Question 4587
 (i) x^4 + 5x^3 + x^2 - 25 =0
     x^3(x + 5) + x^2 - 25 =0
     x^3(x + 5) + (x+5)(x-5) =0
     (x+5)(x^3 + x - 5) = 0
     So, x = -5 or x^3 + x - 5= 0
   However,unfortunately,it is impossible to factor out 
   x^3 + x - 5 =0 unless you use the Cartan's formula about the cubic
  polynomial to find the roots. 
  If you go to quickmath.com you can get the complicated
  solutions of 1 real and two conjugate complex numbers.
  But I think it is beyond what your level of learning.
  [Maybe you only need one root.]
  

 (ii) 3x^(3/2) + 7x^(1/3) - 6 = 0.
  Let u = x^(1/3), we have
   3 u^2 + 7 u - 6 = 0.
  Factor out: (3u + 2)(u -3) =0
  So, u = -2/3 or 3.
 
 Hence, x^(1/3) = -2/3 , x = -(2/3)^3 = -8/27 
 or x^(1/3) = 3 , x = 27.
 Plz check by yourself.

 Kenny