Since 3×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 x1+0·x3y + 0·x3y2 + 0·xy3—y1 = x1 — y1 = product. .. Product=6a1 — 10a3x — 22a2 x2 + 46ax3 — 20x1. (3.) Multiply 2a3 — 3ab2 + 5b3 by 2a2 — 562 Here the coefficients of a2 in the multiplicand, and a in the multiplier, are each zero; hence, Hence 4a5 16a3 b2 + 10 a2 b3 + 15ab1 — 2565 = product. The coefficient of a being zero in the product, causes that term to disappear. Or, ax-(a+b)x2+(a+b+c)x3—(a+b+c)x1+(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 ne 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. ambp a ba = аааа = aaa = a3 = a2-4 ..... Generally aaaaa..... to m factors; aa=aaa.... to n factors .... to (p-q) factors 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 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 monomia.. Divide each of the terms of the dividend separately by the divisor, and conLect the quotients with their respective signs. EXAMPLES. (1.) Divide 6a2 x1y* — 12a3 x3y®+15a 25 y3 by 3a2x2 y2. (3.) Divide "+1—x"+2+x"+3—2"+1 by x". Ans. (a+b)+2(a+b)—3. (5.) Divide 12a1y*—16a” y3+20ao y1—28a7y3 by—4a* y3. CASE III. Ans. 3y+a 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, aud 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. EXAMPLES. (1.) Divide a1-4a3 x+6a2 x2—4a x3+x1 by a2—2a x+x2. -2a3 x+5a2 x2—4a x3 -2a3 x+4a2x2-2a x3 a2x2—2a x3+x* a2x2—2a x3+x^ Arranging the terms according to the descending powers of x, we have x2—2a x+a2) x1—4a x3+6a2 x2—4a3x+a1 (x2—2a x+a2 x1—2a x3+ a2x2 —2ax3+5a2x2-4a3 x a2x2—2a3 x+a* a2 x2—2a3 x+a*. (2.) Divide x1+x2 y2+y1 by x2+xy+y2. x2+xy+y2) x'+x2y2+y1 (x2—xy+y2 (3.) Divide a3-a3 b2+2a2 b3—a b1+b3 by a2—ab+b2. a2—ab+b2) a3—a3 b2+2a2 b3—a b1+b3 (a3+a2b—ab2+· a3—a1b + a3 b2 2 a2-ab+ Arranging the terms according to powers of b, we get bra ba2) b3—a b1+2a2 b3—a3 b2+a3 (b3+a2b+ -a1b+a baba b3—a b1+ a2b3 a2 b3—a3 b2+a3 —1b+ai The results we have obtained in these two arrangements are apparently different; but their equivalence will be established as follows: (1) (a2—ab+b2) (a2+a2b—ab2) = a3—a3b2+2a2b3—a bꞌ Add remainder = (3.) Divide 12z'—192 by 3x—6. (4.) Divide 6—6y by 2x2-2y3. a-3a*b2+3a2 b1—b® by a3—3a2 b+3a b2—b3. (5) Divide (6.) Divide (7.) Divide (8) Divide a b' by a3+a2b+ab2+b3. (11.) Divide 482'—76a x2—64a2 x+105a3 by 2x—3a. (1.) Divide 2a”—6a2 ba+6a® b2”—2b3 by a"—b". —4a2 ba+6a" bTMa 2ab2b3 2a" bt—b |