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Addition.

SECTION II.

Addition.

21. Addition consists in finding the quantity equivalent to the aggregate or sum of several different quantities.

22. Problem. To find the sum of any given quantities.

Solution. The following solution requires no demonstration.

The quantities to be added are to be written after each other with the proper sign between them, and the polynomial thus obtained can be reduced to its simplest form by art. 20.

EXAMPLES

9x.

9.

20 2.

1. Find the sum of a and a.

Ans. 2 a. 2. Find the sum of 11 x and 9x.

Ans. 20 2. 3. Find the sum of 11 x and

Ans. 2 x. 4. Find the sum of 11 x and 9 x.

Ans.

2x. 5. Find the sum of 11 x and

Ans. 6. Find the sum of a and - 6.

Ans. a

- b. 7. Find the sum of - 6f, 9f, 13f, and — Sf. Ans. 8f. 8. Find the sum of — 126, — 46, and 13b. Ans. - 36.

9. Find the sum of vit ar — ab, ab- vī + xy, a x + xy - 4 ab, vatna- x and xy + xy + ax.

Ans. 2 vit 3 ał -4 ab + 4xy – I.

Subtraction.

10. Find the sum of 7x2 - 6 Na+ 523 % +3-g
--22-3

-8-g
- 12+ V-3232-1+78
- 2x2 +31 +373 % -1-8

22 +8 Ve-5237+9+g Ans. 4x2 +31 +2+5g.

SECTION III.

Subtraction.

23. Subtraction consists in finding the difference between two quantities.

24. Problem. To subtract one quantity from another.

Solution. Let A denote the aggregate of all the positive terms of the quantity to be subtracted, and B the aggregate of all its negative terms; then A - B is the quantity to be subtracted, and let C denote the quantity from which it it is to be taken. If A alone be taken from C, the remainder C

A is as much too small as the quantity subtracted is too large, that is, as much as A is larger than A - B. The required remainder is, consequently, obtained by increasing C- A by the excess of A above A. -B, that is, by B, and it is thus found to be C- A + B.

The same result would be obtained by adding to C the quantity A-B, with its signs reversed, so as to make it - A+B. Hence,

To subtract one quantity from another, change the signs of the quantity to be subtracted from t to ,

Multiplication of Monomials.

and from - to t, and add it with its signs thus reversed to the quantity from which it is to be taken.

EXAMPLES.

a

C.

a

5 a.

12 a. 20 a.

.5 a.

1. From
take btc.

Ans. a -b-
2. From
take - 6.

Ans. a+b. 3. From 5 a take

Ans. 10 a. 4. From na take

Ans. 5. From - 19 a take

Ans. a. 6. From 12 take 7.

Ans. 19. 7. From 2 take 5.

Ans. 7. 8. From -11

- 20.

Ans. 9. 9. From 3 a - 17b - 106 + 13 a - 2 a take 6b-82-6.- 2a+30+9a - 5 h.

Ans. 15 a - 32b-3d+5 h. 10. Reduce 32 a +36 — (5a +176) to its simplest form.

Ans. 27 a

146. 11. Reduce a +6—(2 a — 3b) — (5a +76) (- 13 a + 2b) to its simplest form. Ans. 7a-56.

take

SECTION IV.

Multiplication.

25. Problem. To find the continued product of several monomials.

Solution. The required product is indicated by writing the given monomials after each other with the sign of multiplication between them, and thus a monomial is formed, which is the continued product of all the factors of the given

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.

EXAMPLES.

1. Multiply a b by cde.

Ans. abcde. 2. Find the continued product of 3 ab, 2 c d, and efg.

Ans. 6 abcdefg. 3. Multiply am by an.

Ans. amtn. 4. Find the continued product of 5 a3, a?, 7 as, and 3 a6.

Ans. 105 221. 5. Multiply 7 a3 b2 by 10 a b5c2d. Ans. 70 a4 b7c2d.

6. Find the continued product of 5 a 64, a2 b8, and 4 a 63 c.

Ans. 20 a6 615 c. 7. Find the continued product of am bp c9, an brcs, and am tnb.

Ans. a2m +2n bp+r+1c9+8.

26. Problem. To find the product of two polynomials.

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 A - C 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

(A -C) B=AB - BC, and that of (A

C) by D must be

(A C) D=AD - CD; and, therefore, the required product is, by art. 24,

(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 D; but the negative terms of the product, as - BC and - 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 - D by A. Hence,

The product of two polynomials is obtained by multiplying each term of one factor by each term of the other, as in art. 25, 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.

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