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ContentTheory of Equation
Theory of EquationNewton's Method of Divisors459 To discover the integral roots of an equation.ExampleTo ascertain if 5 be a root of π₯4β6π₯3+86π₯2β176π₯+105=0 If 5 be a root it will divide 105. Add the quotient to the next coefficient. Result, β155. If 5 be a root it will divide β155. Add the quotient to the next coefficient; and so on. If the number tried be a root, the divisions will be effectible to the end, and the last quotient will be β1, or βπ0, if π0 be not unity.5460 In employing this method, limits of the roots may first be found, and divisors chosen between those limits. 461 Also, to lessen the number of trial divisors, take any integer π; then any divisor π of the last term can be rejected if πβπ does not divide π(π).)10521β1765)β155β31865)5511β65)β5β1 In practice take π=+1 and β1. To find whether any of the roots determined as above are repeated, divide π(π₯) by the factors corresponding to them, and then apply the method of divisors to the resulting equation. ExampleTake the equation π₯6+2π₯5β17π₯4β26π₯3+88π₯2+72π₯β144=0 Putting π₯=1, we find π(1)=β24. The divisors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 24, β― The values of πβπ (since π=1) are therefore 0, 1, 2, 3, 5, 7, 8, 11, 15, 23, β― Of these last numbers only 1, 2, 3, and 8 will divide 24. Hence 2, 3, 4, and 9 are the only divisors of 144 which it is of use to try. The only integral roots of the equation will be found to be Β±2 and Β±3. 462 If π(π₯) and πΉ(π) have common roots, they are contained in the greatest common measure of π(π₯) and πΉ(π). 463 If π(π₯) has for its roots π, π(π), π, π(π) amongst others; then the equations π(π₯)=0 and π{π(π₯)}=0 have the common roots π and π. 464 But, if all the roots occur in pairs in this way, these equations coincide. For example, suppose that each pair of roots, π and π, satisfies the equation π + π=2π. We may then assume π β π=2π§. Therefore π(π§+π)=0. This equation involves only even powers of π§, and may be solved for π§2. 465 Otherwise, Let ππ=π§; then π(π₯) is divisible by (π₯βπ)(π₯βπ)=π₯2 β2ππ₯+π§. Perform the division until a remainder is obtained of the form ππ₯+π. where π, π only involve π§.The equations π=0, π=0 determine π§, by (462); and π and π are found from π + π=2π, ππ=π§. Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveΒ©sideway ID: 210800009 Last Updated: 8/9/2021 Revision: 0 Ref: References
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