CUBIC EQUATIONS. 579 



<p 2 = 10 3 . 3 3 . 13 2 , it suffices to write 2 = 10 . 3 and 4> = \/10 . 3. Hence the 

 value of b in this equation is 10 x 3 or 30, and since q — — 105, b + q = — 75 

 = — 3a 2 , .-. a = 5 and the roots of the equation are 5 + \/30, 5 — \/30, — 10. 



Example 3. Let x 3 — I5x + 4 = 0, here q = — 15, r = — 4 and 2 = — 4<? 3 

 - 27r 2 = 13068. This is twice divisible by 2, the quotients being 6534 and 3267, 

 then 3 times divisible by 3, the quotients being 1089, 363, 121, and the last 

 number is the square of 11 .-. <p 2 = 2 2 . o 3 . II 2 , which reduces itself, by ex- 

 punging the square factors, to 3. But we have seen that b is what <p 2 becomes 

 after the omission of all square factors, .*. b = 3 .-. b + q = 3 — 15 = — 12 = — 

 3a 2 .-. a = 2, and the roots are 2 ±\/3, — 4. 



Example 4. Let x 3 — 80a: + 200 = 0. Here q = — 80, r = — 200 

 .-. </> 2 = — 4^ 3 — 27r 2 = 968000. First divide by 10 2 which leaves quotient 

 9680, then 4 times by 2, the quotients being 4840, 2420, 1210, 605, which, divided 

 by 5, leaves 121 or ll 2 . Hence <f> 2 = 10 2 . 2 4 . 5 . II 2 , and omitting the square 

 factors, we find b = 5. Then b + q = 5 — 80 = — 75 = — Sa 2 .-. a = 5 and the 

 roots are 5 ± a/5, — 10. 



Example 5. Let x 3 — 30 x + 36 = 0. Here q = — 30, r = — 36 and <t> 2 comes 

 out = 73008. Divide 4 times by 2, and the last quotient is 4563. Then divide 

 3 times by 3, and the last quotient 169 is seen to be the square of 13. Hence 

 <p 2 = 2\ 3 3 . 13 2 , and omitting the square factors, b = 3 whence b + q = 3 — 30 

 = — 27 = — 3« 2 , whence a = 3, and the roots are 3 =b a/3, — 6, When the co- 

 efficients are large numbers, the trouble of solving numerical equations naturally 

 increases. But this is a mere affair of arithmetic, and the principles of solution 

 remain unaltered. I will give an example or two with large coefficients. 



Example 6. Let x 3 — 1456 x — 456 = 0. Here q = — 1456, r = 456 and 

 <p 2 = — 4^ 3 — 27r 2 comes out 12340892992, a number which is 6 times divisible 

 by 2. Continuing the reduction, we finally obtain = 8. 719 -s/373, whence 

 b = 373 .-. £ + q = 373 — 1456 = — 1083 =—3a 2 whence a 2 = 361 .-. a = 19. 



Hence the roots are 19 ± a/373, — 38. 



Let us verify this 



the root— 38, 





- 38 3 = - 54872 



1456 



- 456 = - 456 



38 



- 55328 



11648 





4368 



+ 55328 



Therefore the equation is satisfied. 



Example 7. Let x 3 — 160a; + 504 = 0. Here q = — 160, r = — 504. There- 

 fore — 4<? 3 — 27r 2 comes out = 9525568. This number is 3 times divisible by 4, 

 and then it is found to be divisible by 13. The quotient is 11449, which a table 

 of squares shows to be the square of 107. Hence <f> 2 = 2 6 . 13 . 107 2 , whence 

 omitting the square factors 6 = 13. And b + q= 13 — 160= — 147 = — 3a a 



