This is also true of two polynomials; as an illustration of which, see Example 7, Art. 61.

64. In the multiplication of two polynomials, when the partial products do not contain similar terms, if there be m terms in the multiplicand, and n terms in the multiplier, the number of terms in the product will be mXn. Thus, in Example 6, Art. 61, there are 3 terms in the multiplicand, 2 in the multiplier, and 3×2—6 in the product.

65. If the partial products contain similar terms, the number of terms in the reduced product will evidently be less than mn; see Examples 7 to 18 inclusive, Art. 61.

66. When the multiplication of two polynomials, indicated by a parenthesis, as (m+n)(p—q), is actually performed, the expression is said to be expanded, or developed.

DIVISION.

67. Division, in Algebra, is the process of finding how many times one algebraic quantity is contained in

another.

Or, having the product of two factors, and one of them given, Division teaches the method of finding the other.

The quantity by which we divide is called the divisor; the quantity to be divided, the dividend; the result of the operation, the quotient.

68. In division, as in multiplication, there are four things to be considered, viz.: the sign; the coefficient; the exponent; and the literal part.

From the foregoing illustration, we derive the following

Rule of the Signs.-Like signs in the divisor and dividend give plus in the quotient; unlike signs give minus.

70. The rule of the coëfficients, the rule of the exponents, and the rule of the literal part, may all be derived from the solution of a single example.

Required to find how often 2a2 is contained in 6a3b.

Since division is the reverse of multiplication, the quotient multiplied by the divisor, must produce the dividend; hence, to obtain this quotient, it is obvious,

1st. That the coëfficient of the quotient must be such a number, that when multiplied by 2 the product shall be 6; therefore, to obtain it, we divide 6 by 2. Hence, the

Rule of the Coefficients.-Divide the coefficient of the dividend by the coefficient of the divisor.

2d. The exponent of a in the quotient must be such a number, that when 2, the exponent of a in the divisor, is added to it, the sum shall be 5; that is, it must be 3, or 5-2. Hence, the

Rule of the Exponents.-Subtract the exponent of any letter in the divisor from the exponent of the same letter in the dividend for its exponent in the quotient.

3d. The letter b, which is a factor of the dividend, but not of the divisor, must be in the quotient. Hence, the

Rule of the Literal Part.- Write, in the quotient, every letter found in the dividend, and not in the divisor.

71. The preceding rules, taken together, give the following

GENERAL RULE FOR DIVIDING ONE MONOMIAL BY ANOTHER.

1. Prefix the proper sign, on the principle that like signs give plus, and unlike signs give minus.

2. Divide the coefficient of the dividend by that of the divisor.

3. Subtract the exponent of the divisor from that of the dividend, when the same letter or letters occur in both.

4. Annex any letter found in the dividend but not in the divisor.

1. Divide 4a5 by 2a2 and by -2a2. Ans. 2a3 and —2a3.

NOTE. In the following examples, the quantities included within the parenthesis are to be considered together, as a single quantity.

9. Divide (a+b)3 by (a+b)2..

Ans. (a+b).

. Ans. (m-n).

72. It is evident that one monomial can not be divided by another in the following cases:

1st. When the coëfficient of the dividend is not exactly divisible by the coëfficient of the divisor.

2d. When the same literal factor has a greater exponent in the divisor than in the dividend.

3d. When the divisor contains one or more literal factors not found in the dividend.

In each of these cases the division is to be indicated by a fraction. See Art. 119.

73. It has been shown, Art. 53, that any product is multiplied by multiplying either of its factors; hence, conversely, any dividend will be divided by dividing either of its factors.

Thus, 6x9÷3-2X9; or, 6X3=18.

74. Division of Polynomials by Monomials.—In multiplying a polynomial by a monomial, we multiply each term. of the multiplicand by the multiplier. Hence, conversely, we have the following

RULE FOR DIVIDING A POLYNOMIAL BY A MONOMIAL.

Divide each term of the dividend by the divisor, according to the rule for the division of monomials.

NOTE. Place the divisor on the left, as in arithmetic.

4.

5.

10a2z-15%2-25z by 5z..

Ans. 2a2-3z-5.

3ab+12abx-9a2b by -3ab. Ans. -1-4x+3a. 5x3y3—40a2x2y2+25a1xy by —5xy.

8. 3a(x+y)+c2(x+y) by x+y. Ans. 3a+c2(x+y).

9.

(b+c)(b—c)2—(b—c)(b+c)2 by (b+c)(b−c). Ans. (b−c)—(b+c)——2c.

10. b2c(m+n)—bc2(m+n) by bc(m+n)

Ans. b-c.

DIVISION OF ONE POLYNOMIAL BY ANOTHER.

75. To deduce a rule for the division of polynomials, we shall first form a product, and then reverse the operation.

The dividend, or product, and the divisor, being given, (Art. 67), it is now required to find the quotient, or the other factor.

This dividend has been formed by multiplying the divisor by the several terms of the quotient, and adding the partial products together. These several unknown terms, constituting the quotient, we are now to find.

Arranging the dividend and divisor according to the decreasing powers of the letter a, it is plain that the division of a5, the first term of the dividend, by a3, the first term of the divisor, will give a2, the first term of the quotient.

If we subtract from the dividend a5-5a4b, which is the product of the divisor a3-5a2b by a2, the first term of the quotient, the remainder +2a1b—11a3b2+5a2b3, will be the product of the divisor by the other terms of the quotient.

The first term +2a1b of the 1st remainder, is the product of the 1st term a3 of the divisor by the 1st of the remaining unknown terms of the quotient; therefore, we shall obtain the 2d term of the required quotient, by dividing +2a4b by a3; this gives +2ab.

Multiplying the divisor by +2ab, and subtracting the product, we have a 2d remainder, which is the product of the divisor by the remaining term or terms of the quotient; hence, the division of the 1st term a3b2 of this 2d remainder, by the 1st term a3 of the divisor, must give the 3d term of the quotient, which is found to be -62.

The remainder zero, shows that the quotient a2+2ab-b2 is exact, since the subtraction of the three partial products has exhausted the dividend.