"Roots" would imply that
(x^3 + 125) / (x +5) = 0
Factoring the top as a sum of cubes, we get
(x + 5)(x^2 - 5x + 25) So we have
(x + 5)(x^2 - 5x + 25) / (x + 5) = 0 Only the numerator can make this zero
So setting both factors to 0, we have
(x + 5 ) = 0 so x =-5, but we must reject this since it makes the denominator of the original function undefined.
The other two roots are non-real...using the on-site solver, we have
x2−5×x+25=0⇒{x=−(5×√3×i−5)2x=(5×√3×i+5)2}⇒{x=−(−52+4.330127018925i)x=52+4.330127018925i}
So...we have no real solutions to this "rational" function
