1 76 On extracting the Roots of Equations. 



Example. — Given the function ^x^+ \5x*-—6x+lS to trans- 

 form it into another m x — S. 



Operation. 

 4 + 15- 6+ 15 (3 



+ 27 

 39+75 

 4 + 51 + 1924-240. Therefore, 



4a^+15x--6x = 4(x-3)H51(x-3)- + 192(j:-3) + 240. 



Roots of Equations. 



By the application of this method of transforming algebraic 

 functions, we may easily discover the initial values of all the 

 roots of any given equation, by attending to the following pro- 

 perty, first given by Descartes. 



« Every equation (if impossible roots be allowed to make up 

 the nmnber) exhibits as many changes of signs as it has posi- 

 tive roots, and as many continuations of the same sign as it 

 has negative roots." 



Therefore, if, in diminishing the root of an equation by n of 

 any denomination of figures, and if the transformed equation 

 in x — n has the same number of changes of signs as the ori- 

 ginal equation in x, but if in diminishing the root of the trans- 

 formed equation in x—n by unity, a certain number of changes 

 of sio'ns disappear ; then n will be the initial value of as many 

 roots of the original equation as the number of changes which 

 are lost in the last transformation in x-~n — \. 



Example. — Transform the equation x*— 5x^ + 3x- — 9.r + 4 = 

 by dmiinishing its root by unity, the root of the transfonned 

 equation bv 3, and the second transformed equation by unity. 

 " X, 1- 5-1- 3- 9+ 4 (1 



— 4 



- 3- 1 



_ 2— 4- 10 

 x-l, 1- 1- 6- 14 — 6 (3 



+ 2 



+ 5+0 



+ 8 + 15- 14 

 a--4, 1 + 11 + 39+ 31—4 8(1 



+ 12 



+ 13 + 51 



+ 14 + 64+ 82 

 x—5, 1 + 15 + 78 + 146 + 34. 



In passing from the equation in x to that in x— 1 three 

 changes of signs have disappeared ; therefore between and 1, 

 there is at least one root, and may be three, and one change 

 of signs remain. In passuig from the equation in x— 1 to that 



in 



