, or from the law of formation.

If all the terms of a polynomial have a common factor, that factor is a factor of the polynomial; and the other factor may be found by dividing the polynomial by the common factor.

EXAMPLES.

1. Resolve 3a2b2+3ab3+3ab2c into factors.

Ans. 3ab (a+b+c).

2. Resolve 5a4b2-10a3b3—5a2b1-5a2b2 into factors.

Ans. 5a2b2(a2-2ab-b2-1).

3. Resolve 6a2b2c3-12ab2c3mx-18ab2c3y into factors.
Ans. 6ab2c3 (a-2mx-3y).

4. Resolve 7a3b2-7a2b3-7a2b2c into factors.
5. Resolve 8a2bc+12ab2c-16abc2 into factors.
6. Resolve 10ab2cmx-5ab2cy+5ab2c into factors.

87. When two terms of a trinomial are perfect squares, and the third term is twice the product of their square roots, the trinomial will be the square of the sum or difference of these roots, Arts. 66 and 67, and may be resolved into factors accordingly.

EXAMPLES.

1. Resolve a2-2ab+b2 into factors. 2. Resolve a2+4ab+4b2 into factors.

3. Resolve a2-6ab+9b2 into factors.

Ans. (a—b) (a—b).

Ans. (a+2b) (a+2b).

4. Resolve 9a2-24ab+1662 into factors.
5. Resolve 25a*—60a2b3+3666 into factors.
6. Resolve 4m2n2-4mn+1 into factors.

7. Resolve 49a*b*—168a3b3+144a2b2 into factors.
8. Resolve n3+2n2+n into three factors.

9. Resolve 16a+b2-24a2bmx+9m2x2 into factors. 10. Resolve m1n2+2m3n3+m2nt into three factors.

88. If a binomial consists of two squares connected by the minus sign, it must be equal to the product of the sum and difference of the square roots of the two terms, Art. 69, and may be resolved into factors accordingly.

EXAMPLES.

1. Resolve 4a2-962 into factors. Ans. (2a+3b) (2a-3b). 2. Resolve 9a2b2-16a2c2 into factors.

3. Resolve a3x-9ax3 into three factors.
4. Resolve a1-b1 into three factors.
5. Resolve a6-b6 into its factors.
6. Resolve as-b8 into four factors.
7. Resolve 1-2 into two factors.
8. Resolve 4 into two factors.

89. If the two terms of a binomial are both powers of the same degree, it may generally be resolved into factors according to the principles of Arts. 82-84.

EXAMPLES.

1. Resolve a3-73 into its factors. Ans. (a2+ab+b2) (a—b). 2. Resolve a3+63 into its factors.

3. Resolve a6-66 into four factors.

4. Resolve a3-863 into its factors.

5. Resolve 8a3-1 into its factors.
6. Resolve 8a3-863 into three factors.
7. Resolve 1+2763 into its factors.
8. Resolve 8a3+2763 into its factors.
9. Resolve a16-616 into five factors.

CHAPTER VI.

GREATEST COMMON DIVISOR.-LEAST COMMON MULTIPLE.

90. A common divisor of two quantities is a quantity which will divide them both without a remainder. Thus 2ab is a common divisor of 6a2b2x and 10a3b3y.

91. A prime factor is one that can not be resolved into any other factors. It is, therefore, divisible only by itself and unity. Thus the quantity 2a2-2ab is the product of the three prime factors 2, a, and a-b.

92. The greatest common divisor of two quantities is the greatest quantity which will divide each of them without a remainder. It is the continued product of all the prime factors which are common to both. The term greatest here refers to the degree of a quantity, or of its leading term, and not to its arithmetical value.

93. When both quantities can be resolved into prime factors by methods already explained, the greatest common divisor may be found by the following

RULE.

Resolve both quantities into their prime factors. The continued product of all those factors which are common to both, will be the greatest common divisor required.

EXAMPLES.

1. Find the greatest common divisor of 4a2bx and 6ab2x3. Resolving into factors, we have

4a2bx=2α × 2a xbxx.

6ab2x3-2a3b × b × × × × ×x.

The common factors are 2a, b, and x. Hence the greatest

common divisor is 2abx.

2. Find the greatest common divisor of 4am2+4bm2 and San+3bn.

Resolving into factors, we have

4am2+4bm2=2m×2m (a+b).

3an+3bn=3n(a+b).

Hence a+b is the greatest common divisor.

3. Find the greatest common divisor of 23-33 and x2-y2. x3—y3=(x—y) (x2+xy+y2). x2-y2=(x-y)(x+y).

Hence x-y is the greatest common divisor.

4. Find the greatest common divisor of 35a2bmx2 and 42am2x3.

5. Find the greatest common divisor of 3a2x-6abx+3b2x and 4a2y-4b2y.

6. Find the greatest common divisor of 9mx2— 6mx + m and 9nx2-n.

7. Find the greatest common divisor of 12a2-36ab+27b2 and 8a2-1862.

94. When the given quantities can not be resolved into prime factors by inspection, the greatest common divisor may be found by applying the following principle:

The greatest common divisor of two quantities is the same with the greatest common divisor of the least quantity, and their remainder after division.

To prove this principle, let the greatest of the two quantities be represented by A, and the least by B. Divide A by B; let the entire part of the quotient be represented by Q, and the remainder by R. Then, since the dividend must be equal to the product of the divisor by the quotient, plus the remainder, we shall have A=QB+R.

Now every number which will divide B will divide QB; and every number which will divide R and QB will divide R+QB, or A. That is, every number which is a common divisor of B and R is a common divisor of A and B.

Again: every number which will divide A and B will di