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Multiplication of Polynomials.

monomials. But, as the order of the factors may be changed at pleasure, the numerical factors may all be united in one. product.

Hence the coefficient of the product of given monomials is the product of their coefficients.

The different powers of the same letter may also be brought together, and since, by art. 6, each exponent denotes the number of times which the letter occurs as a factor in the corresponding term, the number of times which it occurs as a factor in the product must be equal to the sum of the exponents.

Hence every letter which is contained in any of the given factors must be written in the product, with an exponent equal to the sum of all its exponents in the different factors.

29. EXAMPLES.

1. Multiply a b by cde.

Ans. abcde.

2. Find the continued product of 3 ab, 2 cd, and eƒ g.

3. Multiply am by a".

Ans. 6 abcdefg.

Ans. am+n.

4. Find the continued product of 5 a3, a7, 7 a5, and 3 ao.

5. Multiply 7 a3 b2 by 10 a b5 c2 d.

6. Find the continued product of

4 abs c.

Ans. 105 a21.

Ans. 70 a4 b7 c2 d.

5 a3 b4, a2 b8, and Ans. 20 a6 b15 c.

7. Find the continued product of am br cq, an br c3, and am+n b. Ans. a2m+2 n bp +r+1 cq+s

30. Problem. To find the product of two poly

nomials.

Multiplication of Polynomials.

Solution. Denote the aggregate of all the positive terms ⚫ of one factor by A and of the other by B, and those of their negative terms respectively by C and D; and, then, the factors are A C and B-D.

Now if AC is multiplied by B it is taken as many times too often as there are units in D; so that the required product must be the product of A-C by B, diminished by the product of A-C by D; that is,

(A — C') (B — D) = (A — C') B — (A — C) D. Again, by similar reasoning, the product of A-C by B, that is, of B by A-C, must be

(AC) BAB—BC,

and that of (AC) by D must be

(A — C) D = AD — CD ;

and, therefore, the required product is, by art. 26,

(A — C) (B — D) = AB — BC— AD + CD.

The positive terms of this product, AB and CD, are obtained from the product of the positive terms A and B, or from that of the negative terms C and negative terms of the product, as -BC and

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D; but the

- AD, are

obtained from the product of the negative term of one factor by the positive term of the other, as ― C by B or Hence,

-D by A.

The product of two polynomials is obtained by multiplying each term of one factor by each term of the other, as in art. 28, and the product of two terms which have the same sign is to be affected with the sign+, while the product of two terms which have contrary signs is to be affected by the sign

The result is to be reduced as in art. 20.

Multiplication of Polynomials.

31. EXAMPLES.

1. Multiply x2 + y2 by x+y.

Ans. x3 + x2 y+x y2+y3.

2. Multiply + x y +7 a x by a x +5 a x. Ans. 6 a x6+6 a x2 y6 +42 a2 x2.

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9. Find the continued product of -a2b, c2 e, -a, -e3 x2, c,-2ax, -3abex, -7, and b3 3.

10. Find the continued product of 7 a b x, b2 x7, — 2 b, — 3, and — 5 a b3 x5.

11. Multiply a+b by c+d.

Ans. 42 a5 b5 c3 e5 x7.

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Ans. -210 a9 b7 x15.

Ans. acad+bc+bd.

12. Multiply a3b2-c by a2-b3.

a2

Ans. a5a3 b3+ a2 b2a2 c-b5+ b3 c.

13. Multiply a+b+c by a + b—c.

Ans. a2+2ab+b2 — c2.

14. Multiply 2-3x-7 by x-2.

Ans. x3-5 x2 −x + 14.

15. Multiply a2 + a1+a6 by a2 — 1. Ans. a8. a2.

16. Multiply 8 ao b3 + 36 ao b4 + 54 a7 b5 + 27 a6 b6 by 8 a9b3-36 a8 b4 + 54 a7 b5. - 27 a6 b6.

Ans. 64 a18 b6-432 a16 b8 +972 a14 b10. 729 a12 b12.

Product of Sum and Difference; of Homogeneous Quantities.

17. Find the continued product of 3x+2y, 2x-3y, −x+y, and −2 x — y.

Ans. 12x4-16 x3 y —13 x2 y2 + 11 x y3 +6 y2.

18. Multiply a + b by a
a+b b.

Ans. a2 — b2.

19. Multiply 2 a3 x+7a2 x5 by 2 a3 x — 7 a2 x5.

Ans. 4 a6 x2 - 49 a4 x10.

32. Corollary. The continued product of several monomials is, as in examples 8 and 9, positive, when the number of negative factors is even; and it is negative, as in example 10, when the number of negative factors is odd.

33. Corollary. The product of the sum of two numbers by their difference is, as in examples 18 and 19, equal to the difference of their squares.

34. Theorem. The product of homogeneous polynomials is also homogeneous, and the degree of the product is equal to the sum of the degrees of the factors.

Demonstration. For the number of factors in each term of the product is equal to the sum of the numbers of factors in all the terms from which it is obtained; and, therefore, by art. 15, the degree of each term of the product is equal to the sum of the degrees of the factors. Thus, in example 16, the degree of each factor is 12, and that of the product is 12+12 or 24.

Division of Monomials.

SECTION V.

Division

35. Problem. To divide one monomial by another.

Solution. Since the dividend is the product of the divisor and quotient, the quotient must be obtained by suppressing in the dividend all the factors of the divisor which are explicitly contained in the dividend, and simply indicating the division with regard to the remaining factors of the divisor. Hence, from art. 28,

Suppress the greatest common factor of the numerical coefficients.

Suppress each letter of the divisor or dividend in the term in which it has the least exponent, and retain it in the other term, giving it an exponent equal to the difference of its exponents in the two terms. But when a letter occurs in only one term, it is to be retained in that term, with its exponent unchanged.

The required quotient is, then, equal to the quotient of the remaining portion of the dividend divided by that of the divisor, and may be indicated as in art. 8; or, when the divisor is reduced to unity, the quotient is simply equal to the remaining portion of the dividend.

The sign of the quotient must, from art. 30, be the same as that of the divisor when the dividend is positive, and it must be the reverse of that of the divisor when the dividend is negative; whence we readily obtain the rule.

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