370 Bangma's Method of solving Equations. 



powers of numerical equations* (translated from O. S. Bang- 

 ma's Dutch Algebra) would be interesting to your readers, and 

 worthy of insertion in your valuable publication, I shall feel 

 happy in having communicated it to you. 



I am, gentlemen, respectfully yours, 



J. ROWBOTHAM. 



560 -14 

 546 ,,,-28 



504 -1 ^ 



To find the roots of the higher powers of equations, we 

 shall begin with the following cubic equation : 



x 3 — 14.r*— 20r= —546. 

 Bring all the terms of the equation to one side and arrange 

 them according to the power of x, as follows : 

 .r 3 —14.r 9 — 29x4 546 = 0. 

 For x substitute three successive terms of a series of na- 

 tural numbers; as, —1, 0, and 1, by which the following re- 

 sults will be obtained : 



560, 546, 504. 

 By subtracting the first of these numbers from the second, 

 and the second from the third, we have 

 — 14 and —42. 

 Again : by subtracting the first number from the second, 

 we have —28. 



The preceding numbers may be arranged as follows : 

 — 1 

 

 1 



The lower numbers 1, 504, —42, —28, and the constant 

 number 6, serve to find the positive roots of cubic equations. 

 The upper numbers —1, 560, —14, —28, and the con- 

 stant number 6, serve to find the negative roots f. 

 To find the positive roots, write in a line, 

 1, 504, —42, —28. 



* This method of solving equations will be further illustrated (should we 

 not be too limited for space) in A Practical System of Algebra, by Mr. 

 Peter Nicholson and myself, which work is now in the press, and will be 

 published in a short time. 



f The preceding method is applicable to all the higher powers, with this 

 difference only, — that we must take for x so many successive terms of the 

 series of natural numbers, as the highest exponent of x contains units ; 

 thus, For cubic equations 0, 1, 2; or —1, 0, 1. 



For biquadratic 0, 1, 2, 3; or — 1, 0, 1, 2. 



For 5th power 0, 1, 2, 3. 4; or — 2, —1, 0, 1, 2, &c. 



The constant number, mentioned above, will be formed; 

 For the 3d power, by 1 X 2x3=6. 

 For the 4th power, 1 X 2x3x4=24. 

 For the 5th power, 1x2x3x4x5=120, &c. 



Now, 



