A Treatise on AlgebraHarper & brothers, 1846 - 346 páginas |
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Página 31
... have 3abc5def But according to the principle stated in the preceding Article , this result may be written 3 × 5abcdef , or 15abcdef . Hence we deduce the following RULE . Multiply the coefficients of the two terms together.
... have 3abc5def But according to the principle stated in the preceding Article , this result may be written 3 × 5abcdef , or 15abcdef . Hence we deduce the following RULE . Multiply the coefficients of the two terms together.
Página 32
... Hence , number 3 showing that a enters three if the same letters are found in two monomials which are to be multiplied together , the expression for the product may be abbre- viated by adding the exponents of the same letters . Thus ...
... Hence , number 3 showing that a enters three if the same letters are found in two monomials which are to be multiplied together , the expression for the product may be abbre- viated by adding the exponents of the same letters . Thus ...
Página 33
... Hence we deduce the following RULE . Multiply each term of the multiplicand separately by the multiplier , and add together the products . EXAMPLES . Multiply 3a + 2b a2 + 2x + 1 By 4a 4x xy 3y2 + 5xy + 2 3x3 + xy + 2y2 5x2y Product ...
... Hence we deduce the following RULE . Multiply each term of the multiplicand separately by the multiplier , and add together the products . EXAMPLES . Multiply 3a + 2b a2 + 2x + 1 By 4a 4x xy 3y2 + 5xy + 2 3x3 + xy + 2y2 5x2y Product ...
Página 34
... hence if - a is repeated b times , it will make - ba orab . That is , a minus quantity multiplied by a plus quan- tity gives minus . Thirdly , if a is to be multiplied by — b , we may resolve it into the preceding case , for 3 ...
... hence if - a is repeated b times , it will make - ba orab . That is , a minus quantity multiplied by a plus quan- tity gives minus . Thirdly , if a is to be multiplied by — b , we may resolve it into the preceding case , for 3 ...
Página 35
... hence we have 483016 + 10 which is equal to 12. This result is obviously correct , for 8-5 is equal to 3 , and 6-2 is equal to 4 ; that is , it was required to multiply 3 by 4 , the result of which is 12 as found above . ( 54. ) The ...
... hence we have 483016 + 10 which is equal to 12. This result is obviously correct , for 8-5 is equal to 3 , and 6-2 is equal to 4 ; that is , it was required to multiply 3 by 4 , the result of which is 12 as found above . ( 54. ) The ...
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according to Art algebraic arithmetical arithmetical progression binomial cents Charles Anthon coefficients Completing the square contrary sign cube root denotes derived polynomial Divide the number dividend divisible equal roots equation whose roots equation x³ example exponent expression Extract the square factors find the square find the values following RULE fourth power fourth root fraction geometrical geometrical progression given equation greater greatest common divisor Hence infinite series irreducible fraction last term less letters taken method miles monomial mth degree multiplied negative nth root number of permutations number of terms obtain order of differences preceding Prob problem proportion PROPOSITION quadratic equations quotient ratio real roots Reduce remainder represent Required the square Required the sum result second degree second term Sheep extra square root subtract suppose surd theorem unity unknown quantity variation whence whole number
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Página 189 - ... the product of the two, plus the square of the second. In the third case, we have (a + b) (a — 6) = a2 — b2. (3) That is, the product of the sum and difference of two quantities is equal to the difference of their squares.
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Página 43 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.