138 RESOLUTION OF SOLVABLE EQUATIONS 
By the substitution of these values, equation (25) becomes 
1156 x 34 4 
5B, gis | 6265333? — 2886277 x 13600357 t = 
PLP6 ag 334 
Pe Hiss ; 39254397600889 — 39254397600889 i =*0, 
$13. As an additional verification, the equation 
x + 10x23 — 8022 + 1452 — 480 = 0 
may be taken. Here, by §9, 
g=_—lA=4,P—e2=7,FR + 2&2 = 25, 
Therefore 
A= 2° x 3? x 7 x 29, B= —2 x 5. kK ieee 
DD re DEK DS > eee 
A= — 2 x 34 x,141, By = 24 x 3 x WK eee 
Dis = 2F x 38 5 She 19: 
B+ D2 = 22 x 292 x 14281, 
B+ D? = 28 x 32 x 5 x 338016989, 
Al— D2 e2z = 0, 
A?— Di ctz = 24 x 38x 5 x 7 x 17? x 277. 
By the substitution of these values, equation (25) becomes 
Ae NO eee ET xX As? teas {277 x 14281? 
+ 5° x 7 x 338016989 — 2° x 3 x 141 x 2393 x 14281 |= 0. 
The Trinomial Quintic x + pax + ps = 0. 
§14. In this case, by (6) and (10), g = 0,and& = 0. Therefore, 
_ he (@ — ¢? z) 
I ae 
Dy AAT eae Therefore, by (12), 
; tes 
oe a0he i BaP) 4+ 15a? z. (26) 
a 
Also, by §3, B = = | —absYe Dheco(0—g2)}. Therefore, by (13), 
& 
Shec 
a2 
ps = - (02 — g*z) + 44acz. (27) 
