An Elementary Treatise on Algebra: To which are Added Exponential Equations and Logarithms |
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Page 9
... Problem . To find the continued product of sev- eral monomials . Solution . The required product is indicated by writing the given monomials after each other with the sign of multi- plication between them , and thus a monomial is formed ...
... Problem . To find the continued product of sev- eral monomials . Solution . The required product is indicated by writing the given monomials after each other with the sign of multi- plication between them , and thus a monomial is formed ...
Page 10
... cq , an brcs , and am + nb . Ans . a2m + 2n bp + r + 1cq + 8 . 26. Problem . To find the product of two polyno- mials . Multiplication of Polynomials . Solution . Denote the aggregate of 10- [ CH . I. § IV . ALGEBRA .
... cq , an brcs , and am + nb . Ans . a2m + 2n bp + r + 1cq + 8 . 26. Problem . To find the product of two polyno- mials . Multiplication of Polynomials . Solution . Denote the aggregate of 10- [ CH . I. § IV . ALGEBRA .
Page 13
... 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 . 2 Division of Monomials . SECTION V. Division . 30. Problem CH . I. § IV . ] 13 MULTIPLICATION .
... 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 . 2 Division of Monomials . SECTION V. Division . 30. Problem CH . I. § IV . ] 13 MULTIPLICATION .
Page 14
... 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 con- tained ...
... 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 con- tained ...
Page 17
... - 1e by 2 ab - 3c5dh . Ans . 7a - 9b4c - 2d - 2ch - 1 . 14. Divide -3 am by 2 am+nbc-1 . Ans . -a - 6-1c . Division of Polynomials . 34. Problem . To divide one 2 * CH . I. § V. ] 17 DIVISION . Vinculum, parenthesis, and bar (17),
... - 1e by 2 ab - 3c5dh . Ans . 7a - 9b4c - 2d - 2ch - 1 . 14. Divide -3 am by 2 am+nbc-1 . Ans . -a - 6-1c . Division of Polynomials . 34. Problem . To divide one 2 * CH . I. § V. ] 17 DIVISION . Vinculum, parenthesis, and bar (17),
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Common terms and phrases
3d root 94 become zero approximate values arithmetical progression binomial Binomial Theorem coefficient commensurable roots common denominator continued fraction continued product Corollary courier deficient terms denote distance gone dividend equal roots equal to zero Examples of putting factor Find the 3d Find the continued Find the greatest Find the square Find the sum Free the equation gallons Geometrical Progression given equation given number gives greatest common divisor Hence imaginary roots last term least common multiple less logarithm merator monomials negative roots number of real number of terms obtained polynomial positive roots preceding article Problem proportion putting Questions quantities in example Questions into Equations quotient radical quantities ratio real root reduced remainder required equation required number Scholium Solution Solve the equation square root subtracted suppressed tained Theorem third three equations tity unity unknown quan unknown quantity whence wine
Popular passages
Page 47 - In any proportion the terms are in proportion by Composition and Division ; that is, the sum of the first two terms is to their difference, as the sum of the last two terms is to their difference.
Page 54 - There is a number consisting of two digits, the second of which is greater than the first, and if the number be divided by the sum of its digits, the quotient is 4...
Page 149 - Multiply the divisor, thus increased, by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 197 - Problem. To find the last term of an arithmetical progression when its first term, common difference, and number of terms are known. Solution. In this case a, r, and n are supposed to be known, and I is to be found.
Page 262 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.
Page 62 - A term may be transposed from one member of an equation to the other by changing its sign.
Page 44 - Arrange the terms in the statement so that the causes shall compose one couplet, and the effects the other, putting ( ) in the place of the required term. II. If the required term be an extreme, divide the product of the means by the given extreme ; if the required farm be a mean, divide the product of the extremes by the given mean.
Page 46 - Likewise, the sum of the antecedents is to their difference, as the sum of the consequents is to their difference.
Page 99 - What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes §; but the denominator being doubled, and the numerator increased by 2, the value becomes f?
Page 206 - The sum of the squares of the extremes of four numbers in arithmetical progression is 200, and the sum of the squares of the means is 136. What are the numbers ? Ans.