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Theory of Equation
 Newton's Method of Divisors
  Example
  Example
 Sources and References

Theory of Equation

Newton's Method of Divisors

459 To discover the integral roots of an equation.

Example

To ascertain if 5 be a root of π‘₯4βˆ’6π‘₯3+86π‘₯2βˆ’176π‘₯+105=0 If 5 be a root it will divide 105. Add the quotient to the next coefficient. Result, βˆ’155. If 5 be a root it will divide βˆ’155. 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 βˆ’1, or βˆ’π‘0, if 𝑝0 be not unity.
 5)105
    21
  βˆ’176
5)βˆ’155
   βˆ’31
    86
  5)55
    11
    βˆ’6
  5)βˆ’5
    βˆ’1
460 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‘š; then any divisor π‘Ž of the last term can be rejected if π‘Žβˆ’π‘š does not divide 𝑓(π‘š).
In practice take π‘š=+1 and βˆ’1.
To find whether any of the roots determined as above are repeated, divide 𝑓(π‘₯) by the factors corresponding to them, and then apply the method of divisors to the resulting equation.

Example

Take the equation π‘₯6+2π‘₯5βˆ’17π‘₯4βˆ’26π‘₯3+88π‘₯2+72π‘₯βˆ’144=0 Putting π‘₯=1, we find 𝑓(1)=βˆ’24. The divisors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 24, β‹― The values of π‘Žβˆ’π‘š (since π‘š=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 𝑓(π‘₯) and 𝐹(𝑋) have common roots, they are contained in the greatest common measure of 𝑓(π‘₯) and 𝐹(𝑋). 463 If 𝑓(π‘₯) has for its roots π‘Ž, πœ™(π‘Ž), 𝑏, πœ™(𝑏) amongst others; then the equations 𝑓(π‘₯)=0 and 𝑓{πœ™(π‘₯)}=0 have the common roots π‘Ž and 𝑏. 464 But, if all the roots occur in pairs in this way, these equations coincide. For example, suppose that each pair of roots, π‘Ž and 𝑏, satisfies the equation π‘Ž + 𝑏=2π‘Ÿ. We may then assume π‘Ž βˆ’ 𝑏=2𝑧. Therefore 𝑓(𝑧+π‘Ÿ)=0. This equation involves only even powers of 𝑧, and may be solved for 𝑧2. 465 Otherwise, Let π‘Žπ‘=𝑧; then 𝑓(π‘₯) is divisible by (π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘)=π‘₯2 βˆ’2π‘Ÿπ‘₯+𝑧. Perform the division until a remainder is obtained of the form 𝑃π‘₯+𝑄. where 𝑃, 𝑄 only involve 𝑧.
The equations 𝑃=0, 𝑄=0 determine 𝑧, by (462); and π‘Ž and 𝑏 are found from π‘Ž + 𝑏=2π‘Ÿ, π‘Žπ‘=𝑧.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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ID: 210800009 Last Updated: 8/9/2021 Revision: 0 Ref:

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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