CASE III. 13. When both multiplicand and multiplier are compound quantities. Multiply each term of the multiplicand, in succession, by each term of the multiplier, and the sum of these partial products will give the complete product. (6) Multiply 4a3-5a2b—8ab2+2b3 by 2a2—3ab-4b2. 4a3- 5a2b- 8ab2+ 2b3 2a2- 3ab- 4b2 8a5-10a1b-16a3b2+ 4a2b3 -12ab+15a3b2+24a2b3— 6ab1 -16a3b2+20a2b3+32ab1-865 8a5—22a1b—17a3b2+48a2b3+26ab1—8b5 = product. ́ (7) Multiply a'b—ab' by h'k—hk'. a'b-ab' h'k-hk' a'bh'k-ab'h'k -a'bhk'+ab'hk' a'bh'k—ab'h'k—a'bhk'+ab'hk' = product. G (8) Multiply +xm1y+xm_2y2+x-3y3+, &c., by x+y. 2m+xml+xm¬2y2+xTM−3+y3+..... (9) Multiply x2+y2 by x2—y2. (11) Multiply 5a1-2ab+4a2b2 by a3-4a2b+2b3. (16) Multiply x2+y2+z2—xy—xz-yz by x+y+z. MULTIPLICATION BY DETACHED COEFFICIENTS. 14. In many cases the powers of the quantity or quantities in the multiplication of polynomials may be omitted, and the operation performed by the coefficients alone; for the same powers occupy the same vertical columns, when the polynomials are arranged according to the successive powers of the letters; and these successive powers, generally increasing or decreasing by a common difference, are readily supplied in the final product. Since x3×x=x, the highest power of x is 4, and decreases successively by unity, while that of y increases by unity; hence the product is x2 + 0·x3 y + 0·x23 y2 + 0·x y3 — y1 = x1— y1 = product. .. Product 6a1 — 10a3x — 22a2 x2 + 46ax3 — 20x1. (3.) Multiply 2a3 — 3ab2 + 5b3 by 2a2 — 5b2 Here the coefficients of a2 in the multiplicand, and a in the multiplier, are each zero; hence, 4 + 0 −16 + 10 + 15 — 25 Hence 4a3 - 16a3 b2 + 10 a2 b3 + 15ab1. - 2565 = product. The coefficient of a1 being zero in the product, causes that term to dis Or, ax-(a+b)x2+(a+b+c)x3—(a+b+c)x2+(a+b+c)x3—(b+c)x® +cx2. DIVISION. 15. THE object of algebraic division is to discover one of the factors of a given product, the other factor being given; and as multiplication is divided into three cases, so in like manner division is also divided into the three following cases. (1) When both dividend and divisor are monomials. (2) When the dividend is a polynomial, and the divisor a monomial. (3) When both dividend and divisor are polynomials. CASE I. 16. When both dividend and divisor are monomials. Write the divisor under the dividend, in the form of a fraction; cancel like quantities in both divisor and dividend, and suppress the greatest factor common to the two coefficients. 17. Powers of the same quantity are divided by subtracting the exponent of the divisor from that of the dividend, and writing the remainder as the exponent of the quotient. From this reasoning it follows that every quantity whose exponent is 0, is equal to 1. But we may subtract 5, the greater exponent, from 3, the less, and affect the difference with the sign —; hence a3 For more information on negative exponents, see a subsequent article. 18. In multiplication, the product of two terms, having the same sign, is affected with the sign +; and the product of two terms, having different signs, is affected with the sign -; hence we may conclude, (1.) That if the term of the dividend have the sign+, and that of the divisor the sign+, the resulting term of the quotient must have the sign +. (2.) That if the term of the dividend have the sign +, and that of the divisor the sign -> the resulting term of the quotient must have the sign (3.) That if the term of the dividend have the sign -> and that of the divisor the sign +, the resulting term of the quotient must have the sign (4.) That if the term of the dividend have the sign and that of the divisor the sign —, the resulting term of the quotient must nave the sign + . -. 19. When the dividend is a polynomial, and the divisor a monomial. Divide each of the terms of the dividend separately by the divisor, and connect the quotients with their respective signs. EXAMPLES. (1.) Divide 6a2 x1y — 12a3 x3 y + 15a1 x5 y3 by 3a2 x2 y2. 6a2 x4y-12a3 x3 y° +15a1 x5 y3 = 2x2 y1-4a x y1+5a2 x3y. 3a2 x2 y2 Ans. x-xx3—x1. (3.) Divide "+1—x2+2+x1+3—x2+1 by x2. Ans. (a+b)2+2 (a+b)—3. (5.) Divide 12a1y-16a y+20ay-28a7 y3 by-4a1y3. CASE III. Ans. 3y+4a y2-5a2y+7a3. 20. When both dividend and divisor are polynomials. 1. Arrange the dividend and divisor according to the powers of the same letter in both. 2. Divide the first term of the dividend by the first term of the divisor, and the result will be the first term in the quotient, by which multiply all the terms in the divisor, and subtract the product from the dividend. 3. Then to the remainder annex as many of the remaining terms of the dividend as are necessary, and find the next term in the quotient as before. |