2d. If no two of the partial products are similar, there will be no reduction among the terms of the entire product: hence, The total number of terms in the entire product will be equal to the number of terms in the multiplicand multiplied by the number of terms in the multiplier.

This is evident, since each term of the multiplier will produce as many terms as there are terms in the multiplicand. Thus, in example 16th, there are three terms in the multiplicand and two in the multiplier: hence, the number of terms in the product is equal to 3 × 2 = 6.

3d. Among the different terms of the product, there are always some which cannot be reduced with any others. For, let us consider the product with reference to any letter common to the multiplicand and multiplier. Then, the irreducible terms are,

1st. The term produced by the multiplication of the two terms of the multiplicand and multiplier which contain the highest exponent of this letter; and the term produced by the multiplication of the two terms which contain the lowest exponent of this letter. For, these two partial products will contain this letter, affected with a higher and lower exponent than either of the other partial products, and consequently, they cannot be similar to any of them. This remark, the truth of which is deduced from the law of the exponents, will be very useful in division.

If we examine the multiplicand and multiplier, with reference to a, we see that the product of 5a4b2 by a2b, must be irreducible; also, the product of 2ab3 by ab2. If we consider the letter b, we see that the product of ab4 by - ab2, must be irreducible, also that of 3a2b by a2b.

46. We will apply the rules for the multiplication of algebraic quantities in the demonstration of the following theorems.

THEOREM I.

The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second.

Let a denote one of the quantities and b the other: then

Now, we have from known principles,

(a + b)2 = (a + b) x (a + b) = a2 + 2ab + b2,

which result is the enunciation of the theorem in the language of Algebra.

To apply this result to finding the square of the binomial

Also,

also,

(6a2b+ 9ab3) = 36a8b2 + 108a5b4 + 81a2b6 ;
(8a3+7acb)2 =.

THEOREM II.

The square of the difference between two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second.

Let a represent one of the quantities and b the other: then their difference.

a-b=

Now, we have from known principles,

2

(a - b)2= (a - b) x (a - b) = a2

2ab + b2,

which is the algebraic enunciation of the theorem. To apply this to an example, we have

168a365 + 144a266.

Also,

(7a2b2 - 12ab3) = 49a4b4
(4a363 - 7c2d3)2 =

THEOREM III.

The product of the sum of two quantities multiplied by their

difference, is equal to the difference of their squares.

Let the quantities be denoted by a and b.

Then, a+b = their sum, and a

We have, from known principles,

b=

their difference.

(a + b) x (a – b) = a2 – b2,

which is the algebraic enunciation of the theorem. To apply this principle to an example, we have

Also,

(8a3 + 7ab2) × (8a3 7ab2) = 64a6 49a264.
(9ac + 7ab5) × (9ac - 7ab5) =

47. By considering the last three results, it will be perceived that their composition, or the manner in which they are formed from the multiplicand and multiplier, is entirely independent of any particular values that may be attributed to the letters a and b which enter the two factors.

The manner in which an algebraic product is formed from its two factors, is called the law of this product; and this law remains always the same, whatever values may be attributed to the letters which enter into the two factors.

Of factoring Polynomials.

48. A given polynomial may often be resolved into two factors by mere inspection. This is generally done by selecting all the factors common to every term of the polynomial for one factor, and writing what remains of each term within a parenthesis for the other factor.

1. Take, for example, the polynomial

ab + ac;

in which, it is plain, that a is a factor of both terms: hence

ab + ac = a (b + c).

2. Take, for a second example, the polynomial

ab2c + 5ab3 + ab2c2.

It is plain that a and b are factors of all the terms: hence

ab2c + 5ab3 + ab2c2 = ab2 (c + 5b + c2),

3. Take the polynomial 25a4

30a3b + 15a2b2; it is evident

that 5 and a2 are factors of each of the terms. We may, therefore, put the polynomial under the form

5a2 (5a2 - 6ab + 362).

4. Find the factors of 3a2b + 9a2c + 18a2xy.

Ans. 3a2 (b+3c + 6xy).

5. Find the factors of 8a2cx - 18асх2 + 2ac5y 30a6c9x.

Ans. 2ac (4ax 9x2 + c2y - 15a5c8x).

6. Find the factors of 24a2b2cx - 30a8b5c6y + 36a7b&cd+6abc. Ans. 6abc (4abx - 5a7bc5y + 6ab7d + 1).

7. Find the factors of a2 + 2ab + b2.

Ans. (a + b) x (a + b).

(a - b) x (a - b). 6a3b + 8a2b5

16ab2

11. Find the factors of the polynomial 15abc2 3bc2 + 9a3b5c6 12db6c2.

12. Find the factors of the polynomial 25a6bc6 30abc+d

5ac4

60ac6.

13. Find the factors of the polynomial 42a2b2

Ans. 7ab (6ab

7abcd + 7abd. cd + d).

DIVISION.

49. DIVISION, in Algebra, explains the method of finding from two given quantities, a third quantity, which multiplied by the first shall produce the second.

The first of the given quantities is called the divisor: the second, the dividend; and the third, or quantity sought, the quotient. Let us first consider the case of two monomials, and divide 35a3b2c by 7ab.

The division may be indicated thus,

Now, since the quotient must be such a quantity as multiplied by the divisor will produce the dividend, the co-efficient of the quotient multiplied by 7 must give 35, the co-efficient of the dividend; hence, the new co-efficient 5 is found by dividing 35 by 7. Again, the exponent of any letter, as a, in the quotient, added to the exponent of the same letter in the divisor, must give the exponent of this letter in the dividend: hence, the exponent in the quotient is found by subtracting the exponent in the divisor from that in the dividend. Thus, the exponent of

a is 3-1 = 2, and of b, 2 − 1 = 1,

and since c is not found in the divisor, there is nothing to be subtracted from its exponent.

50. Hence, for the division of monomials, we have the following

RULE.

I. Divide the co-efficient of the dividend by the co-efficient of the divisor, for a new co-efficient.

II. Write after this co-efficient, all the letters of the dividend, and affect each with an exponent equal to the excess of its exponent in the dividend over that in the divisor.

51. It follows from the preceding rule that the exact division of monomials will be impossible.

1st. When the co-efficient of the dividend is not divisible by that of the divisor.

2d. When the exponent of the same letter is greater in the divisor than in the dividend.

3d. When the divisor contains one or more letters which are not found in the dividend.

When either of these three cases occurs, the quotient remains under the form of a monomial fraction; that is, a monomial expression, necessarily containing the algebraic sign of division. Such expressions may frequently be reduced.

Here an entire monomial cannot be obtained for a quotient; for, 12 is not divisible by 8, and moreover, the exponent of c is less in the dividend than in the divisor. But the expression can be reduced, by dividing the numerator and denominator by the factors 4, a2, b, and c, which are common to both the terms of the fraction.