232 SCIENCE PROGRESS 



recorded series in the three previous sub-sections, or, if these are 

 not at hand, by the easily remembered process of division 

 itself. The following well-known ones will suffice : 



A. o= 5 + 2x — x*. 



Newton's famous equation. In this form it is not suitable 

 for obtaining the positive root by simple division. But as 

 the root evidently lies between 2 and 3, transfer the origin by 

 dividing by — 2 and we have 



0=1 — loy — 6y* — y l . 



To reduce the coefficient of the second term to unity we 

 may either divide throughout by 10 (which is a convenient 



divisor) ; or may put y= — z ; or may put y= ioz and divide 



by 100 (sub-section (3) ). By the last, 



0= *oi — z — 6z* — ioz'. 



Then, either by direct division of by 0+60'+ 100 5 , or 

 from the series, 



z = 01 — 6('oooi) + (72 — 10) ("oi) 3 — 5(216 — 6o)( , oi) 4 . . . 



= "oi — '0006+ "000062 — '0000078+ . . . ; 

 and x= 2 + ioz= 2'i, 2 '094, 2*09462, 2 '094542 . . . , 



the last convergent being right to five figures, though we have 

 considered only four terms of the invert. The root is 

 2*09455148 . . . 



To investigate the two possible negative roots of the original 



equation, put x= — -y and divide by 5. We get 



0=1 -y+-2.y*; 



but as -^ > J -, the roots are not real — sub-section (2). 

 8 27 



B. o = 5 — 2x — 3x* + x l . 



To find the least root of this (negative), divide by 

 0+2, which gives = 11— 22y + gy i — y l . The root is 

 Xi = — 1*3300587 ... To obtain the second (lesser positive) 

 root, divide this last equation by — 3, or, which is the same 

 thing, the original equation by — 1 ; whence we obtain the 

 transformation = 1 — $y -f- y* (see section 2 (8)), which 

 gives the root x 2 =1*2016376 . . . For the third root, divide 



