Theory of Equation

Theory of Numbers

400 General form of a rational integral equation of the 𝑛^th degree. 𝑝₀𝑥^𝑛+𝑝₁𝑥^𝑛−1+𝑝₂𝑥^𝑛−2+⋯+𝑝_𝑛−1𝑥+𝑝_𝑛 The left side will be designated 𝑓(𝑥) in the following summary. 401 If 𝑓(𝑥) be divided by 𝑥−𝑎, the remainder will be 𝑓(𝑎). By assuming 𝑓(𝑥)=𝑃(𝑥−𝑎)+𝑅. 402 If 𝑎 be a root of the equation 𝑓(𝑥)=0, then 𝑓(𝑎)=0. 403 To compute 𝑓(𝑎) numerically; divide 𝑓(𝑥) by 𝑥−𝑎, and the remainder will be 𝑓(𝑎). 404

Example

To find the value of 4𝑥⁶−3𝑥⁵+12𝑥⁴−𝑥²+10 when 𝑥=2.

 |4−3+12+ 0− 1+  0+ 10
2|  8+10+44+88+174+348
----------------------
  4+5+22+44+87+174+358

Thus 𝑓(2)=358 405 If 𝑎,𝑏,𝑐, ⋯, 𝑘 be the roots of the equation 𝑓(𝑥)=0; then, by (401) and (402), 𝑓(𝑥)=𝑝₀(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)⋯(𝑥−𝑘) By multiplying out the last equation, and equating coefficients with equation (400), considering 𝑝₀=1, the following results are obtained:- 406 −𝑝₁=the sum of all the roots of 𝑓(𝑥). 𝑝₂=the sum of the products of the roots taken two at a time. −𝑝₃=the sum of the products of the roots taken three at a time. ⋯ (−1)^𝑟𝑝_𝑟=the sum of the products of the roots taken three at 𝑟 a time. ⋯ (−1)^𝑛𝑝_𝑛=product of all roots. 407 The number of roots of 𝑓(𝑥) is equal to the degree of the equation. 408 Imaginary roots must occur in pairs of the form 𝛼+𝛽−1, 𝛼−𝛽−1 The quadratic factor corresponding to these roots will then have real coefficients; for it will be 𝑥²−2𝛼𝑥+𝛼²+𝛽²405, 226 409 If 𝑓(𝑥) be of an odd degree, it has at least one real root of the opposite sign to 𝑝_𝑛. Thus 𝑥²−1=0 has at least one positive root. 410 if 𝑓(𝑥) be of an even degree, and 𝑝_𝑛 negative, there is at least one positive and one negative root. Thus 𝑥⁴−1 has +1 and −1 for roots. 411 If several terms at the beginning of the equation are of one sign, and all the rest of another, there is one, and only one, positive root. Thus 𝑥⁵+2𝑥⁴+3𝑥³+𝑥²−5𝑥−4=0 has only one positive root. 412 If all the terms are positive there is no positive root. 413 If all the terms of an even order are of one sign, and all the rest are of another sign, there is no negative root. 414 Thus 𝑥⁴−𝑥³+𝑥²−𝑥+1=0 has no negative root. 415 If all the indices are even, and all the terms of the same sign, there is no real root; and if all the indices are odd, and all the terms of the same sign, there is no real root but zero.
Thus 𝑥⁴+𝑥²+1=0 has no real root, and 𝑥⁵+𝑥³+𝑥=0 has no real root but zero. In this last equation there is no absolute term, because such a term would involve the zero power of 𝑥, which is even, and by hypothesis is wanting.

Sources and References

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